{"id":36163,"date":"2022-11-26T17:31:00","date_gmt":"2022-11-26T12:01:00","guid":{"rendered":"https:\/\/tsboardsolutions.com\/?p=36163"},"modified":"2022-12-03T14:19:32","modified_gmt":"2022-12-03T08:49:32","slug":"maths-2b-ellipse-important-questions-short-answer-type","status":"publish","type":"post","link":"https:\/\/tsboardsolutions.com\/maths-2b-ellipse-important-questions-short-answer-type\/","title":{"rendered":"TS Inter Second Year Maths 2B Ellipse Important Questions"},"content":{"rendered":"

Students must practice these Maths 2B Important Questions<\/a> TS Inter Second Year Maths 2B Ellipse Important Questions to help strengthen their preparations for exams.<\/p>\n

TS Inter Second Year Maths 2B Ellipse Important Questions<\/h2>\n

Question 1.
\nFind the equation of the ellipse with focus at (1, -1), e = \\(\\frac{2}{3}\\) and directrix as x + y + 2 = 0. [(TS) Mar. ’19; (AP) May ’16]
\nSolution:
\n\"TS
\nGiven that focus S = (1, -1)
\nEquation of the directrix is x + y + 2 = 0
\nEccentricity, e = \\(\\frac{2}{3}\\)
\nLet P(x, y) be any point on the ellipse.
\nSince \u2018P\u2019 lies on the ellipse then
\n\"TS
\n\u21d2 9x2<\/sup> + 9y2<\/sup> – 18x + 18y + 18 – 2x2<\/sup> – 2y2<\/sup> – 4xy – 8x – 8y – 8 = 0
\n\u21d2 7x2<\/sup> – 4xy + 7y2<\/sup> – 26x + 10y + 10 = 0
\n\u2234 The equation of the ellipse is 7x2<\/sup> + 7y2<\/sup> – 4xy – 26x + 10y + 10 = 0<\/p>\n

Question 2.
\nFind the equation of the ellipse in the standard form such that the distance between the foci is 8 and the distance between the directrices is 32. [(AP) May ’17]
\nSolution:
\nLet the equation of the ellipse is
\n\\(\\frac{x^2}{a^2}+\\frac{y^2}{b^2}\\) = 1 (a > b) ………(1)
\nGiven that the distance between the foci = 8
\n2ae = 8
\nae = 4 ……(2)
\nThe distance between directrices = 32
\n\\(\\frac{2a}{e}\\) = 32
\n\\(\\frac{a}{e}\\) = 16 ……….(3)
\nNow (2) \u00d7 (3)
\nae . \\(\\frac{a}{e}\\) = 4 . 16
\n\u21d2 a2<\/sup> = 64
\n\u21d2 a = 8
\nWe know that b2<\/sup> = a2<\/sup>(1 – e2<\/sup>)
\n\u21d2 b2<\/sup> = a2<\/sup> – a2<\/sup>e2<\/sup>
\n\u21d2 b2<\/sup> = (8)2<\/sup> – (4)2<\/sup>
\n\u21d2 b2<\/sup> = 64 – 16
\n\u21d2 b2<\/sup> = 48
\n\u2234 The equation of the ellipse is \\(\\frac{x^2}{64}+\\frac{y^2}{48}=1\\)<\/p>\n

\"TS<\/p>\n

Question 3.
\nFind the equation of the ellipse in the standard form whose distance between foci is ‘2’ and the length of the latus rectum is \\(\\frac{15}{2}\\). [(AP) Mar. ’18; (TS) ’15]
\nSolution:
\nLet the equation of the ellipse is
\n\\(\\frac{x^2}{a^2}+\\frac{y^2}{b^2}=1\\) (a > b) ……..(1)
\nGiven that distance between foci = 2
\n\u21d2 2ae = 2
\n\u21d2 ae = 1
\nLength of latus rectum = \\(\\frac{15}{2}\\)
\n\u2234 \\(\\frac{2 b^2}{a}=\\frac{15}{2}\\)
\nb2<\/sup> = \\(\\frac{15a}{4}\\)
\nWe know that b2<\/sup> = a2<\/sup>(1 – e2<\/sup>)
\n\u21d2 b2<\/sup> = a2<\/sup> – a2<\/sup>e2<\/sup>
\n\u21d2 \\(\\frac{15a}{4}\\) = a2<\/sup> – (1)
\n\u21d2 15a = 4a2<\/sup> – 4
\n\u21d2 4a2<\/sup> – 15a – 4 = 0
\n\u21d2 4a2<\/sup> – 16a + a – 4 = 0
\n\u21d2 4a(a – 4) + 1(a – 4) = 0
\n\u21d2 (a – 4) (4a + 1) = 0
\n\u21d2 a = 4, a = \\(\\frac{-1}{4}\\)
\nSince ‘a’ is always positive then a = 4
\nIf a = 4, then b2<\/sup> = \\(\\frac{15(4)}{4}\\) = 15
\n\u2234 The equation of the ellipse is \\(\\frac{x^2}{16}+\\frac{y^2}{15}=1\\)<\/p>\n

Question 4.
\nFind the equation of the ellipse referred to its major and minor axes as the coordinate axes X, Y respectively with latus rectum of length 4, and distance between foci 4\u221a2. [May & Mar. ’19 (AP); Mar. ’18 (TS)]
\nSolution:
\nLet the equation of ellipse be
\n\\(\\frac{x^2}{a^2}+\\frac{y^2}{b^2}=1\\) (a > b)
\nby given data length of the latus rectum is
\n\\(\\frac{2 \\mathrm{~b}^2}{\\mathrm{a}}\\) = 4
\n\u21d2 b2<\/sup> = 2a ……..(1)
\ndistance between foci is
\n2ae = 4\u221a2
\n\u21d2 ae = 2\u221a2
\n\u21d2 a2<\/sup>e2<\/sup> = 8 …….(2)
\nNow (1) \u21d2 a2<\/sup>(1 – e2<\/sup>) = 2a
\n\u21d2 a2<\/sup> – a2<\/sup>e2<\/sup> = 2a
\n\u21d2 a2<\/sup> – 8 = 2a
\n\u21d2 a2<\/sup> – 2a – 8 = 0
\n\u21d2 a2<\/sup> – 4a + 2a – 8 = 0
\n\u21d2 a(a – 4) + 2(a – 4) = 0
\n\u21d2 (a – 4) (a + 2) = 0
\n\u21d2 a = -2, 4
\nSince a \u2260 -2 \u21d2 a = 4 and
\nb2<\/sup> = 2a = 2(4) = 8
\n\u2234 Required ellipse be \\(\\frac{x^2}{16}+\\frac{y^2}{8}=1\\)<\/p>\n

Question 5.
\nFind the length of the major axis, minor axis, latus rectum, eccentricity, coordinates of centre, foci, and the equations of directrices of the ellipse 9x2<\/sup> + 16y2<\/sup> = 144. [(TS) Mar. ’20, ’16; May ’19, ’17; (AP) Mar. ’17, ’15, ’14]
\nSolution:
\nGiven the equation of the ellipse is 9x2<\/sup> + 16y2<\/sup> = 144
\n\u21d2 \\(\\frac{x^2}{16}+\\frac{y^2}{9}=1\\)
\nComparing with \\(\\frac{x^2}{a^2}+\\frac{y^2}{b^2}=1\\)
\nwe get a = 4, b = 3 (a > b)
\nThe length of the major axis = 2a = 2(4) = 8
\nThe length of the minor axis = 2b = 2(3) = 6
\n\"TS<\/p>\n

Question 6.
\nFind the radius of the circle passing through the foci of an ellipse 9x2<\/sup> + 16y2<\/sup> = 144 and having the least radius.
\nSolution:
\nGiven ellipse is 9x2<\/sup> + 16y2<\/sup> = 144
\n\u21d2 \\(\\frac{x^2}{16}+\\frac{y^2}{9}=1\\)
\nHere a2<\/sup> = 16 \u21d2 a = 4
\nb2<\/sup> = 9 \u21d2 b = 3
\nEccentricity e = \\(\\sqrt{\\frac{a^2-b^2}{a^2}}=\\sqrt{\\frac{16-9}{16}}=\\frac{\\sqrt{7}}{4}\\)
\nSS’ = Diameter of the circle = 2ae
\n\u21d2 Diameter of the circle = 2\u221a7
\n\u21d2 2r = 2\u221a7
\n\u21d2 r = \u221a7
\ni.e., Radius of the circle CS = \u221a7<\/p>\n

\"TS<\/p>\n

Question 7.
\nThe distance of a point on the ellipse x2<\/sup> + 3y2<\/sup> = 6 from its centre is equal to 2. Find the eccentric angles.
\nSolution:
\nGiven the equation of the ellipse is x2<\/sup> + 3y2<\/sup> = 6
\n\u21d2 \\(\\frac{x^2}{6}+\\frac{3 y^2}{6}=1\\)
\n\u21d2 \\(\\frac{x^2}{6}+\\frac{y^2}{2}=1\\)
\na = \u221a6, b = \u221a2, a > b
\n\"TS
\nLet P(a cos \u03b8, b sin \u03b8) = (\u221a6 cos \u03b8, \u221a2 sin \u03b8] be a point on the ellipse.
\nCentre C = (0, 0)
\nGiven that CP = 2
\n\u21d2 CP2<\/sup> = 4
\n\u21d2 (\u221a6 cos \u03b8)2<\/sup> +(\u221a2 sin \u03b8)2<\/sup> = 4
\n\u21d2 6 cos2<\/sup>\u03b8 + 2 sin2<\/sup>\u03b8 = 4
\n\u21d2 4 cos2<\/sup>\u03b8 + 2 cos2<\/sup>\u03b8 + 2 sin2<\/sup>\u03b8 = 4
\n\u21d2 4 cos2<\/sup>\u03b8 + 2(cos2<\/sup>\u03b8 + sin2<\/sup>\u03b8) = 4
\n\u21d2 4 cos2<\/sup>\u03b8 + 2 = 4
\n\u21d2 4cos2<\/sup>\u03b8 = 2
\n\u21d2 cos2<\/sup>\u03b8 = \\(\\frac{1}{2}\\)
\n\u21d2 cos \u03b8 = \\(\\pm \\frac{1}{\\sqrt{2}}\\)
\nIf cos \u03b8 = \\(\\frac{1}{\\sqrt{2}}\\) \u21d2 \u03b8 = \\(\\frac{\\pi}{4}, \\frac{7 \\pi}{4}\\)
\nIf cos \u03b8 = \\(\\frac{-1}{\\sqrt{2}}\\) \u21d2 \u03b8 = \\(\\frac{3\\pi}{4}, \\frac{5 \\pi}{4}\\)
\n\u2234 Eccentric angles are \u03b8 = \\(\\frac{\\pi}{4}, \\frac{3 \\pi}{4}, \\frac{5 \\pi}{4}, \\frac{7 \\pi}{4}\\)<\/p>\n

Question 8.
\nShow that the condition for a straight line y = mx + c may be a tangent to the ellipse \\(\\frac{\\mathbf{x}^2}{\\mathbf{a}^2}+\\frac{\\mathbf{y}^2}{\\mathbf{b}^2}\\) = 1 is c2<\/sup> = a2<\/sup>m2<\/sup> + b2<\/sup>. (May ’06, ’02)
\nSolution:
\nGiven equation of the ellipse is \\(\\frac{\\mathbf{x}^2}{\\mathbf{a}^2}+\\frac{\\mathbf{y}^2}{\\mathbf{b}^2}\\) = 1
\n\"TS
\nSuppose y = mx + c ……….(1)
\nis a tangent to the ellipse \\(\\frac{x^2}{a^2}+\\frac{y^2}{b^2}=1\\)
\nLet P(x1<\/sub>, y1<\/sub>) be the point of contact.
\nThe equation of the tangent at \u2018P\u2019 is S1<\/sub> = 0
\n\u21d2 \\(\\frac{\\mathrm{xx}_1}{\\mathrm{a}^2}+\\frac{\\mathrm{y} \\mathrm{y}_1}{\\mathrm{~b}^2}-1=0\\) ……..(2)
\nNow, (1) & (2) represent the same line.
\n\"TS<\/p>\n

Question 9.
\nFind the equation of the tangents to 9x2<\/sup> + 16y2<\/sup> = 144, which makes equal intercepts on the coordinate axes. [(TS) May ’15]
\nSolution:
\nGiven the equation of the ellipse is 9x2<\/sup> + 16y2<\/sup> = 144
\n\u21d2 \\(\\frac{x^2}{16}+\\frac{y^2}{9}=1\\)
\nHere a = 4, b = 3, a > b
\nLet the equation of the tangent which makes equal intercepts on co-ordinate axes is
\nx + y = k ……(1)
\n\u21d2 y = -x + k
\nComparing with y = mx + c, we get
\nm = -1, c = k
\nSince eq (1) is a tangent to the given ellipse then
\nc2<\/sup> = a2<\/sup>m2<\/sup> + b2<\/sup>
\n\u21d2 (k)2<\/sup> = 16(-1)2<\/sup> + 9
\n\u21d2 k2<\/sup> = 16 + 9
\n\u21d2 k2<\/sup> = 25
\n\u21d2 k = \u00b15
\nSubstitute the value of \u2018k\u2019 in eq (1).
\n\u2234 The equation of the tangent is x + y = \u00b1 5
\n\u21d2 x + y \u00b1 5 = 0<\/p>\n

\"TS<\/p>\n

Question 10.
\nFind the equation of the tangent to the ellipse 2x2<\/sup> + y2<\/sup> = 8, which makes an angle \u03c0\/4 with the x-axis. [(AP) May ’19]
\nSolution:
\nGiven ellipse is 2x2<\/sup> + y2<\/sup> = 8
\n\u21d2 \\(\\frac{x^2}{4}+\\frac{y^2}{8}=1\\) ……..(1)
\nComparing (1) with \\(\\frac{x^2}{a^2}+\\frac{y^2}{b^2}=1\\), we get
\na2<\/sup> = 4, b2<\/sup> = 8
\nGiven that \u03b8 = \\(\\frac{\\pi}{4}\\)
\nThe slope of the tangent is m = tan \u03b8
\n= tan \\(\\frac{\\pi}{4}\\)
\n= 1
\n\u2234 The equation of the tangent to the given ellipse is
\ny = mx \u00b1 \\(\\sqrt{a^2 m^2+b^2}\\)
\n\u21d2 y = 1 . x \u00b1 \\(\\sqrt{4(1)+8}\\)
\n\u21d2 y = x \u00b1 \u221a12
\n\u21d2 x – y \u00b1 2\u221a3 = 0<\/p>\n

Question 11.
\nFind the equation of the tangents to the ellipse 2x2<\/sup> + y2<\/sup> = 8 which are
\n(i) parallel to x – 2y – 4 = 0
\n(ii) perpendicular to x + y + 2 = 0 [(AP) May ’19, ’17; (TS) May ’19, Mar. ’17]
\nSolution:
\nGiven the equation of the ellipse is 2x2<\/sup> + y2<\/sup> = 8
\n\u21d2 \\(\\frac{x^2}{4}+\\frac{y^2}{8}=1\\)
\nHere a2<\/sup> = 4, b2<\/sup> = 8
\n(i) Given the equation of the straight line is x – 2y – 4 = 0
\nThe equation of the tangent parallel to the line x – 2y – 4 = 0 is
\nx – 2y + k = 0 ……..(1)
\n\u21d2 2y = x + k
\n\u21d2 y = \\(\\frac{x}{2}+\\frac{k}{2}\\)
\nComparing with y = mx + c, we get
\nm = \\(\\frac{1}{2}\\), c = \\(\\frac{k}{2}\\)
\nSince eq. (1) is a tangent to the given ellipse then
\n\"TS
\nSubstitute the value of \u2018k\u2019 in eq. (1)
\n\u2234 The equation of the tangent is x – 2y \u00b1 6 = 0
\n(ii) Given the equation of the straight line is x + y + 2 = 0
\nThe equation of the tangent \u22a5r to x + y + 2 = 0 is
\nx – y + k = 0 ………(2)
\ny = x + k
\nComparing with y = mx + c we get
\nm = 1, c = k
\nSince eq. (2) is a tangent to the given ellipse then
\nc2<\/sup> = a2<\/sup>m2<\/sup> + b2<\/sup>
\n\u21d2 k2<\/sup> = 4(1)2<\/sup> + 8
\n\u21d2 k2<\/sup> = 12
\n\u21d2 k = \u00b1 2\u221a3
\nSubstitute the value of \u2018k\u2019 in eq. (2)
\n\u2234 The equation of the tangent is x – y \u00b1 2\u221a3 = 0<\/p>\n

Question 12.
\nFind the equation of tangent and normal to the ellipse 9x2<\/sup> + 16y2<\/sup> = 144 at the end of the latus rectum in the 1st quadrant. [(AP) Mar. ’15]
\nSolution:
\nGiven the equation of the ellipse is 9x2<\/sup> + 16y2<\/sup> = 144
\n\u21d2 \\(\\frac{x^2}{16}+\\frac{y^2}{9}=1\\)
\nHere a = 4, b = 3, a > b
\n\"TS
\n\"TS<\/p>\n

Question 13.
\nA circle of radius 4, is concentric with the ellipse 3x2<\/sup> + 13y2<\/sup> = 78. Prove that the common tangent is inclined to the major axis at angle \u03c0\/4. [(AP) May ’18]
\nSolution:
\nGiven the equation of the ellipse is 3x2<\/sup> + 13y2<\/sup> = 78
\n\u21d2 \\(\\frac{x^2}{26}+\\frac{y^2}{6}=1\\)
\nHere a2<\/sup> = 26, b2<\/sup> = 6, a > b
\n\"TS
\nCentre of the ellipse C = (0, 0)
\nCentre of the circle C = (0, 0) (concentric)
\ngiven that the radius of the circle r = 4
\nThe equation of a circle of radius 4 is concentric with the given ellipse, then
\nx2<\/sup> + y2<\/sup> = (4)2<\/sup>
\n\u21d2 x2<\/sup> + y2<\/sup> = 16
\nThe equation of a tangent to the given ellipse is y = mx \u00b1 \\(\\sqrt{a^2 m^2+b^2}\\)
\n\u21d2 y = mx \u00b1 \\(\\sqrt{26 m^2+6}\\) ……..(1)
\nSince (1) is a tangent to the circle x2<\/sup> + y2<\/sup> = 16, then r = d
\n\"TS
\nSquaring on both sides
\n\u21d2 16(m2<\/sup> + 1) = 26m2<\/sup> + 6
\n\u21d2 16m2<\/sup> + 16 = 26m2<\/sup> + 6
\n\u21d2 10m2<\/sup> – 10 = 0
\n\u21d2 10m2<\/sup> = 10
\n\u21d2 m = 1
\nHere, m is the slope of the tangent, then
\nm = tan \u03b8
\n\u21d2 1 = tan \u03b8
\n\u21d2 \u03b8 = \\(\\frac{\\pi}{4}\\)
\n\u2234 A common tangent is inclined to the major axis at an angle \\(\\frac{\\pi}{4}\\).<\/p>\n

\"TS<\/p>\n

Question 14.
\nIf the normal at one end of the latus rectum of the ellipse \\(\\frac{x^2}{a^2}+\\frac{y^2}{b^2}=1\\) passes through one end of the minor axis, then show that e4<\/sup> + e2<\/sup> = 1. (e is the eccentricity of the ellipse) [(TS) May ’17, ’14]
\nSolution:
\n\"TS
\nGiven equation of the ellipse is \\(\\frac{x^2}{a^2}+\\frac{y^2}{b^2}=1\\)
\nLet \u2018L\u2019 be the one end of the latus rectum of \\(\\frac{x^2}{a^2}+\\frac{y^2}{b^2}=1\\)
\nThen the coordinates of L = (ae, \\(\\frac{b^2}{a}\\))
\n\u2234 The equation of the normal at \u2018L\u2019 is
\n\"TS
\nSince eq.(1) passes through one end of the minor axis B'(0, -b) then
\n\"TS<\/p>\n

Question 15.
\nThe tangent and normal to the ellipse x2<\/sup> + 4y2<\/sup> = 4 at a point P(\u03b8) on it meets the major axis in Q and R respectively. If 0 < \u03b8 < \\(\\frac{\\pi}{2}\\) and QR = 2 then show that \u03b8 = cos-1<\/sup>(2\/3). [May ’16 (AP) May ’14]
\nSolution:
\n\"TS
\nGiven the equation of the ellipse is x2<\/sup> + 4y2<\/sup> = 4
\n\u21d2 \\(\\frac{x^2}{4}+\\frac{y^2}{1}=1\\)
\nHere a = 2, b = 1
\nLet \u2018\u03b8\u2019 be the eccentric angle of \u2018P\u2019 then
\nP(a cos \u03b8, b sin \u03b8) = (2 cos \u03b8, sin \u03b8)
\nThe equation of the tangent at \u2018P\u2019 is
\n\"TS
\n\"TS
\n\"TS<\/p>\n

Question 16.
\nFind the eccentricity, coordinates of foci, length of latus rectum, and equations of directrices of the ellipse 9x2<\/sup> + 16y2<\/sup> – 36x + 32y – 92 = 0. [(TS) May & Mar. ’18, ’15]
\nSolution:
\nGiven equation of the ellipse is 9x2<\/sup> + 16y2<\/sup> – 36x + 32y – 92 = 0
\n\u21d2 (9x2<\/sup> – 36x) + (16y2<\/sup> + 32y) – 92 = 0
\n\u21d2 9(x2<\/sup> – 4x) + 16(y2<\/sup> + 2y) – 92 = 0
\n\u21d2 9(x2<\/sup> – 2 . 2x + 22<\/sup> – 22<\/sup>) + 16(y2<\/sup> + 2 . 1 . y + 12<\/sup> – 12<\/sup>) – 92 = 0
\n\u21d2 9((x – 2)2<\/sup> – 4) + 16((y + 1)2<\/sup> – 1) – 92 = 0
\n\u21d2 9(x- 2)2<\/sup> – 36 + 16(y + 1)2<\/sup> – 16 – 92 = 0
\n\u21d2 9(x – 2)2<\/sup> + 16 (y + 1)2<\/sup> – 144 = 0
\n\u21d2 9(x – 2)2<\/sup> + 16(y + 1)2<\/sup> = 144
\n\"TS
\n\"TS<\/p>\n

Question 17.
\nFind the length of the major axis, minor axis, latus rectum, eccentricity, coordinates of centre, foci, and the equations of directrices of the ellipse 4x2<\/sup> + y2<\/sup> – 8x + 2y + 1 = 0. [(AP) May ’18]
\nSolution:
\nGiven equation of the ellipse is 4x2<\/sup> + y2<\/sup> – 8x + 2y + 1 = 0
\n\u21d2 (4x2<\/sup> – 8x) + (y2<\/sup> + 2y) + 1 = 0
\n\u21d2 4(x2<\/sup> – 2x) + (y2<\/sup> + 2y) + 1 = 0
\n\u21d2 4((x)2<\/sup> – 2x . 1 + 12<\/sup> – 12<\/sup>) + ((y)2<\/sup> + 2 . y . 1 + 12<\/sup> – 12<\/sup>) + 1 = 0
\n\u21d2 4((x – 1)2<\/sup> – 1) + ((y + 1)2<\/sup> – 1) + 1 = o
\n\u21d2 4(x – 1)2<\/sup> – 4 + (y + 1)2<\/sup> – 1 + 1 = 0
\n\u21d2 4(x – 1)2<\/sup> + (y + 1)2<\/sup> = 4
\n\u21d2 \\(\\frac{4(x-1)^2}{4}+\\frac{(y+1)^2}{4}=1\\)
\n\u21d2 \\(\\frac{(x-1)^2}{1}+\\frac{(y-(-1))^2}{4}=1\\)
\nComparing with \\(\\frac{(\\mathrm{x}-\\mathrm{h})^2}{\\mathrm{a}^2}+\\frac{(\\mathrm{y}-\\mathrm{k})^2}{\\mathrm{~b}^2}=1\\)
\nwe get h = 1, k = -1, a = 1, b = 2, a < b
\nThe length of the major axis = 2b = 2(2) = 4
\nThe length of the minor axis = 2a = 2(1) = 2
\nThe length of latus rectum = \\(\\frac{2 a^2}{b}=\\frac{2 \\cdot 1}{2}\\) = 1
\n\"TS<\/p>\n

Question 18.
\nFind the eccentricity, coordinates of foci, length of latus rectum, and equations of directrices of the ellipse 3x2<\/sup> + y2<\/sup> – 6x – 2y – 5 = 0. [(AP) May ’15]
\nSolution:
\nGiven equation of the ellipse is 3x2<\/sup> + y2<\/sup> – 6x – 2y – 5 = 0
\n\u21d2 (3x2<\/sup> – 6x) + (y2<\/sup> – 2y) – 5 = 0
\n\u21d2 3(x2<\/sup> – 2x) + (y2<\/sup> – 2y) – 5 = 0
\n\u21d2 3((x)2<\/sup> – 2 . 1 . x + 12<\/sup> – 12<\/sup>) + ((y)2<\/sup> – 2 . 1 . y + 12<\/sup> – 12<\/sup>) – 5 = 0
\n\u21d2 3((x – 1)2<\/sup> – 1) + ((y – 1)2<\/sup> – 1) – 5 = 0
\n\u21d2 3(x – 1)2<\/sup> – 3 + (y – 1)2<\/sup> – 1 – 5 = 0
\n\u21d2 3(x – 1)2<\/sup> + (y – 1)2<\/sup> – 9 = 0
\n\u21d2 3(x – 1)2<\/sup> + (y – 1)2<\/sup> = 9
\n\u21d2 \\(\\frac{(x-1)^2}{9 \/ 3}+\\frac{(y-1)^2}{9}=1\\)
\n\u21d2 \\(\\frac{(x-1)^2}{3}+\\frac{(y-1)^2}{9}=1\\)
\n\"TS<\/p>\n

Question 19.
\nFind the equation of an ellipse in the form \\(\\frac{(x-h)^2}{a^2}+\\frac{(y-k)^2}{b^2}=1\\), given that centre (0, -3), e = \\(\\frac{2}{3}\\), semi-minor axis 5.
\nSolution:
\nThe equation of the ellipse is
\n\\(\\frac{(x-h)^2}{a^2}+\\frac{(y-k)^2}{b^2}=1\\) ………(1)
\nGiven that Centre C(h, k) = (0, -3)
\nEccentricity e = \\(\\frac{2}{3}\\)
\nCase (i): If a > b, the length of the semi-minor axis b = 5
\nWe know that b2<\/sup> = a2<\/sup>(1 – e2<\/sup>)
\n25 = a2<\/sup>(1 – \\(\\frac{4}{9}\\))
\n\"TS
\n\"TS<\/p>\n

\"TS<\/p>\n

Question 20.
\nFind the equation of the ellipse in the form \\(\\frac{(x-h)^2}{a^2}+\\frac{(y-k)^2}{b^2}=1\\), given that centre (2, -1) one end of the major axis (2, -5), e = \\(\\frac{1}{3}\\).
\nSolution:
\nThe equation of the ellipse is
\n\\(\\frac{(x-h)^2}{a^2}+\\frac{(y-k)^2}{b^2}=1\\) …….(1)
\n\"TS
\nGiven that
\nCentre C(h, k) = (2, -1)
\nOne end of the major axis B = (2, -5)
\nEccentricity e = \\(\\frac{1}{3}\\)
\nIn C, B, the x-coordinates are equal, then the major axis is parallel to the y-axis, i.e., a < b
\n\"TS<\/p>\n

Question 21.
\nProve that the equation of the chord joining the points ‘\u03b1’ and ‘\u03b2’ on the ellipse \\(\\frac{x^2}{a^2}+\\frac{y^2}{b^2}=1\\) is \\(\\frac{x}{a} \\cos \\left(\\frac{\\alpha+\\beta}{2}\\right)+\\frac{y}{b} \\sin \\left(\\frac{\\alpha+\\beta}{2}\\right)\\) = \\(\\cos \\left(\\frac{\\alpha-\\beta}{2}\\right)\\) [(TS) May ’16]
\nSolution:
\nGiven equation of the ellipse is \\(\\frac{x^2}{a^2}+\\frac{y^2}{b^2}=1\\)
\nGiven points on the ellipse is P(a cos \u03b1, b sin \u03b1), Q(a cos \u03b2, b sin \u03b2)
\n\"TS
\n\"TS
\n\"TS<\/p>\n

Question 22.
\nShow that the foot of the perpendicular drawn from centre to any tangent to the ellipse lies on the curve (x2<\/sup> + y2<\/sup>) = a2<\/sup>x2<\/sup> + b2<\/sup>y2<\/sup>.
\nSolution:
\nLet, the equation of the ellipse is
\n\\(\\frac{x^2}{a^2}+\\frac{y^2}{b^2}=1\\) (a > b)
\nLet, P(x1<\/sub>, y1<\/sub>) be the foot of the perpendicular drawn from the centre on any tangent to the ellipse.
\n\"TS
\n\"TS<\/p>\n

Question 23.
\nShow that the point of intersection of the perpendicular tangent to an ellipse lies on a circle. [(AP) Mar. ’16]
\nSolution:
\n\"TS
\nLet the equation of the ellipse be
\nS = \\(\\frac{x^2}{a^2}+\\frac{y^2}{b^2}-1\\) = 0
\nLet P(x1<\/sub>, y1<\/sub>) be the point of intersection of \u22a5r tangent drawn to the ellipse.
\nLet y = mx \u00b1 \\(\\sqrt{a^2 m^2+b^2}\\) be a tangent to the ellipse S = 0 passing through \u2018P\u2019, then
\ny1<\/sub> = mx1<\/sub> \u00b1 \\(\\sqrt{a^2 m^2+b^2}\\)
\ny1<\/sub> – mx1<\/sub> = \u00b1 \\(\\sqrt{a^2 m^2+b^2}\\)
\nSquaring on both sides,
\n\"TS
\nIf m1<\/sub>, m2<\/sub> are the slopes of the tangents through \u2018P\u2019 then m1<\/sub>, m2<\/sub> are the roots of (1).
\nSince the tangents through \u2018P\u2019 are \u22a5r then
\nm1<\/sub>m2<\/sub> = -1
\n\u21d2 \\(\\frac{y_1^2-b^2}{x_1^2-a^2}=-1\\)
\n\u21d2 \\(y_1^2-b^2=-x_1^2+a^2\\)
\n\u21d2 \\(\\mathrm{x}_1^2+\\mathrm{y}_1^2=\\mathrm{a}^2+\\mathrm{b}^2\\)
\n\u2234 P lies on x2<\/sup> + y2<\/sup> = a2<\/sup> + b2<\/sup> which is the circle with centre as the origin, the centre of the ellipse.<\/p>\n

Question 24.
\nShow that the locus of the feet of the perpendiculars drawn from foci to any tangent of the ellipse is the auxiliary circle. [(AP) Mar. ’19, ’17]
\nSolution:
\n\"TS
\nLet the equation of the ellipse be
\nS = \\(\\frac{x^2}{a^2}+\\frac{y^2}{b^2}-1\\) = 0
\nLet P(x1<\/sub>, y1<\/sub>) be the foot of the \u22a5r drawn from either of the foci to a tangent.
\nThe equation of the tangent to the ellipse S = 0 is
\ny = mx \u00b1 \\(\\sqrt{a^2 m^2+b^2}\\) ……..(1)
\nThe equation to the perpendicular from either focus (\u00b1ae, 0) on this tangent is
\n\"TS
\n\u2234 \u2018P\u2019 lies on x2<\/sup> + y2<\/sup> = a2<\/sup> which is a circle with centre as the origin, the centre of the ellipse.<\/p>\n

\"TS<\/p>\n

Question 25.
\nProve that the equation of an ellipse in the standard form is \\(\\frac{x^2}{a^2}+\\frac{y^2}{b^2}=1\\). (Mar. ’95)
\nSolution:
\n\"TS
\nLet \u2018S\u2019 be the focus, \u2018e\u2019 be the eccentricity and l = 0 be the directrix of the ellipse.
\nLet \u2018P\u2019 be a point on the ellipse.
\nLet M, Z be the projections (foot of the \u22a5rs) of P, S on the directrix L = 0 respectively.
\nLet \u2018M\u2019 be the projection of \u2018P\u2019 on \u2018SZ\u2019.
\nSince e < 1 we can divide SZ both internally and externally in the ratio e : 1.
\nLet A, A’ be the points of division of SZ in the ratio e : 1 internally and externally respectively.
\nLet AA’ = 2a
\nLet \u2018C\u2019 be the midpoint of AA’.
\nPoints A, A’ lies on the ellipse and \\(\\frac{\\mathrm{SA}}{\\mathrm{AZ}}=\\frac{\\mathrm{SA}^{\\prime}}{\\mathrm{ZA}^{\\prime}}=\\frac{\\mathrm{e}}{1}\\)
\nNow \\(\\frac{S A}{A Z}=\\frac{e}{1}\\)
\n\u21d2 SA = e AZ
\n\u21d2 CA – CS = e (CZ – CA) ……..(1)
\nAlso \\(\\frac{\\mathrm{SA}^{\\prime}}{\\mathrm{ZA}^{\\prime}}=\\frac{\\mathrm{e}}{1}\\)
\n\u21d2 SA’ = e ZA’
\nCS + CA’ = e (CA’ + CZ) ……..(2)
\nNow (1) + (2)
\n\u21d2 CA – CS + CS + CA’ = e (CZ – CA) + e(CZ + CA\u2019)
\n\u21d2 CA + CA’ = e (CZ – CA) + e (CZ + CA\u2019)
\n\u21d2 CA + CA’ = e (2CZ – CA + CA’)
\nSince ‘C’ is the midpoint of AA’ then CA = CA’
\nCA + CA = e (2CZ – CA + CA)
\n\u21d2 2CA = 2e CZ
\n\u21d2 CA = e CZ
\n\u21d2 a = e CZ
\n\u21d2 CZ = \\(\\frac{a}{e}\\)
\n\u2234 Equation of the directrix is x = \\(\\frac{a}{e}\\)
\n(1) – (2)
\n\u21d2 CA – CS – CS + CA’ = e (CZ – CA) + e(CZ + CZ)
\n\u21d2 CA – CS – CS – CA’ = e (CZ – CA – CA – CZ)
\n\u21d2 CA – 2CS – CA’ = e(-CA – CA’)
\nSince \u2018C\u2019 is the midpoint at AA’ then CA = CA’
\nCA – 2CS – CA = e(-CA – CA)
\n\u21d2 -2CS = e(-2CA)
\n\u21d2 CS = e CA
\n\u21d2 CS = eQ
\n\u2234 Co-ordinates of focus \u2018S\u2019 are (ae, 0)
\nTake CS, the principal axis of the ellipse as the X-axis, and QS \u22a5r to the CX as Y-axis then S = (ae, 0) and the ellipse is in standard form.
\nLet P = (x, y)
\nNow PM = NZ = CZ – CN = \\(\\frac{a}{e}\\) – x
\n\u2018P\u2019 lies on the ellipse then \\(\\frac{SP}{PM}\\) = e
\nSP = e . PM
\n\"TS<\/p>\n

Question 26.
\nIf the length of the latus rectum is equal to half of its minor axis of an ellipse in the standard form, then find the eccentricity of the ellipse. (May, Mar. ’10)
\nSolution:
\nLet the equation of the ellipse is \\(\\frac{x^2}{a^2}+\\frac{y^2}{b^2}=1\\) (a > b)
\nThe length of the latus rectum = \\(\\frac{2 b^2}{a}\\)
\nLength of minor axis = 2b
\nGiven that the length of the latus rectum is equal to half of its minor axis then
\n\"TS<\/p>\n

Question 27.
\nFind the eccentricity of the ellipse (in the standard form), if the length of the latus rectum is equal to half of its major axis.
\nSolution:
\nLet the equation of the ellipse is
\n\\(\\frac{x^2}{a^2}+\\frac{y^2}{b^2}=1\\) (a > b)
\nThe length of the latus rectum = \\(\\frac{2 b^2}{a}\\)
\nLength of the major axis = 2a
\nGiven that the length of the latus rectum is equal to half of its major axis then
\n\"TS<\/p>\n

Question 28.
\nFind the equation of tangent and normal to the ellipse x2<\/sup> + 8y2<\/sup> = 33 at (-1, 2). [(TS) Mar. ’20; May ’16]
\nSolution:
\nGiven the equation of the ellipse is x2<\/sup> + 8y2<\/sup> = 33
\nLet the given point P(x1<\/sub>, y1<\/sub>) = (-1, 2)
\nThe equation of the tangent at ‘P’ is S1<\/sub> = 0
\n\u21d2 xx1<\/sub> + 8yy1<\/sub> = 33
\n\u21d2 x(-1) + 8y(2) = 33
\n\u21d2 -x + 16y = 33
\n\u21d2 x – 16y + 33 = 0
\nSlope of the tangent at P’ is m = \\(\\frac{-1}{-16}=\\frac{1}{16}\\)
\nSlope of the normal at ‘P’ = \\(\\frac{-1}{m}=\\frac{-1}{\\frac{1}{16}}\\) = -16
\nThe equation of the normal at ‘P’ is
\ny – y1<\/sub> = \\(\\frac{-1}{m}\\)(x – x1<\/sub>)
\n\u21d2 y – 2 = -16(x + 1)
\n\u21d2 y – 2 = -16x – 16
\n\u21d2 16x + y + 14 = 0<\/p>\n

\"TS<\/p>\n

Question 29.
\nFind the value of ‘k’ if 4x + y + k = 0 is tangent to the ellipse x2<\/sup> + 3y2<\/sup> = 3.
\nSolution:
\nGiven the equation of the ellipse is x2<\/sup> + 3y2<\/sup> = 3
\n\u21d2 \\(\\frac{x^2}{3}+\\frac{y^2}{1}=1\\)
\nHere a2<\/sup> = 3, b2<\/sup> = 1
\nGiven the equation of the straight line is
\n4x + y + k = 0 ……..(1)
\ny = -4x – k
\nComparing with y = mx + c we get
\nm = -4, c = -k
\nSince eq. (1) is a tangent to the given ellipse then c2<\/sup> = a2<\/sup>m2<\/sup> + b2<\/sup>
\n\u21d2 (-k)2<\/sup> = 3(-4)2<\/sup> + 1
\n\u21d2 k2<\/sup> = 49
\n\u21d2 k = \u00b17<\/p>\n

Question 30.
\nFind the condition for the line x cos \u03b1 + y sin \u03b1 = p to be a tangent to the ellipse \\(\\frac{x^2}{a^2}+\\frac{y^2}{b^2}=1\\). [(AP) Mar. ’20, ’14]
\nSolution:
\nGiven equation of the ellipse is \\(\\frac{x^2}{a^2}+\\frac{y^2}{b^2}=1\\)
\nGiven the equation of the straight line is x cos \u03b1 + y sin \u03b1 = p
\ny sin \u03b1 = -x cos \u03b1 + p
\n\"TS<\/p>\n

Question 31.
\nIf the length of the major axis of an ellipse is three times the length, of its minor axis, then find the eccentricity of the ellipse.
\nSolution:
\nIn an ellipse
\nLength of major axis = 2a
\nLength of minor axis = 2b
\nGiven that length of the major axis = 2(length of the minor axis)
\n\"TS<\/p>\n

Question 32.
\nFind the equation of the auxiliary circle of the ellipse 9x2<\/sup> + 16y2<\/sup> = 144.
\nSolution:
\nGiven ellipse is 9x2<\/sup> + 16y2<\/sup> = 144
\n\u21d2 \\(\\frac{x^2}{16}+\\frac{y^2}{9}=1\\)
\nHere a2<\/sup> = 16, b2<\/sup> = 9 (a > b)
\n\u2234 The equation of an auxiliary circle is x2<\/sup> + y2<\/sup> = a2<\/sup>
\n\u21d2 x2<\/sup> + y2<\/sup> = 16<\/p>\n

Question 33.
\nFind the equation of the director circle of the ellipse 9x2<\/sup> + 25y2<\/sup> = 225.
\nSolution:
\nGiven ellipse is 9x2<\/sup> + 25y2<\/sup> = 225
\n\u21d2 \\(\\frac{x^2}{25}+\\frac{y^2}{9}=1\\)
\nHere a2<\/sup> = 25, b2<\/sup> = 9
\n\u2234 Equation of director circle is x2<\/sup> + y2<\/sup> = a2<\/sup> + b2<\/sup>
\n\u21d2 x2<\/sup> + y2<\/sup> = 25 + 9
\n\u21d2 x2<\/sup> + y2<\/sup> = 34<\/p>\n

\"TS<\/p>\n

Question 34.
\nIf P(x, y) is any point on the ellipse \\(\\frac{x^2}{a^2}+\\frac{y^2}{b^2}=1\\) (a > b) whose foci are \u2018S\u2019 and S’ then show that SP + S’P is a constant. (Mar. ’13, May, Mar. ’93)
\nSolution:
\n\"TS
\nThe equation of the ellipse is given as
\n\\(\\frac{x^2}{a^2}+\\frac{y^2}{b^2}=1\\) (a > b)
\nLet S, S’ be the foci and ZM, Z’M’ be the corresponding directrices.
\nJoin SP and S’P.
\nDraw PN \u22a5r to X-axis and M’MP \u22a5r to the two directrices.
\nBy the definition of the ellipse,
\n\\(\\frac{\\mathrm{SP}}{\\mathrm{PM}}\\) = e
\n\u21d2 SP = e PM
\n= e NZ
\n= e (CZ – CN)
\n= e(\\(\\frac{a}{e}\\) – x)
\nand \\(\\frac{\\mathrm{S}^{\\prime} \\mathrm{P}}{\\mathrm{PM}^{\\prime}}\\) = e
\nS’P = e PM’
\n= e(NZ’)
\n= e(CN + CZ’)
\n= e(x + \\(\\frac{a}{e}\\))
\nNow SP + S’P
\n= \\(e\\left(\\frac{a}{e}-x\\right)+e\\left(x+\\frac{a}{e}\\right)\\)
\n= a – ex + ex + a
\n= 2a (constant)
\n= length of the major axis<\/p>\n

Question 35.
\nThe orbit of the Earth is an ellipse with eccentricity \\(\\frac{1}{60}\\) with the sun at one of its foci, the major axis being approximately 186 \u00d7 106<\/sup> miles in length. Find the shortest and longest distance of the earth from the sun.
\nSolution:
\nAssume that the orbit of the earth is
\n\\(\\frac{x^2}{a^2}+\\frac{y^2}{b^2}=1\\) (a > b)
\nSince the major axis is 186 \u00d7 106 miles
\n2a = 186 \u00d7 106<\/sup> miles
\n\u2234 a = 93 \u00d7 106<\/sup> miles
\nIf \u2019e\u2019 be the eccentricity of the orbit, then
\ne = \\(\\frac{1}{60}\\)
\nwe know that the longest and shortest distances from the earth from the sun are respectively.
\na + ae and a – ae.
\nThe longest distance = a + ae
\n= a(1 + e)
\n= 93 \u00d7 106<\/sup> (1 + \\(\\frac{1}{60}\\))
\n= 9455 \u00d7 104<\/sup> miles
\nThe shortest distance = a – ae
\n= a(1 – e)
\n= 93 \u00d7 106<\/sup> (1 – \\(\\frac{1}{60}\\))
\n= 9145 \u00d7 104<\/sup> miles<\/p>\n

Question 36.
\nA man running on a race course notices that the sum of the distances between the two flag posts is always 10 m and the distance between the flag posts is 8m. Find the equation of the race course traced by the man.
\nSolution:
\nLet S, S’ be the two flag posts on the x-axis, so that ‘O’ is the midpoint of S, S’.
\n\"TS
\nGiven that, the distance between the two flag posts = 8
\n\u21d2 2ae = 8
\n\u21d2 ae = 4
\n\u2234 S(ae, 0) = (4, 0)
\nS'(-ae, 0) = (-4, 0)
\nLet P(x, y) be any point on the locus given that,
\nthe sum of the distances of two flag posts = 10
\nSP + S’P = 10
\nSP = 10 – S’P
\n\"TS
\n\"TS<\/p>\n

Question 37.
\nFind the equation of the ellipse in the standard form, if it passes through the points (-2, 2) and (3, -1). [Mar. ’17 (TS)]
\nSolution:
\nLet the equation of the ellipse in the standard form is
\n\\(\\frac{x^2}{a^2}+\\frac{y^2}{b^2}=1\\) (a > b) ……(1)
\nSince equation (1) passes through the point (-2, 2) then
\n\"TS
\n\"TS
\n\"TS<\/p>\n

\"TS<\/p>\n

Question 38.
\nFind the equation of tangent and normal to the ellipse x2<\/sup> + 2y2<\/sup> – 4x + 12y + 14 = 0 at (2, -1).
\nSolution:
\nGiven equation of the ellipse is x2<\/sup> + 2y2<\/sup> – 4x + 12y + 14 = 0
\nLet the given point P(x1<\/sub>, y1<\/sub>) = (2, -1)
\nThe equation of the tangent at \u2018P\u2019 is S1<\/sub> = 0
\n\u21d2 xx1<\/sub> + 2yy1<\/sub> – 2(x + x1<\/sub>) + 6 (y + y1<\/sub>) + 14 = 0
\n\u21d2 x(2) + 2y(-1) – 2(x + 2) + 6(y – 1) + 14 = 0
\n\u21d2 2x – 2y – 2x – 4 + 6y – 6 + 14 = 0
\n\u21d2 4y + 4 = 0
\n\u21d2 y + 1 = 0
\nThe slope of the tangent at \u2018P\u2019 is m = \\(\\frac{-0}{1}\\) = 0
\nThe slope of the normal at \u2018P\u2019 is \\(\\frac{-1}{\\mathrm{~m}}=\\frac{-1}{0}\\)
\n\u2234 The equation of the normal at \u2018P\u2019 is
\ny – y1<\/sub> = \\(\\frac{-1}{m}\\)(x – x1<\/sub>)
\n\u21d2 y + 1 = \\(\\frac{-1}{0}\\) (x – 2)
\n\u21d2 0 = -x + 2
\n\u21d2 x – 2 = 0<\/p>\n

Question 39.
\nFind the equations of tangents to the ellipse 2x2<\/sup> + 3y2<\/sup> = 11 at the points whose ordinate is 1. [(TS) Mar. ’19, ’16]
\nSolution:
\nGiven the equation of the ellipse is 2x2<\/sup> + 3y2<\/sup> = 11
\ngiven that y = 1
\n2x2<\/sup> + 3 = 11
\n\u21d2 2x2<\/sup> = 8
\n\u21d2 x2<\/sup> = 4
\n\u21d2 x = \u00b12
\nPoints of contact P(2, 1) & Q(-2, 1)
\nCase (i): P(2, 1)
\nThe equation of the tangent is S1<\/sub> = 0
\n\u21d2 2xx1<\/sub> + 3yy1<\/sub> = 11
\n\u21d2 2x(2) + 3y(1) = 11
\n\u21d2 4x + 3y = 11
\nCase (ii): Q(-2, 1)
\nThe equation of the tangent is S1<\/sub> = 0
\n\u21d2 2xx1<\/sub> + 3yy1<\/sub> = 11
\n\u21d2 2x(-2) + 3y(1) = 11
\n\u21d2 -4x + 3y = 11
\n\u21d2 4x – 3y + 11 = 0<\/p>\n

Question 40.
\nFind the equations of the tangents to the ellipse x2<\/sup> + 2y2<\/sup> = 3 drawn from the point (1, 2) and also find the angle between them.
\nSolution:
\nGiven the equation of the ellipse is x2<\/sup> + 2y2<\/sup> = 3
\n\"TS
\nLet the given point P(x1<\/sub>, y1<\/sub>) = (1, 2)
\nThe equation of the pair of tangents to the ellipse x2<\/sup> + 2y2<\/sup> = 3 drawn from the point P(1, 2) is S . S11<\/sub> = \\(\\mathrm{S}_1{ }^2\\)
\n\u21d2 (x2<\/sup> + 2y2<\/sup> – 3) \\(\\left(x_1{ }^2+2 y_1{ }^2-3\\right)\\) = (xx1<\/sub> + 2yy1<\/sub> – 3)2<\/sup>
\n\u21d2 (x2<\/sup> + 2y2<\/sup> – 3)[(1)2<\/sup> + 2(2)2<\/sup> – 3)] = [x(1) + 2y(2) – 3]2<\/sup>
\n\u21d2 (x2<\/sup> + 2y2<\/sup> – 3) (1 + 8 – 3) = (x + 4y – 3)2<\/sup>
\n\u21d2 (x2<\/sup> + 2y2<\/sup> – 3) (6) = (x + 4y – 3)2<\/sup>
\n\u21d2 6x2<\/sup> + 12y2<\/sup> – 18 = x2<\/sup> + 16y2<\/sup> + 9 + 8xy – 24y – 6x
\n\u21d2 5x2<\/sup> – 8xy – 4y2<\/sup> + 6x + 24y – 27 = 0
\nConsider 5x2<\/sup> – 8xy – 4y2<\/sup> = 0
\n\u21d2 5x2<\/sup> – 10xy + 2xy – 4y2<\/sup> = 0
\n\u21d2 5x(x – 2y) + 2y(x – 2y) = 0
\n\u21d2 (x – 2y) (5x + 2y) = 0
\n\u21d2 x – 2y = 0, 5x + 2y = 0
\n\u2234 5x2<\/sup> – 8xy – 4y2<\/sup> + 6x + 24y – 27 = (x – 2y + l) (5x + 2y + k)
\nComparing the coefficient of x on both sides, we get ‘
\nk + 5l = 6 ………(1)
\nComparing the coefficient of y on both sides, we get
\n-2k + 2l = 24
\nk – l = -12 ………(2)
\nSolve (1) & (2)
\n\"TS<\/p>\n

Question 41.
\nFind the condition for the line lx + my + n = 0 to be a tangent to the ellipse \\(\\frac{\\mathbf{x}^2}{a^2}+\\frac{y^2}{b^2}=1\\). [(AP) May ’15]
\nSolution:
\n\"TS
\nGiven equation of the ellipse is \\(\\frac{\\mathbf{x}^2}{a^2}+\\frac{y^2}{b^2}=1\\)
\nSuppose lx + my + n = 0 ……..(1)
\nis a tangent to the ellipse \\(\\frac{\\mathbf{x}^2}{a^2}+\\frac{y^2}{b^2}=1\\)
\nLet P(x1<\/sub>, y1<\/sub>) be the point of contact.
\nThe equation of the tangent at ‘P’ is S1<\/sub> = 0
\n\\(\\frac{x_1}{a^2}+\\frac{y_1}{b^2}-1=0\\) ……..(2)
\nNow (1) & (2) represent the same line.
\n\"TS<\/p>\n

\"TS<\/p>\n

Question 42.
\nFind the condition for the line lx + my + n = 0 to be normal to the ellipse \\(\\frac{x^2}{a^2}+\\frac{y^2}{b^2}=1\\). [Mar. ’01, ’91]
\nSolution:
\n\"TS
\nGiven the equation of the ellipse is
\n\\(\\frac{x^2}{a^2}+\\frac{y^2}{b^2}=1\\) …….(1)
\nSuppose lx + my + n = 0 is a normal to the ellipse \\(\\frac{x^2}{a^2}+\\frac{y^2}{b^2}=1\\)
\nLet P(a cos \u03b8, b sin \u03b8) be a point on the ellipse.
\n\u2234 The equation of the normal at P(\u03b8) is
\n\\(\\frac{a x}{\\cos \\theta}-\\frac{b y}{\\sin \\theta}=a^2-b^2\\) ……..(2)
\nNow (1) & (2) represent the same line.
\n\"TS
\n\"TS<\/p>\n

Question 43.
\n‘C’ is the centre, AA’ and BB’ are the major axis and minor axis of the ellipse \\(\\frac{x^2}{a^2}+\\frac{y^2}{b^2}=1\\). If PN is the ordinate of a point P on the ellipse, then show that \\(\\frac{(P N)^2}{\\left(A^{\\prime} N\\right)(A N)}=\\frac{(B C)^2}{(C A)^2}\\)
\nSolution:
\nLet P(\u03b8) = P(a cos \u03b8, b sin \u03b8) be a point on the ellipse \\(\\frac{x^2}{a^2}+\\frac{y^2}{b^2}=1\\)
\n\u2234 PN = b sin \u03b8, CN = a cos \u03b8 and CA = CA’ = a, CB = CB’ = b
\nLHS
\n\"TS<\/p>\n

Question 44.
\nIf a tangent to the ellipse \\(\\frac{x^2}{a^2}+\\frac{y^2}{b^2}=1\\) (a > b) meets its major axis and minor axis at M and N respectively, then prove that \\(\\frac{a^2}{(\\mathrm{CM})^2}+\\frac{b^2}{(\\mathrm{CN})^2}=1\\) where c is the centre of the ellipse. [(TS) May ’18; (AP) Mar. ’18]
\nSolution:
\nLet P(\u03b8) = P (a cos \u03b8, b sin \u03b8)
\nbe a point on the ellipse \\(\\frac{x^2}{a^2}+\\frac{y^2}{b^2}=1\\)
\nThe equation of tangent at P(\u03b8) is
\n\\(\\frac{x \\cos \\theta}{a}+\\frac{y \\sin \\theta}{b}=1 \\Rightarrow \\frac{x}{\\left(\\frac{a}{\\cos \\theta}\\right)}+\\frac{y}{\\left(\\frac{b}{\\sin \\theta}\\right)}=1\\)
\nIt meets the major axis (x-axis) and minor axis (y-axis) at M and N respectively.
\n\"TS<\/p>\n

Question 45.
\nIf PN is the ordinate of a point P on the ellipse \\(\\frac{x^2}{a^2}+\\frac{y^2}{b^2}=1\\) and the tangent at P meets the x-axis at T, then show that (CN) (CT) = a2<\/sup> where C is the centre of the ellipse.
\nSolution:
\nGiven equation of the ellipse is \\(\\frac{x^2}{a^2}+\\frac{y^2}{b^2}=1\\)
\nLet P(\u03b8) = (a cos \u03b8, b sin \u03b8) be a point on the ellipse \\(\\frac{x^2}{a^2}+\\frac{y^2}{b^2}=1\\)
\n\"TS
\nThe equation of the tangent at P(\u03b8) is
\n\\(\\frac{x}{a} \\cos \\theta+\\frac{y}{b} \\sin \\theta=1\\)
\n\u21d2 \\(\\frac{x}{\\frac{a}{\\cos \\theta}}+\\frac{y}{\\frac{b}{\\sin \\theta}}=1\\)
\nThis tangent meets the x-axis at T, then x-intercept CT = \\(\\frac{a}{\\cos \\theta}\\)
\nThe ordinate of P is PN = b sin \u03b8
\nThe absissa of P is CN = a cos \u03b8
\nLHS = CN . CT
\n= a cos \u03b8 . \\(\\frac{a}{\\cos \\theta}\\)
\n= a2<\/sup>
\n= RHS
\n\u2234 CN . CT = a2<\/sup><\/p>\n

\"TS<\/p>\n

Question 46.
\nShow that the equation of the normal to the ellipse \\(\\frac{x^2}{a^2}+\\frac{y^2}{b^2}=1\\) at P(x1, y1) is \\(\\frac{a^2 x}{x_1}-\\frac{b^2 y}{y_1}=a^2-b^2\\). (May ’98)
\nSolution:
\n\"TS
\n\"TS<\/p>\n

Question 47.
\nShow that the equation of the tangent at P(\u03b8) on the ellipse S = 0 is \\(\\frac{x}{a} \\cos \\theta+\\frac{y}{b} \\sin \\theta=1\\). (Mar. ’99)
\nSolution:
\n\"TS
\nLet the equation of the ellipse be
\nS = \\(\\frac{x^2}{a^2}+\\frac{y^2}{b^2}-1\\) = 0
\nGiven point P = (a cos \u03b8, b sin \u03b8)
\nThe equation of the tangent at \u2018P\u2019 is S1 = 0
\n\"TS<\/p>\n

Question 48.
\nS and T are the foci of an ellipse and B is one end of the minor axis. If STB is an equilateral triangle, then find the eccentricity of the ellipse. [(AP) Mar. ’20]
\nSolution:
\n\"TS
\nLet \\(\\frac{x^2}{a^2}+\\frac{y^2}{b^2}=1\\) (a > b) be an ellipse, whose foci are \u2018S\u2019 and \u2018T\u2019.
\n\u2018B\u2019 is one end of the minor axis such that STB is an equilateral triangle.
\nThen SB = ST = TB.
\nWe have S = (ae, 0), T = (-ae, 0) and B = (0, b)
\nConsider SB = ST
\n\u21d2 SB2<\/sup> = ST2<\/sup>
\n\u21d2 (ae – 0)2<\/sup> + (0 – b)2<\/sup> = (ae + ae)2<\/sup> + (0 – 0)2<\/sup>
\n\u21d2 a2<\/sup>e2<\/sup> + b2<\/sup> = (2ae)2<\/sup>
\n\u21d2 a2<\/sup>e2<\/sup> + a2<\/sup>(1 – e2<\/sup>) = 4a2<\/sup>e2<\/sup>
\n\u21d2 e2<\/sup> + (1 – e2<\/sup>) = 4e2<\/sup>
\n\u21d2 4e2<\/sup> = 1
\n\u21d2 e2<\/sup> = \\(\\frac{1}{4}\\)
\n\u21d2 e = \\(\\frac{1}{2}\\)
\n\u2234 Eccentricity of the ellipse, e = \\(\\frac{1}{2}\\)<\/p>\n

Question 49.
\nIf the ends of the major axis of an ellipse are (5, 0) and (-5, 0). Find the equation of the ellipse in the standard form if its focus lies on the line 3x – 5y – 9 = 0.
\nSolution:
\nLet the equation of the ellipse in the standard form is \\(\\frac{x^2}{a^2}+\\frac{y^2}{b^2}=1\\) (a > b)
\nGiven that the ends of the major axis of an ellipse are A = (a, 0) = (5, 0)
\nA’ = (-a, 0) = (-5, 0)
\n\u2234 a = 5
\nGiven the equation of the straight line is 3x – 5y – 9 = 0
\nFocus \u2018S\u2019 = (ae, 0)
\nSince focus \u2018S\u2019 (ae, 0) lies on the line 3x – 5y – 9 = 0 then
\n3(ae) – 5(0) – 9 = 0
\n\u21d2 3ae – 9 = 0
\n\u21d2 ae = 3
\nWe know that b2<\/sup> = a2<\/sup>(1 – e2<\/sup>)
\n= a2<\/sup> – a2<\/sup>e2<\/sup>
\n= 52<\/sup> – 32<\/sup>
\nb2<\/sup> = 16
\n\u2234 The equation of the ellipse is \\(\\frac{x^2}{25}+\\frac{y^2}{16}=1\\)
\n\u21d2 16x2<\/sup> + 25y2<\/sup> = 400<\/p>\n

\"TS<\/p>\n

Question 50.
\nFind the coordinates of the points on the ellipse x2<\/sup> + 3y2<\/sup> = 37 at which the normal is parallel to the line 6x – 5y = 2.
\nSolution:
\nGiven the equation of the ellipse is x2<\/sup> + 3y2<\/sup> = 37
\n\"TS
\nLet P(a cos \u03b8, b sin \u03b8) be a point on the ellipse
\nThe equation of the tangent at P(\u03b8) is
\n\\(\\frac{x}{a} \\cos \\theta+\\frac{y}{b} \\sin \\theta=1\\)
\nThe slope of the tangent at P is
\n\"TS
\nGiven the equation of the straight line is 6x – 5y = 2
\nSlope of the line is m = \\(\\frac{-6}{-5}=\\frac{6}{5}\\)
\nSince the normal is parallel to the line 6x – 5y = 2, the slopes are equal.
\n\"TS
\nSince tan \u03b8 is positive, then P = (5, 2), (-5, -2).<\/p>\n","protected":false},"excerpt":{"rendered":"

Students must practice these Maths 2B Important Questions TS Inter Second Year Maths 2B Ellipse Important Questions to help strengthen their preparations for exams. TS Inter Second Year Maths 2B Ellipse Important Questions Question 1. 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