Maths 2B Important Questions<\/a> TS Inter Second Year Maths 2B Ellipse Important Questions to help strengthen their preparations for exams.<\/p>\nTS 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
\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
\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>\nQuestion 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<\/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>\nQuestion 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>\nQuestion 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<\/p>\nQuestion 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<\/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
\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>\nQuestion 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
\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<\/p>\nQuestion 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<\/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>\nQuestion 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
\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>\nQuestion 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
\n<\/p>\nQuestion 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
\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
\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<\/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
\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
\nSince eq.(1) passes through one end of the minor axis B'(0, -b) then
\n<\/p>\nQuestion 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
\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
\n
\n<\/p>\nQuestion 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
\n<\/p>\nQuestion 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<\/p>\nQuestion 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<\/p>\nQuestion 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
\n<\/p>\n<\/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
\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<\/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
\n
\n<\/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
\n<\/p>\nQuestion 23.
\nShow that the point of intersection of the perpendicular tangent to an ellipse lies on a circle. [(AP) Mar. ’16]
\nSolution:
\n
\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
\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> = a