{"id":36358,"date":"2022-11-28T15:56:58","date_gmt":"2022-11-28T10:26:58","guid":{"rendered":"https:\/\/tsboardsolutions.com\/?p=36358"},"modified":"2022-12-03T14:18:27","modified_gmt":"2022-12-03T08:48:27","slug":"maths-2b-hyperbola-important-questions-short-answer-type","status":"publish","type":"post","link":"https:\/\/tsboardsolutions.com\/maths-2b-hyperbola-important-questions-short-answer-type\/","title":{"rendered":"TS Inter Second Year Maths 2B Hyperbola Important Questions Short Answer Type"},"content":{"rendered":"

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

TS Inter Second Year Maths 2B Hyperbola Important Questions Short Answer Type<\/h2>\n

Question 1.
\nOne focus of a hyperbola is located at the point (1, -3) and the corresponding directrix is the line y = 2. Find the equation of the hyperbola if its eccentricity is \\(\\frac{3}{2}\\).
\nSolution:
\n\"TS
\nGiven that focus S = (1, -3)
\nThe equation of the directrix is y – 2 = 0
\nEccentricity e = \\(\\frac{3}{2}\\)
\nLet P(x, y) be any point on the hyperbola.
\nSince P(x, y) lies on the hyperbola then SP
\n\\(\\frac{SP}{PM}\\) = e
\nSP = e . PM
\n\u21d2 \\(\\sqrt{(x-1)^2+(y+3)^2}=\\frac{3}{2}\\left|\\frac{0 \\cdot x+y \\cdot 1-2}{\\sqrt{0^2+1^2}}\\right|\\)
\n\u21d2 \\(\\sqrt{(x-1)^2+(y+3)^2}=\\frac{3}{2}|y-2|\\)
\nSquaring on both sides
\n\u21d2 (x – 1)2<\/sup> + (y + 3)2<\/sup> = \\(\\frac{9}{4}\\)(y – 2)2<\/sup>
\n\u21d2 x2<\/sup> + 1 – 2x + y2<\/sup> + 9 + 6y = \\(\\frac{9}{4}\\)(y2<\/sup> + 4 – 4y)
\n\u21d2 4x2<\/sup> + 4 – 8x + 4y2<\/sup> + 24y + 36 = 9y2<\/sup> – 36y + 36
\n\u21d2 4x2<\/sup> – 5y2<\/sup> – 8x + 60y + 4 = 0
\n\u2234 The equation of the hyperbola is 4x2<\/sup> – 5y2<\/sup> – 8x + 60y + 4 = 0.<\/p>\n

\"TS<\/p>\n

Question 2.
\nFind the centre, foci, eccentricity, equation of the directrices, and length of the latus rectum of the hyperbola x2<\/sup> – 4y2<\/sup> = 4. [(AP) Mar. ’20, ’18, ’16; May ’16; (TS) Mar. ’19]
\nSolution:
\nGiven the equation of the hyperbola is x2<\/sup> – 4y2<\/sup> = 4
\n\u21d2 \\(\\frac{x^2}{4}-\\frac{4 y^2}{4}=1\\)
\n\u21d2 \\(\\frac{x^2}{4}-\\frac{y^2}{1}=1\\)
\nHere a = 2, b = 1
\nIt is a hyperbola.
\nCentre = (0, 0)
\n\"TS
\n\"TS<\/p>\n

Question 3.
\nFind the centre, foci, eccentricity, equation of the directrices, and length of the latus rectum of the hyperbola 16y2<\/sup> – 9x2<\/sup> = 144. [(AP) Mar. ’18, May ’17]
\nSolution:
\nGiven the equation of the hyperbola is 16y2<\/sup> – 9x2<\/sup> = 144
\n\u21d2 \\(\\frac{y^2}{9}-\\frac{x^2}{16}=1\\)
\n\u21d2 \\(\\frac{x^2}{16}-\\frac{y^2}{9}=-1\\)
\nHere a = 4, b = 3
\nIt is a conjugate hyperbola.
\nCentre C = (0, 0)
\n\"TS<\/p>\n

Question 4.
\nFind the centre, foci, eccentricity, equation of the directrices, and length of the latus rectum of the hyperbola 5x2<\/sup> – 4y2<\/sup> + 20x + 8y = 4.
\nSolution:
\nGiven equation of the hyperbola is 5x2<\/sup> – 4y2<\/sup> + 20x + 8y = 4
\n\u21d2 (5x2<\/sup> + 20x) + (-4y2<\/sup> + 8y) = 4
\n\u21d2 5(x2<\/sup> + 4x) – 4(y2<\/sup> – 2y) = 4
\n\u21d2 5(x2<\/sup> + 2 . 2x + 22<\/sup> – 22<\/sup>) – 4(y2<\/sup> – 2 . 1 . y + 12<\/sup> – 12<\/sup>) = 4
\n\u21d2 5((x + 2)2<\/sup> – 4) – 4((y – 1)2<\/sup> – 1) = 4
\n\u21d2 5(x + 2)2<\/sup> – 20 – 4(y – 1)2<\/sup> + 4 = 4
\n\u21d2 5(x + 2)2<\/sup> – 4(y – 1)2<\/sup> = 20
\n\"TS
\nwe get h = -2; k = 1; a = 2; b = \u221a3
\nIt is a hyperbola.
\nCentre C = (h, k) = (-2, 1)
\n\"TS
\n\u21d2 x = -2 \u00b1 \\(\\frac{4}{3}\\)
\n\u21d2 3x = -6 \u00b1 4
\n\u21d2 3x = -10 or 3x = -2
\n\u21d2 3x + 10 = 0 or 3x + 2 = 0
\nThe length of the latus rectum
\n\"TS<\/p>\n

Question 5.
\nIf the line lx + my + n = 0 is a tangent to the hyperbola \\(\\frac{x^2}{a^2}-\\frac{y^2}{b^2}=1\\), then show that a2<\/sup>l2<\/sup> – b2<\/sup>m2<\/sup> = n2<\/sup>. (May ’07)
\nSolution:
\n\"TS
\nGiven equation of the hyperbola is \\(\\frac{x^2}{a^2}-\\frac{y^2}{b^2}=1\\)
\nLet the line lx + my + n = 0 ………(1)
\nbe a tangent to the given hyperbola at P(x1<\/sub>, y1<\/sub>)
\n\u2234 The equation of the tangent at P(x1<\/sub>, y1<\/sub>) is S1<\/sub> = 0
\n\\(\\frac{\\mathrm{xx}_1}{\\mathrm{a}^2}-\\frac{\\mathrm{yy}_1}{\\mathrm{~b}^2}-1=0\\)
\nNow (1) and (2) represent the same line.
\n\"TS
\nSince P(x1<\/sub>, y1<\/sub>) lies on line (1) then
\nlx1<\/sub> + my1<\/sub> + n = 0
\n\\(l\\left(\\frac{-\\mathrm{a}^2 l}{\\mathrm{n}}\\right)+\\mathrm{m}\\left(\\frac{\\mathrm{b}^2 \\mathrm{~m}}{\\mathrm{n}}\\right)+\\mathrm{n}=0\\)
\n\u21d2 -a2<\/sup>l2<\/sup> + b2<\/sup>m2<\/sup> + n2<\/sup> = 0
\n\u21d2 n2<\/sup> = a2<\/sup>l2<\/sup> – b2<\/sup>m2<\/sup>
\n\u21d2 a2<\/sup>l2<\/sup> – b2<\/sup>m2<\/sup> = n2<\/sup><\/p>\n

\"TS<\/p>\n

Question 6.
\nFind the equations of the tangents to the hyperbola x2<\/sup> – 4y2<\/sup> = 4 which are (i) parallel (ii) perpendicular to the line x + 2y = 0. [Mar. ’19 (AP) (TS) ’15 (TS); Mar. ’14; May ’14, ’13]
\nSolution:
\nGiven the equation of the hyperbola is x2<\/sup> – 4y2<\/sup> = 4
\n\u21d2 \\(\\frac{x^2}{4}-\\frac{y^2}{1}=1\\)
\nGiven the equation of the straight line is x + 2y = 0
\n(i) The equation of the tangent parallel to the line x + 2y = 0 is x + 2y + k = 0 ………(1)
\n2y = -x – k
\ny = \\(\\frac{-\\mathrm{x}}{2}-\\frac{\\mathrm{k}}{2}\\)
\nComparing with y = mx + c,
\nwe get m = \\(\\frac{-1}{2}\\), c = \\(\\frac{-k}{2}\\)
\nSince equation (1) is a tangent to the given hyperbola then
\nc2<\/sup> = a2<\/sup>m2<\/sup> – b2<\/sup>
\n\u21d2 \\(\\left(\\frac{-k}{2}\\right)^2=4\\left(\\frac{-1}{2}\\right)^2-1\\)
\n\u21d2 \\(\\frac{k^2}{4}=\\frac{4}{4}-1\\)
\n\u21d2 k2<\/sup> = 0
\n\u21d2 k = 0
\nSubstitute the value of k in equation (1)
\nx + 2y + 0 = 0
\n\u21d2 x + 2y = 0
\n\u2234 No parallel tangents to x + 2y = 0.
\n(ii) The equation of the tangent perpendicular to line x + 2y = 0 is
\n2x – y + k = 0
\n\u21d2 y = 2x + k ……..(2)
\nComparing with y = mx + c,
\nwe get m = 2, c = k.
\nSince equation (2) is a tangent to the given hyperbola then
\nc2<\/sup> = a2<\/sup>m2<\/sup> – b2<\/sup>
\n\u21d2 k2<\/sup> = 4(2)2<\/sup> – (1)2<\/sup> = 16 – 1
\n\u21d2 k2<\/sup> = 15
\n\u21d2 k = \u00b1\u221a15
\nSubstitute the value of k in equation (2)
\ny = 2x \u00b1 \u221a15
\n\u2234 The required perpendicular tangents are y = 2x \u00b1 \u221a15<\/p>\n

Question 7.
\nFind the equations of the tangents to the hyperbola 3x2<\/sup> – 4y2<\/sup> =12 which are (i) Parallel and (ii) Perpendicular to the line y = x – 7. [(TS) Mar. ’20, May ’18, ’15; (AP) Mar. ’17, ’15, May ’16, ’15]
\nSolution:
\nGiven the equation of the hyperbola is 3x2<\/sup> – 4y2<\/sup> = 12
\n\u21d2 \\(\\frac{x^2}{4}-\\frac{y^2}{3}=1\\)
\nHere a2<\/sup> = 4, b2<\/sup> = 3
\nGiven the equation of the straight line is y = x – 7
\n\u21d2 x – y – 7 = 0
\n(i) The equation of the tangent parallel to the line x – y – 7 = 0 is
\nx – y + k = 0 ………(i)
\ny = x + k
\nComparing with y = mx + c,
\nwe get m = 1, c = k
\nSince equation (1) is a tangent to the given hyperbola then
\nc2<\/sup> = a2<\/sup>m2<\/sup> – b2<\/sup>
\n\u21d2 k2<\/sup> = 4(1)2<\/sup> – 3
\n\u21d2 k2<\/sup> = 1
\n\u21d2 k = \u00b11
\nSubstitute the value of k in equation (1)
\n\u2234 The required parallel tangents are x – y \u00b1 1 = 0
\n(ii) The equation of the tangent perpendicular to the line x – y – 7 = 0 is
\nx + y + k = 0 …….(2)
\n\u21d2 y = -x – k
\nComparing with y = mx + c,
\nwe get m = -1, c = -k
\nSince equation (2) is a tangent to the given hyperbola then
\nc2<\/sup> = a2<\/sup>m2<\/sup> – b2<\/sup>
\n(-k)2<\/sup> = 4(-1)2<\/sup> – 3
\n\u21d2 k2<\/sup> = 4 – 3
\n\u21d2 k2<\/sup> = 1
\n\u21d2 k = \u00b11
\nSubstitute the value of k in equation (2)
\n\u2234 The required perpendicular tangents are x + y \u00b1 1 = 0.<\/p>\n

Question 8.
\nFind the equations of the tangents drawn to the hyperbola 2x2<\/sup> – 3y2<\/sup> = 6 through (-2, 1).
\nSolution:
\nGiven the equation of the hyperbola is 2x2<\/sup> – 3y2<\/sup> = 6
\n\u21d2 \\(\\frac{x^2}{3}-\\frac{y^2}{2}=1\\)
\nHere a2<\/sup> = 3, b2<\/sup> = 2
\nLet, the given point P(x, y) = (-2, 1)
\nLet, the equation of the tangent to the hyperbola is
\ny = mx \u00b1 \\(\\sqrt{a^2 m^2-b^2}\\)
\n\u21d2 y = mx \u00b1 \\(\\sqrt{3 m^2-2}\\) ……..(1)
\nSince, this tangent passes through the point P(-2, 1)
\ny = m(-2) \u00b1 \\(\\sqrt{3 m^2-2}\\)
\n\u21d2 1 = -2m \u00b1 \\(\\sqrt{3 m^2-2}\\)
\n\u21d2 1 + 2m = \u00b1\\(\\sqrt{3 m^2-2}\\)
\nSquaring on both sides
\n(1 + 2m)2<\/sup> = (\u00b1\\(\\sqrt{3 m^2-2}\\))2<\/sup>
\n\u21d2 1 + 4m2<\/sup> + 4m = 3m2<\/sup> – 2
\n\u21d2 m2<\/sup> + 4m + 3 = 0
\n\u21d2 m2<\/sup> + 3m + m + 3 = 0
\n\u21d2 m(m + 3) + 1(m + 3) = 0
\n\u21d2 (m + 3)(m + 1) = 0
\n\u21d2 m + 3 = 0 (or) m + 1 = 0
\n\u21d2 m = -3 (or) m = -1
\nCase (i): If m = -1, then the required tangents are from (1)
\ny = (-1)x \u00b1 \\(\\sqrt{3(-1)^2-2}\\)
\n\u21d2 y = -x \u00b1 \\(\\sqrt{3-2}\\)
\n\u21d2 y = -x \u00b1 1
\n\u21d2 x + y \u00b1 1 = 0
\nSince, the point (-2, 1) does not lie on the line x + y – 1 = 0
\n\u2234 The tangent is x + y + 1 = 0
\nCase (ii): If m = -3, then the required tangents are from (1)
\ny = (-3)x \u00b1 \\(\\sqrt{3(-3)^2-2}\\)
\n\u21d2 y = -3x \u00b1 \\(\\sqrt{27-2}\\)
\n\u21d2 y = -3x \u00b1 \u221a25
\n\u21d2 3x + y \u00b1 5 = 0
\nSince, the point (-2, 1) does not lie on the line 3x + y – 5 = 0
\n\u2234 The tangent is 3x + y + 5 = 0
\n\u2234 Required tangents are x + y + 1 = 0, 3x + y + 5 = 0<\/p>\n

\"TS<\/p>\n

Question 9.
\nFind the equation of the hyperbola whose asymptotes are the straight lines x + 2y + 3 = 0, 3x + 4y + 5 = 0, and which passes through the point (1, -1).
\nSolution:
\nGiven that the asymptotes of a hyperbola are
\nx + 2y + 3 = 0 ……….(1)
\n3x + 4y + 5 = 0 ……….(2)
\nLet, the given point P = (1, -1)
\nThe equation of the hyperbola whose asymptotes are the straight lines (1) & (2) is
\n(x + 2y + 3) (3x + 4y + 5) + k = 0 ……….(3)
\nSince equation (3) passes through the point P(1, -1) then
\n(1 + 2(-1) + 3) (3(1) + 4(-1) + 5) + k = 0
\n\u21d2 (1 – 2 + 3) (3 – 4 + 5) + k = 0
\n\u21d2 8 + k = 0
\n\u21d2 k = -8
\nSubstitute the value of \u2018k\u2019 in eq. (3)
\n\u2234 The required equation of the hyperbola is (x + 2y + 3) (3x + 4y + 5) – 8 = 0
\n\u21d2 3x2<\/sup> + 4xy + 5x + 6xy + 8y2<\/sup> + 10y + 9x + 12y + 15 – 8 = 0
\n\u21d2 3x2<\/sup> + 10xy + 8y2<\/sup> + 14x + 22y + 7 = 0<\/p>\n

Question 10.
\nProve that the product of the perpendicular distances from any point on a hyperbola to its asymptotes is constant.
\nSolution:
\nLet S = \\(\\frac{x^2}{a^2}+\\frac{y^2}{b^2}-1\\) = 0 be the given hyperbola.
\nLet P = (a sec \u03b8, b tan \u03b8) be any point on S = 0.
\nThe equations of asymptotes of hyperbola S = 0 are \\(\\frac{x}{a}+\\frac{y}{b}=0\\) and \\(\\frac{x}{a}-\\frac{y}{b}=0\\)
\n\u21d2 bx + ay = 0 ……..(1)
\nand bx – ay = 0 ……… (2)
\nLet PM be the length of the perpendicular drawn from P(a sec \u03b8, b tan \u03b8) on line (1).
\n\u2234 PM = \\(\\frac{|b a \\sec \\theta+a b \\tan \\theta|}{\\sqrt{a^2+b^2}}\\)
\nLet PN be the length of the perpendicular drawn from P (a sec \u03b8, b tan \u03b8) on line (2).
\n\"TS
\n\"TS
\n\u2234 The product of the perpendicular distances from any point on a hyperbola to its asymptotes is a constant.<\/p>\n

Question 11.
\nFind the centre, foci, eccentricity, equation of the directrices, and length of the latus rectum of the hyperbola 9x2<\/sup> – 16y2<\/sup> + 72x – 32y – 16 = 0. (May ’01)
\nSolution:
\nGiven equation of the hyperbola is 9x2<\/sup> – 16y2<\/sup> + 72x – 32y -16 = 0
\n\u21d2 (9x2<\/sup> + 72x) +(-16y2<\/sup> – 32y) = 16
\n\u21d2 9(x2<\/sup> + 8x) – 16 (y2<\/sup> + 2y) = 16
\n\u21d2 9(x2<\/sup> + 2 . 4 . x + 42<\/sup> – 42<\/sup>) – 16(y2<\/sup> + 2 . 1 . y + 12<\/sup> – 12<\/sup>) = 16
\n\u21d2 9((x + 4)2<\/sup> – 16) – 16((y + 1)2<\/sup> – 1) = 16
\n\u21d2 9(x + 4)2<\/sup> – 144 – 16(y + 1)2<\/sup> + 16 = 16
\n\u21d2 9(x + 4)2<\/sup> – 16(y + 1)2<\/sup> = 144
\n\"TS
\nwe get h = -4, k = -1, a = 4, b = 3.
\nIt is a hyperbola.
\nCentre C = (h, k) = (-4, -1)
\n\"TS
\nx = -4 \u00b1 \\(\\frac{16}{5}\\)
\n5x = -20 + 16 or 5x = – 20 – 16
\n5k = -4 or 5x = -36
\n5x + 4 = 0 or 5x + 36 = 0
\nLength of latus rectum = \\(\\frac{2 \\mathrm{~b}^2}{\\mathrm{a}}=\\frac{9}{2}\\)<\/p>\n

Question 12.
\nFind the centre, eccentricity, foci, directrices, and the length of the latus rectum of the hyperbola 4x2<\/sup> – 9y2<\/sup> – 8x – 32 = 0.
\nSolution:
\nGiven equation of the hyperbola is 4x2<\/sup> – 9y2<\/sup> – 8x – 32 = 0
\n\u21d2 (4x2<\/sup> – 8x) + (-9y2<\/sup>) = 32
\n\u21d2 4(x2<\/sup> – 2x) – 9y2<\/sup> = 32
\n\u21d2 4((x)2<\/sup> – 2 . 1 . x + 12<\/sup> – 12<\/sup>) – 9y2<\/sup> = 32
\n\u21d2 4((x – 1)2<\/sup> – 1) – 9y2<\/sup> = 32
\n\u21d2 4(x – 1)2<\/sup> – 4 – 9y2<\/sup> = 32
\n\u21d2 4(x – 1)2<\/sup> – 9y2<\/sup> = 36
\n\"TS
\nwe get a = 3, b = 2, h = 1, k = 0
\nIt is a hyperbola.
\nCentre C(h, k) = (1, 0)
\n\"TS<\/p>\n

Question 13.
\nTangents to the hyperbola \\(\\frac{x^2}{a^2}-\\frac{y^2}{b^2}=1\\) make angles \u03b81<\/sub>, \u03b82<\/sub> with transverse axis of a hyperbola. Show that the point of intersection of these tangents lies on the curve 2xy = k(x2<\/sup> – a2<\/sup>), when tan \u03b81<\/sub> + tan \u03b82<\/sub> = k. [(TS) May ’19, ’16; (AP) May ’18]
\nSolution:
\nThe equation of the hyperbola is \\(\\frac{x^2}{a^2}-\\frac{y^2}{b^2}=1\\)
\nLet P(x1<\/sub>, y1<\/sub>) be the point of intersection of the tangents
\n\"TS
\n\"TS<\/p>\n

\"TS<\/p>\n

Question 14.
\nProve that the point of intersection of two perpendicular tangents to the hyperbola \\(\\frac{x^2}{a^2}-\\frac{y^2}{b^2}=1\\) lies on the circle x2<\/sup> + y2<\/sup> = a2<\/sup> – b2<\/sup>. [(TS) May ’17; Mar. ’16]
\nSolution:
\nGiven equation of the hyperbola is \\(\\frac{x^2}{a^2}-\\frac{y^2}{b^2}=1\\)
\nEquation of any tangent to the hyperbola is y = mx \u00b1 \\(\\sqrt{a^2 m^2-b^2}\\)
\nSuppose P(x1<\/sub>, y1<\/sub>) is the point of intersection of a tangent.
\nSince P lies on the tangent 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
\nThis is a quadratic equation in m given the values for m say m1<\/sub> and m2<\/sub>.
\nThe tangents are perpendicular then m1<\/sub>m2<\/sub> = -1
\n\"TS
\n\u2234 P(x1<\/sub>, y1<\/sub>) lies on the circle x2<\/sup> + y2<\/sup> = a2<\/sup> – b2<\/sup><\/p>\n

Question 15.
\nShow that the locus of feet of the \u22a5ars drawn from foci to any tangent of the hyperbola \\(\\frac{x^2}{a^2}-\\frac{y^2}{b^2}=1\\) is the auxiliary circle of the hyperbola.
\nSolution:
\n\"TS
\nLet the equation of hyperbola 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 \u22a5ar drawn from either of the foci to a tangent.
\nThe equation of the tangent to the hyperbola S = 0 is
\ny = mx \u00b1 \\(\\sqrt{a^2 m^2-b^2}\\) …….(1)
\nThe equation to the \u22a5ar from either focus (\u00b1ae, 0) on this tangent is
\ny – y1<\/sub> = \\(\\frac{1}{m}\\)(x – x1<\/sub>)
\n\"TS
\n\"TS
\n\u2234 P lies on x2<\/sup> + y2<\/sup> = a2<\/sup> which is a circle with the centre as the origin, the centre of the hyperbola.<\/p>\n

Question 16.
\nFind the centre, eccentricity, foci, length of latus rectum, and equations of the directrices of the hyperbola 4(y + 3)2<\/sup> – 9(x – 2)2<\/sup> = 1. [(AP) May ’19]
\nSolution:
\nGiven hyperbola is 4(y + 3)2<\/sup> – 9(x – 2)2<\/sup> = 1
\n\u21d2 \\(\\frac{(y+3)^2}{\\frac{1}{4}}-\\frac{(x-2)^2}{\\frac{1}{9}}=1\\)
\n\"TS
\n\"TS<\/p>\n

\"TS<\/p>\n

Question 17.
\nShow that the equation of a hyperbola in the standard form is \\(\\frac{x^2}{a^2}-\\frac{y^2}{b^2}=1\\). [May ’02, ’99, ’98]
\nSolution:
\nProof: Let ‘S’ be the focus, ‘e’ be the eccentricity and L = 0 be the directrix of the hyperbola.
\n\"TS
\nLet ‘P’ be a point on the hyperbola.
\nLet MZ be the projection of PS on the directrix L = 0 respectively.
\nLet ‘N’ be the projection of ‘P’ on SZ.
\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 ‘C’ be the midpoint of AA’.
\nPoints A, A’ lies on the hyperbola then
\n\\(\\frac{\\mathrm{SA}}{\\mathrm{AZ}}=\\frac{\\mathrm{SA}^{\\prime}}{\\mathrm{ZA}^{\\prime}}=\\frac{\\mathrm{e}}{1}\\)
\nNow \\(\\frac{\\mathrm{SA}}{\\mathrm{AZ}}=\\frac{\\mathrm{e}}{1}\\)
\nSA = e AZ
\nCS – CA = e (CA – CZ) ……..(1)
\nand \\(\\frac{\\mathrm{SA}^{\\prime}}{\\mathrm{ZA}^{\\prime}}=\\frac{\\mathrm{e}}{1}\\)
\nSA’ = eZA’
\nCS + CA’ = e (CZ + CA’) ……….(2)
\nNow (1) + (2),
\n(CS – CA) + (CS + CA’) = e (CA – CZ) + e(CA’ + CZ)
\nCS – CA + CS + CA’ = e (CA – CZ + CZ + CA’)
\n2CS – CA + CA’ = e (CA + CA’)
\nSince ‘C’ is the midpoint of AA’ then CA = CA’
\n2CS – CA + CA = e (CA + CA)
\n2CS = 2eCA
\nCS = e CA
\nCS = ae (\u2235 CA = a)
\n\u2234 The coordinates of focus S = (ae, 0)
\nNow (1) – (2),
\n(CS – CA) – (CS + CA’) = e(CA – CZ) – e(CA’ + CZ)
\nCS – CA – CS – CA’ = e(CA – CZ – CA’ – CZ) – CA – CA’ = e (CA – 2CZ – CA’)
\nSince ‘C’ is the midpoint of AA’ then CA = CA’
\n-CA – CA = e(CA – 2CZ – CA)
\n– 2CA = -2e CZ
\nCA = e CZ
\na = e CZ
\nCZ = \\(\\frac{a}{e}\\)
\n\u2234 The equation of the directrix is x = \\(\\frac{a}{e}\\).
\nTake CS, the principal axis of the hyperbola as the X-axis, and CY \u22a5 CS as Y-axis.
\nThen S = (ae, 0) and the hyperbola is in the standard form.
\nLet P = (x, y)
\nNow, PM = ZN = CN – CZ = x – \\(\\frac{a}{e}\\)
\nP lies on the hyperbola
\n\\(\\frac{SP}{PM}\\) = e
\nSP = e PM
\n\\(\\sqrt{(x-a e)^2+(y-0)^2}=e\\left(x-\\frac{a}{e}\\right)\\)
\nSquaring on both sides
\n(x – ae)2<\/sup> + y2<\/sup> = \\(\\mathrm{e}^2\\left(\\mathrm{x}-\\frac{\\mathrm{a}}{\\mathrm{e}}\\right)^2\\)
\n\"TS
\nwhere b2<\/sup> = a2<\/sup>(e2<\/sup> – 1) > 0
\nThe locus of ‘P’ is \\(\\frac{x^2}{a^2}-\\frac{y^2}{b^2}=1\\)
\n\u2234 The equation of the hyperbola is \\(\\frac{x^2}{a^2}-\\frac{y^2}{b^2}=1\\)<\/p>\n

Question 18.
\nShow that the difference in focal distances of any point on the hyperbola is constant
\n(or)
\nThe difference of the focal distances of any point on the hyperbola is constant i.e. if ‘P’ is a point on the hyperbola \\(\\frac{x^2}{a^2}-\\frac{y^2}{b^2}=1\\) with foci S and S’ then S’P – SP = 2a.
\nSolution:
\n\"TS
\nLet P (x, y) be any point on the hyperbola whose centre is the origin ‘C’.
\nFoci are S, S’
\nDirectrices of ZM and Z’M’.
\nLet PN, PM, PM’ be the perpendiculars drawn from ‘P’ upon X-axis and the two directrices respectively.
\nNow, \\(\\frac{SP}{PM}\\) = e
\nSP = e PM
\n= e ZN
\n= e (CN – CZ)
\n= e (x – \\(\\frac{a}{e}\\))
\nAlso = \\(\\frac{\\mathrm{S}^{\\prime} \\mathrm{P}}{\\mathrm{PM}^{\\prime}}=\\mathrm{e}\\)
\nS’P = ePM’
\n= eNZ’
\n= e (CN + CZ’)
\n= e (x + \\(\\frac{a}{e}\\))
\nLHS = S’P – SP
\n= \\(e\\left(x+\\frac{a}{e}\\right)-e\\left(x-\\frac{a}{e}\\right)\\)
\n= ex + a – ex + a
\n= 2a (constant)
\n\u2234 S’P – SP = 2a<\/p>\n

Question 19.
\nShow that the condition for straight line y = mx + c to be a tangent to the hyperbola S = 0 is c2<\/sup> = a2<\/sup>m2<\/sup> – b2<\/sup>. [Mar. ’03]
\nSolution:
\nSuppose y = mx + c ………(1) is a tangent to the hyperbola \\(\\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 P is
\n\\(\\frac{x x_1}{a^2}-\\frac{y y_1}{b^2}-1=0\\) ……..(2)
\nNow (1) and (2) represent the same line
\n\"TS<\/p>\n

Question 20.
\nShow that the equation of the normal at P(\u03b8) on the hyperbola \\(\\frac{x^2}{a^2}-\\frac{y^2}{b^2}=1\\) is \\(\\frac{a x}{\\sec \\theta}+\\frac{b y}{\\tan \\theta}\\) = a2<\/sup> + b2<\/sup>. [Mar. ’06]
\nSolution:
\nThe equation of the tangent at P(\u03b8) is \\(\\frac{x \\sec \\theta}{a}-\\frac{y \\tan \\theta}{b}=1\\)
\nSlope of tangent = \\(\\frac{\\frac{-\\sec \\theta}{a}}{\\frac{-\\tan \\theta}{b}}=\\frac{b \\sec \\theta}{a \\tan \\theta}\\)
\nSlope of normal = \\(-\\frac{a \\tan \\theta}{b \\sec \\theta}\\)
\n\u2234 The equation of the normal at P is
\ny – b tan \u03b8 = \\(-\\frac{a \\tan \\theta}{b \\sec \\theta}\\) (x – a sec \u03b8)
\n\u21d2 by sec \u03b8 – b2<\/sup> sec \u03b8 tan \u03b8 = -ax tan \u03b8 + a2<\/sup> sec \u03b8 tan \u03b8
\n\u21d2 ax tan \u03b8 + by sec \u03b8 = (a2<\/sup> + b2<\/sup>) sec \u03b8 tan \u03b8
\n\u21d2 \\(\\frac{a x}{\\sec \\theta}+\\frac{b y}{\\tan \\theta}=a^2+b^2\\)<\/p>\n

\"TS<\/p>\n

Question 21.
\nIf the line lx + my = 1 is a normal to the hyperbola \\(\\frac{x^2}{a^2}-\\frac{y^2}{b^2}=1\\), then show that \\(\\frac{a^2}{l^2}-\\frac{b^2}{m^2}=\\left(a^2+b^2\\right)^2\\).
\nSolution:
\n\"TS
\nGiven equation of the hyperbola is \\(\\frac{x^2}{a^2}-\\frac{y^2}{b^2}=1\\)
\nLet the line lx + my – 1 = 0 ……….(1) be normal to the given hyperbola at P(\u03b8).
\nThe equation of normal at P(\u03b8) to the given hyperbola is
\n\\(\\frac{a x}{\\sec \\theta}+\\frac{b y}{\\tan \\theta}\\) = a2<\/sup> + b2<\/sup> …….(2)
\nNow (1) & (2) represent the same line.
\n\"TS
\nwhich is the required condition.<\/p>\n

Question 22.
\nShow that the equation \\(\\frac{x^2}{9-c}+\\frac{y^2}{5-c}=1\\) represents
\n(i) an ellipse if ‘c’ is a real constant less than 5.
\n(i) a hyperbola if ‘c’ is any real constant between 5 and 9.
\n(iii) Show that each ellipse in (a) and each hyperbola in (b) has foci at the two points (\u00b12, 0), independent of the value ‘c’.
\nSolution:
\n(i) Given equation is \\(\\frac{x^2}{9-c}+\\frac{y^2}{5-c}=1\\)
\nThis equation represents an ellipse if
\n9 – c > 0 and 5 – c > 0
\n9 > c and 5 > c
\nc < 9 and c < 5
\n\u2234 c < 5
\n(ii) Given equation is \\(\\frac{x^2}{9-c}+\\frac{y^2}{5-c}=1\\)
\nThis equation represents a hyperbola if
\n9 – c > 0 and 5 – c < 0
\n9 > 0 and 5 < c
\nc < 9 and c > 5
\n\u2234 5 < c < 9
\n(iii) In case (i): a2<\/sup> = 9 – c, b2<\/sup> = 5 – c
\nNow, a2<\/sup> – b2<\/sup> = (9 – c) – (5 – c)
\n\u21d2 a2<\/sup> – a2<\/sup>(1 – e2<\/sup>) = 9 – c – 5 + c
\n\u21d2 a2<\/sup>e2<\/sup> = 9 – 5
\n\u21d2 a2<\/sup>e2<\/sup> = 4
\n\u21d2 ae = 2
\n\u2234 foci are (\u00b1ae, 0) = (\u00b12, 0)
\nIn case (ii): a2<\/sup> = 9 – c, b2<\/sup> = c – 5
\nNow, a2<\/sup> + b2<\/sup> = 9 – c + c – 5
\n\u21d2 a2<\/sup> + a2<\/sup>(e2<\/sup> – 1) = 9 – 5
\n\u21d2 a2<\/sup>e2<\/sup> = 4
\n\u21d2 ae = 2
\n\u2234 foci are (\u00b1ae, 0) = (\u00b12, 0)
\nThese are independent of the value of ‘c’<\/p>\n

\"TS<\/p>\n

Question 23.
\nA circle cuts the rectangular hyperbola xy = 1, in the points (x1<\/sub>, y1<\/sub>), r = 1, 2, 3, 4. Prove that x1<\/sub>x2<\/sub>x3<\/sub>x4<\/sub> = y1<\/sub>y2<\/sub>y3<\/sub>y4<\/sub> = 1.
\nSolution:
\nLet the circle be x2<\/sup> + y2<\/sup> = a2<\/sup>
\nSince (t, \\(\\frac{1}{2}\\)) (t \u2260 0) lies on xy = 1
\nThe points of intersection of the circle and the hyperbola are given by
\n\\(t^2+\\frac{1}{t^2}=a^2\\)
\n\u21d2 t4<\/sup> – a2<\/sup>t2<\/sup> + 1 = 0
\n\u21d2 t4<\/sup> + 0 . t3<\/sup> – a2<\/sup>t + 0 . t + 1 = 0
\nIf t1<\/sub>, t2<\/sub>, t3<\/sub>, t4<\/sub> are the roots of the above biquadratic, then
\nproduct of the roots t1<\/sub>t2<\/sub>t3<\/sub>t4<\/sub> = \\(\\frac{1}{1}\\) = 1
\n\"TS<\/p>\n","protected":false},"excerpt":{"rendered":"

Students must practice these Maths 2B Important Questions TS Inter Second Year Maths 2B Hyperbola Important Questions Short Answer Type to help strengthen their preparations for exams. TS Inter Second Year Maths 2B Hyperbola Important Questions Short Answer Type Question 1. One focus of a hyperbola is located at the point (1, -3) and the … Read more<\/a><\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":[],"categories":[26],"tags":[],"yoast_head":"\nTS Inter Second Year Maths 2B Hyperbola Important Questions Short Answer Type - TS Board Solutions<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/tsboardsolutions.com\/maths-2b-hyperbola-important-questions-short-answer-type\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"TS Inter Second Year Maths 2B Hyperbola Important Questions Short Answer Type - TS Board Solutions\" \/>\n<meta property=\"og:description\" content=\"Students must practice these Maths 2B Important Questions TS Inter Second Year Maths 2B Hyperbola Important Questions Short Answer Type to help strengthen their preparations for exams. 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