{"id":8034,"date":"2024-01-24T02:13:34","date_gmt":"2024-01-23T20:43:34","guid":{"rendered":"https:\/\/tsboardsolutions.in\/?p=8034"},"modified":"2024-01-25T17:56:20","modified_gmt":"2024-01-25T12:26:20","slug":"ts-inter-2nd-year-maths-2b-solutions-chapter-3-ex-3b","status":"publish","type":"post","link":"https:\/\/tsboardsolutions.in\/ts-inter-2nd-year-maths-2b-solutions-chapter-3-ex-3b\/","title":{"rendered":"TS Inter 2nd Year Maths 2B Solutions Chapter 3 Parabola Ex 3(b)"},"content":{"rendered":"

Students must practice this TS Intermediate Maths 2B Solutions<\/a> Chapter 3 Parabola Ex 3(b) to find a better approach to solving the problems.<\/p>\n

TS Inter 2nd Year Maths 2B Solutions Chapter 3 Parabola Exercise 3(b)<\/h2>\n

I.<\/span><\/p>\n

Question 1.
\nFind the equations of the tangent and normal to the parabola y2<\/sup> = 6x at the positive end of the latus rectum.
\nSolution:
\nGiven equation of parabola is y2<\/sup> = 6x ……..(1)
\nComparing with y2<\/sup> = 4ax we get 4a = 6
\n\u21d2 a = \\(\\frac{3}{2}\\)
\nLet L be the positive end of the latus rectum of the parabola y2<\/sup> = 6x
\n\u2234 L = (a, 2a) = (\\(\\frac{3}{2}\\), 3)
\nThe equation of tangent to the parabola (1) at (\\(\\frac{3}{2}\\), 3) is yy1<\/sub> – 2a(x + x1<\/sub>) = 0
\ny(3) = 3(x + \\(\\frac{3}{2}\\))
\n\u21d2 3y = 3x + \\(\\frac{9}{2}\\)
\n\u21d2 6y = 6x + 9
\n\u21d2 2x – 2y + 3 = 0 ……(2)
\nThe equation of normal to the parabola (1) at (\\(\\frac{3}{2}\\), 3) is
\n\"TS<\/p>\n

Question 2.
\nFind the equation of tangent and normal to the parabola x2<\/sup> – 4x – 8y + 12 = 0 at (4, \\(\\frac{3}{2}\\)).
\nSolution:
\nGiven equation of parabola is x2<\/sup> – 4x – 8y + 12 = 0 ……..(1)
\nLet P(x1<\/sub>, y1<\/sub>) = P(4, \\(\\frac{3}{2}\\)) be the given point.
\nDifferentiating (1) w.r.t x we get
\n2x – 4 – 8 \\(\\frac{d y}{d x}\\) = 0
\n\u21d2 \\(\\frac{d y}{d x}=\\frac{x-2}{4}\\)
\n\u2234 The slope of the tangent at (4, \\(\\frac{3}{2}\\)) is m = \\(\\frac{4-2}{4}=\\frac{2}{4}=\\frac{1}{2}\\)
\nand slope of the normal at (4, \\(\\frac{3}{2}\\)) is -2
\nThe equation of a tangent to the parabola (1) at (4, \\(\\frac{3}{2}\\)) is y – y1<\/sub> = m(x – x1<\/sub>)
\n\u21d2 y – \\(\\frac{3}{2}\\) = \\(\\frac{1}{2}\\) (x – 4)
\n\u21d2 2y – 3 = x – 4
\n\u21d2 x – 2y – 1 = 0 ………(2)
\nThe equation of normal to the parabola (1) at (4, \\(\\frac{3}{2}\\)) is y – y1<\/sub> = \\(-\\frac{1}{m}\\)(x – x1<\/sub>)
\n\u21d2 y – \\(\\frac{3}{2}\\) = -2(x – 4)
\n\u21d2 2y – 3 = -4(x – 4)
\n\u21d2 4x + 2y – 19 = 0<\/p>\n

\"TS<\/p>\n

Question 3.
\nFind the value of k if the line 2y = 5x + k is a tangent to the parabola y2<\/sup> = 6x.
\nSolution:
\nGiven the equation of the line is 2y = 5x + k.
\n\u21d2 y = \\(\\frac{5}{2} x+\\frac{k}{2}\\) which is of the form y = mx + c where m = \\(\\frac{5}{2}\\) and c = \\(\\frac{k}{2}\\)
\nComparing y2<\/sup> = 6x with y2<\/sup> = 4ax we get
\n4a = 6 \u21d2 a = \\(\\frac{3}{2}\\)
\nThe condition for the line y = mx + c to be a tangent to the parabola y2<\/sup> = 4ax is y = mx + \\(\\frac{a}{m}\\)
\nWhere c = \\(\\frac{a}{m}\\)
\n\u2234 \\(\\frac{k}{2}=\\frac{3 \/ 2}{5 \/ 2}=\\frac{3}{5}\\)
\n\u21d2 k = \\(\\frac{6}{5}\\)<\/p>\n

Question 4.
\nFind the equation of normal to the parabola y2<\/sup> = 4x which is parallel to y – 2x + 5 = 0.
\nSolution:
\nGiven equation of parabola is y2<\/sup> = 4x ………(1)
\nComparing with y2<\/sup> = 4ax we get a = 1
\nGiven y – 2x + 5 = 0, the slope, m = 2
\n\u2234 The slope of the line parallel to the above line is ‘2’.
\nThe equation of normal to the parabola which is parallel to the line is having slope m = 2 is
\ny = mx – 2am – am3<\/sup>
\n= 2x – 4 – 8
\n= 2x – 12
\n\u21d2 2x – y – 12 = 0
\n\u2234 The equation of the required normal is 2x – y – 12 = 0.<\/p>\n

Question 5.
\nShow that the line 2x – y + 2 = 0 is a tangent to the parabola y2<\/sup> = 16x. Find also the point of contact.
\nSolution:
\nGiven line is 2x – y + 2 = 0
\nand slope m = 2, and c = 2.
\nComparing y2<\/sup> = 16x with y2<\/sup> = 4ax
\nwe have a = 4
\nCondition for a line y = mx + c to be a tangent to the parabola y2<\/sup> = 4ax is c = \\(\\frac{a}{m}\\)
\n\u2234 c = 2 and \\(\\frac{\\mathrm{a}}{\\mathrm{m}}=\\frac{4}{2}\\) = 2
\n\u2234 c = \\(\\frac{a}{m}\\)
\nThe point of contact = \\(\\left(\\frac{a}{m^2}, \\frac{2 a}{m}\\right)\\) = (1, 4)<\/p>\n

\"TS<\/p>\n

Question 6.
\nFind the equation of a tangent to the parabola y2<\/sup> = 16x inclined at an angle of 60\u00b0 with its axis and also find the point of contact.
\nSolution:
\nGiven parabola is y2<\/sup> = 16x ……..(1)
\nComparing with y2<\/sup> = 4ax we get 4a = 16
\n\u21d2 a = 4
\nGiven that inclination of the tangent line is 60\u00b0.
\n\u2234 Slope of the tangent is m = tan 60\u00b0 = \u221a3
\n\u2234 The equation of the tangent to the parabola (1) having slope \u221a3 is y = mx + \\(\\frac{a}{m}\\)
\n\u21d2 y = \u221a3x + \\(\\frac{4}{\\sqrt{3}}=\\frac{3 x+4}{\\sqrt{3}}\\)
\n\u21d2 \u221a3y = 3x + 5
\n\u21d2 3x – \u221a3y + 4 = 0
\n\u2234 Point of contact = \\(\\left(\\frac{a}{m^2}, \\frac{2 a}{m}\\right)=\\left(\\frac{4}{3}, \\frac{8}{\\sqrt{3}}\\right)\\)<\/p>\n

II.<\/span><\/p>\n

Question 1.
\nFind the equations of tangents to the parabola y2<\/sup> = 16x which are parallel and perpendicular respectively to the line 2x – y + 5 = 0, also find the coordinates of their points of contact.
\nSolution:
\nGiven equation of parabola y2<\/sup> = 16x …….(1)
\nand comparing with y2<\/sup> = 4ax we have 4a = 16
\n\u21d2 a = 4
\nEquation of any line parallel to the given line 2x – y + 5 = 0 is 2x – y + k = 0 …….(1)
\n\u21d2 y = 2x + k
\n\u2234 m = 2 and c = k
\nif (1) is a tangent to the parabola, then c = \\(\\frac{a}{m}\\)
\n\u21d2 k = 2
\n\u2234 The equation of the required tangent is y = 2x + 2
\n\u21d2 2x – y + 2 = 0
\nPoint of contact = \\(\\left(\\frac{a}{m^2}, \\frac{2 a}{m}\\right)\\) = (1, 4)
\nEquation of any line perpendicular at 2x – y + 5 = 0 is x + 2y + k = 0
\n\u21d2 2y = -x – k ………(2)
\n\u21d2 y = \\(-\\frac{\\mathrm{x}}{2}-\\frac{\\mathrm{k}}{2}\\)
\nHere c = \\(\\frac{k}{2}\\) and m = \\(-\\frac{1}{2}\\)
\nIf (2) is a tangent to the parabola then c = \\(\\frac{a}{m}\\)
\n\u21d2 \\(-\\frac{k}{2}=\\frac{4}{-1 \/ 2}=-8\\)
\n\u21d2 k = 16
\n\u2234 The equation of the required tangent is x + 2y + 16 = 0 ……..(3)
\nPoint of contact = \\(\\left(\\frac{a}{m^2}, \\frac{2 a}{m}\\right)\\)
\n= \\(\\left(\\frac{4}{(1 \/ 4)}, \\frac{8}{-1 \/ 2}\\right)\\)
\n= (16, -16)<\/p>\n

\"TS<\/p>\n

Question 2.
\nIf lx + my + n = 0 is a normal to the parabola y2<\/sup> = 4ax, then show that al3<\/sup> + 2alm2<\/sup> + nm2<\/sup> = 0.
\nSolution:
\nGiven equation of parabola is y2<\/sup> = 4ax ……..(1) and given lx + my + n = 0 is a normal to the parabola y2<\/sup> = 4ax ……(2)
\nEquation of normal at a point P(t) on the parabola y2<\/sup> = 4ax is y + xt = 2at + at3<\/sup>
\n\u21d2 xt + y = 2at + at3<\/sup> …….(3)
\nEquations (2) and (3) represent the same line.
\nEliminating ‘t’ from (2) and (3) we get,
\n\"TS<\/p>\n

Question 3.
\nShow that the equation of the common tangents to the circle x2<\/sup> + y2<\/sup> = 2a2<\/sup> and the parabola y2<\/sup> = 8ax are y = \u00b1(x + 2a). (Mar. ’10, ’09)
\nSolution:
\nGiven equation of parabola is y2<\/sup> = 8ax …….(1) and circle is x2<\/sup> + y2<\/sup> = 2a2<\/sup> ……..(2)
\nThe Centre of the circle is C = (0, 0) and the radius of the circle is r = \u221a2a
\nEquation of the tangent to the parabola (1) having slope ‘m’ is y = mx + \\(\\frac{2a}{m}\\)
\n\u21d2 my = m2<\/sup>x + 2a (\u2235 y2<\/sup> = 8ax is the parabola)
\n\u21d2 m2<\/sup>x – my + 2a = 0 ………(3)
\nLine (3) is also a tangent to circle (2).
\n\u2234 The perpendicular distance from C(0, 0) to line (3) is equal to radius \u221a2a.
\n\u2234 \\(\\left|\\frac{2 a}{\\sqrt{m^4+m^2}}\\right|=\\sqrt{2 a}\\)
\n\u21d2 \\(\\frac{4 a^2}{m^4+m^2}\\) = 2a2<\/sup>
\n\u21d2 m4<\/sup> + m2<\/sup> – 2 = 0
\n\u21d2 m4<\/sup> + 2m2<\/sup> – m2<\/sup> – 2 = 0
\n\u21d2 (m2<\/sup> – 1) (m2<\/sup> + 2) = 0
\n\u21d2 m2<\/sup> = 1 or m2<\/sup> = -2
\n\u21d2 m2<\/sup> = 1 (\u2235 m2<\/sup> \u2260 -2)
\n\u21d2 m = \u00b11
\n\u2234 The equations of the required tangents are y = \\(\\pm x+\\frac{2 a}{\\pm 1}\\)
\n\u21d2 y = \u00b1(x + 2a)<\/p>\n

Question 4.
\nProve that the tangents at the extremities of a focal chord of a parabola intersect at right angles on the directrix.
\nSolution:
\n\"TS
\nLet y2<\/sup> = 4ax be the equation of a parabola and
\nLet P\\(\\left(a t_1^2, 2 a t_1\\right)\\), Q\\(\\left(a t_2^2, 2 a t_2\\right)\\) be two extremeties of the focal chord.
\nSince \\(\\overline{\\mathrm{PQ}}\\) is the focal chord we have t1<\/sub>t2<\/sub> = -1 (standard result)
\nThe equation of tangent to the parabola y2<\/sup> = 4ax at P(t1<\/sub>) is x = \\(\\mathrm{yt}_1+\\mathrm{at}{ }_1^2\\) ………(1)
\nSlope of the tangent (m1<\/sub>) = \\(\\frac{1}{t_1}\\)
\nThe equation of the tangent Q(t2<\/sub>) to the parabola y2<\/sup> = 4ax is x = \\(\\mathrm{yt}_2+\\mathrm{at}_2^2\\) ……..(2)
\nSlope of the tangent (m2<\/sub>) = \\(\\frac{1}{t_2}\\)
\nNow m1<\/sub>m2<\/sub> = \\(\\frac{1}{t_1 t_2}\\) = -1 (\u2235 t1<\/sub>t2<\/sub> = -1)
\n\u2234 The tangents drawn at P, Q are perpendicular to each other.
\nThe point of intersection of tangents (1) and (2) = [at1<\/sub>t2<\/sub>, a(t1<\/sub> + t2<\/sub>)]
\n= [-a, a(t1<\/sub> + t2<\/sub>)]
\nSubstituting this point of intersection on the directrix of the parabola y2<\/sup> = 4ax i.e., x + a = 0.
\nWe have -a + a = 0 and hence the point of intersection of tangents lie on the directrix.
\n\u2234 The tangents drawn at the extremities of the focal chord of a parabola intersect at right angles on the directrix.<\/p>\n

\"TS<\/p>\n

Question 5.
\nFind the condition for the line y = mx + c to be a tangent to the parabola x2<\/sup> = 4ay. (Mar. ’12)
\nSolution:
\nGiven equation of parabola as x2<\/sup> = 4ay …….(1)
\nand given line is mx – y + c = 0 ……..(2)
\nSuppose (2) is a tangent to the parabola (1).
\nLet P(x1<\/sub>, y1<\/sub>) be the point of contact.
\nThen the equation of the tangent to the parabola (1) at (x1<\/sub>, y1<\/sub>) is xx1<\/sub> – 2a(y + y1<\/sub>) = 0
\n\u21d2 xx1<\/sub> – 2ay – 2ay1<\/sub> = 0 ………(3)
\nSince equations (2) and (3) represent the same line, the corresponding coefficients are proportional.
\n\u2234 \\(\\frac{x_1}{m}=\\frac{-2 a}{-1}=\\frac{-2 a y_1}{c}\\)
\n\u21d2 x1<\/sub> = 2am, y1<\/sub> = -c
\nSince P(x1<\/sub>, y1<\/sub>) is a point on x2<\/sup> = 4ay we have \\(\\mathbf{x}_1^2\\) = 4ay1<\/sub>
\n\u21d2 (2am)2<\/sup> = 4a(-c)
\n\u21d2 am2<\/sup> = -c
\n\u21d2 c = -am2<\/sup>
\n\u2234 The condition for the line y = mx + c to be a tangent to the parabola x2<\/sup> = 4ay is am2<\/sup> + c = 0.<\/p>\n

Question 6.
\nThree normals are drawn from (k, 0) to the parabola y2<\/sup> = 8x one of the normal is the axis and the remaining two normals are perpendicular to each other, then find the value of k.
\nSolution:
\nGiven equation of parabola is y2<\/sup> = 8x ………(1)
\nComparing with y2<\/sup> = 4ax we have 4a = 8
\n\u21d2 a = 2
\nThe equation of normal to the parabola (1) having slope ‘m’ is
\ny = mx – 2am – am2<\/sup>
\n\u21d2 y = mx – 4m – 2m3<\/sup>
\nGiven that the normal passes through (k, 0).
\nThen 0 = mk – 4m – 2m3<\/sup>
\n\u21d2 m(2m2<\/sup> + 4 – k) = 0
\n\u21d2 m = 0 or 2m2<\/sup> – k + 4 = 0 ……(2)
\nLet m1<\/sub>, m2<\/sub> be the roots of a quadratic equation (2) then
\nm1<\/sub> + m2<\/sub> = 0, m1<\/sub>m2<\/sub> = \\(\\frac{4-k}{2}\\)
\nGiven that the two normals are perpendicular, we have m1<\/sub>m2<\/sub> = -1
\n\u21d2 \\(\\frac{4-k}{2}\\) = -1
\n\u21d2 k = 6<\/p>\n

Question 7.
\nShow that the locus of the point of intersection of the perpendicular tangents to the parabola y2<\/sup> = 4ax is the directrix x + a = 0.
\nSolution:
\nGiven y2<\/sup> = 4ay ………(1) and Let P(x1<\/sub>, y1<\/sub>) be the point of intersection of perpendicular tangents.
\nAny tangent to the parabola will be of the form y = mx + \\(\\frac{a}{m}\\) of this passes through (x1<\/sub>, y1<\/sub>) then y1<\/sub> = mx1<\/sub> + \\(\\frac{a}{m}\\)
\n\u21d2 m2<\/sup>x1<\/sub> – my1<\/sub> + a = 0 …….(2)
\nLet m1<\/sub>, m2<\/sub> be the slopes of tangents drawn from P(x1<\/sub>, y1<\/sub>) to the parabola (1) which corresponds to the roots of a quadratic equation (2).
\nThen m1<\/sub> + m2<\/sub> = \\(\\frac{y_1}{x_1}\\) and m1<\/sub>m2<\/sub> = \\(\\frac{a}{x_1}\\) …….(3)
\nSince the tangents are perpendicular we have m1<\/sub>m2<\/sub> = -1 and hence from (3),
\n-1 = \\(\\frac{a}{x_1}\\)
\n\u21d2 x1<\/sub> + a = 0
\n\u2234 The Locus of (x1<\/sub>, y1<\/sub>) is x + a = 0 which is the equation of the directrix of the parabola (1).<\/p>\n

\"TS<\/p>\n

Question 8.
\nTwo parabolas have the same vertex and equal length of latus rectum such that their axes are at the right angle. Prove that the common tangents touch each at the end of the latus rectum.
\nSolution:
\n\"TS
\nLet the equations of parabolas having the same vertex and equal length of latus rectum be
\ny2<\/sup> = 4ax …….(1)
\nx2<\/sup> = 4ay …….(2)
\nLet the equation of the tangent to the parabola (1) having slope ‘m’ is
\ny = mx + \\(\\frac{a}{m}\\) …….(3)
\nBut this is also a tangent to x2<\/sup> = 4ay
\n\u2234 c = -am2<\/sup>
\n\u2234 \\(\\frac{a}{m}\\) = -am2<\/sup>
\n\u21d2 m3<\/sup> = -1
\n\u21d2 m = 1
\n\u2234 From (3), we have y = -x – a
\n\u21d2 x + y + a = 0 ………(4)
\nL'(a, -2a) is the end of latus rectum of parabola (1).
\n\u2234 From (4), x + y + a = a – 2a + a = 0
\n\u2234 The common tangents to the parabolas y2<\/sup> = 4ax, x2<\/sup> = 4ay will touch the parabola at the end of the latus rectum.<\/p>\n

Question 9.
\nShow that the foot of the perpendicular from focus to the tangent of parabola y2<\/sup> = 4ax lies on the tangent at the vertex.
\nSolution:
\nGiven parabola is y2<\/sup> = 4ax ……….(1)
\nEquation of any tangent to (1) is of the form y = mx + \\(\\frac{a}{m}\\)
\n\u21d2 my = m2<\/sup>x + a
\n\u21d2 m2<\/sup>x – my + a = 0 ……..(2)
\nEquation of the line perpendicular to the line (2) and passing through S(a, 0) is -m(x – a) + m2<\/sup>(y – 0) = 0
\n\u21d2 x + my – a = 0 ……..(3)
\nFrom (2) and (3)
\n(m2<\/sup> + 1)x = 0
\n\u21d2 x = 0
\nFrom (3), my – a = 0
\n\u21d2 y = \\(\\frac{a}{m}\\)
\n\u2234 The point of intersection of (2) and (3) is (0, \\(\\frac{a}{m}\\))
\n\u2234 The point (0, \\(\\frac{a}{m}\\)) lies on the y-axis which is a tangent at the vertex.<\/p>\n

\"TS<\/p>\n

Question 10.
\nShow that the tangent at one extremity of a focal chord of a parabola is parallel to the normal at the other extremity.
\nSolution:
\nLet y2<\/sup> = 4ax be the given parabola.
\nLet P = \\(\\left(a t_1{ }^2, 2 a t_1\\right)\\) and Q = \\(\\left(a t_2{ }^2, 2 a t_2\\right)\\) be the two extremities of a focal chord. then t1<\/sub>t2<\/sub> = -1
\nThe equation of the tangent to the parabola at P(t1<\/sub>) is x – yt1<\/sub> + \\(\\mathrm{at}_1{ }^2\\) = 0 …….(1)
\nthen m1<\/sub> = \\(\\frac{1}{t_1}\\)
\nThe equation of the normal at Q(t2<\/sub>) is xt2<\/sub> + y – 2at2<\/sub> – \\(\\mathrm{at}_2^3\\) = 0 ……..(2)
\nThe slope of the normal (m2<\/sub>) = -t2<\/sub>
\n= \\(-\\left(\\frac{-1}{t_1}\\right)=\\frac{1}{t_1}\\) (\u2235 t1<\/sub>t2<\/sub> = -1)
\n\u2234 m1<\/sub> = m2<\/sub> and the tangent at P(t1<\/sub>) is parallel to the normal at Q(t2<\/sub>) to the parabola y2<\/sup> = 4ax.
\n\u2234 The tangent at one extremity of a focal chord is parallel to the normal at other extremities.<\/p>\n

III.<\/span><\/p>\n

Question 1.
\nThe normal at a point t1<\/sub> on y2<\/sup> = 4ax meets the parabola again at the point t2<\/sub>. Then prove that t1<\/sub>t2<\/sub> + \\(t_1{ }^2\\) + 2 = 0.
\nSolution:
\nGiven equation of parabola is y2<\/sup> = 4ax ……..(1)
\nLet P = \\(\\left(a t_1^2, 2 a t_1\\right)\\) = P(t1<\/sub>) be any point on (1).
\nThe equation of normal at P(t1<\/sub>) on the parabola (1) is y + xt1<\/sub> = 2at1<\/sub> + \\(\\mathrm{at}_1{ }^3\\)
\n\u21d2 t1<\/sub>x + y = 2at1<\/sub> + \\(\\mathrm{at}_1{ }^3\\) ……..(2)
\nThe normal drawn at P(t1<\/sub>) meets the parabola at Q(t2<\/sub>).
\n\u2234 PQ is the chord of the parabola.
\nThe equation of chord \\(\\overline{\\mathrm{PQ}}\\) of the parabola y2<\/sup> = 4ax is y(t1<\/sub> + t2<\/sub>) = 2x + 2at1<\/sub>t2<\/sub>
\n\u21d2 2x = y(t1<\/sub> + t2<\/sub>) + 2at1<\/sub>t2<\/sub>\u00a0= 0 ……….(3)
\nSince equations (2) and (3) represent the same line, the coefficients of line turns are proportional.
\n\"TS
\n\"TS<\/p>\n

Question 2.
\nFrom an external point, P tangents are drawn to the parabola y2<\/sup> = 4ax and these tangents make angles \u03b81<\/sub>, \u03b82<\/sub> with its axis, such that cot \u03b81<\/sub> + cot \u03b82<\/sub> is a constant ‘d’. Then show that all such P lies on a horizontal line.
\nSolution:
\n\"TS
\nGiven y2<\/sup> = 4ax ………(1) and Let P(x1<\/sub>, y1<\/sub>) be an external point.
\nLet the equation of a tangent to the parabola (1) having slope ‘m’ is y = mx + \\(\\frac{a}{m}\\) …….(2)
\nif (2) passes through P(x1<\/sub>, y1<\/sub>) then y1<\/sub> = mx1<\/sub> + \\(\\frac{a}{m}\\)
\n\u21d2 my1<\/sub> = m2<\/sup>x1<\/sub> + a
\n\u21d2 m2<\/sup>x1<\/sub> – my1<\/sub> + a = 0 …….(3)
\nSince this is a quadratic equation in ‘m’.
\nLet m1<\/sub> and m2<\/sub> be the roots. Then
\nm1<\/sub> + m2<\/sub> = \\(\\frac{y_1}{x_1}\\) and m1<\/sub>m2<\/sub> = \\(\\frac{\\mathrm{a}}{\\mathrm{x}_1}\\)
\nGiven that the tangents are making angles \u03b81<\/sub>, \u03b82<\/sub> with its axis we have m1<\/sub> = tan \u03b81<\/sub> and m2<\/sub> = tan \u03b82<\/sub>.
\nAlso given that cot \u03b81<\/sub> + cot \u03b82<\/sub> = d
\n\"TS
\n\u2234 The Locus of P(x1<\/sub>, y1<\/sub>) is y = ad which is a line parallel to X-axis.
\n\u2234 P(x1<\/sub>, y1<\/sub>) lies on a horizontal line.<\/p>\n

\"TS<\/p>\n

Question 3.
\nShow that the common tangents to the circle 2x2<\/sup> + 2y2<\/sup> = a2<\/sup> and the parabola y2<\/sup> = 4ax intersect at the focus of the parabola y2<\/sup> = -4ax.
\nSolution:
\nGiven equation of circle is x2<\/sup> + y2<\/sup> = \\(\\frac{a^2}{2}\\) ……..(1)
\nand the given equation of parabola is y2<\/sup> = 4 ax ……..(2)
\nLet y = mx + c be a common tangent to (1) and (2).
\nIf y = mx + c is a tangent to (1) then by the condition of tangency
\nc2<\/sup> = \\(\\frac{a^2}{2}\\) (1 + m2) ………(3)
\nAlso if y = mx + c is a tangent to the parabola y2<\/sup> = 4ax then
\nc = \\(\\frac{a}{m}\\) …….(4)
\n\u2234 From (3) and (4) we have \\(\\frac{a^2}{m^2}=\\frac{a^2}{2}\\left(1+m^2\\right)\\)
\n\u21d2 m4<\/sup> + m2<\/sup> = 2
\n\u21d2 m4<\/sup> + m2<\/sup> – 2 = 0
\n\u21d2 m4<\/sup> + 2m2<\/sup> – m2<\/sup> – 2 = 0
\n\u21d2 (m2<\/sup> + 2) (m2<\/sup> – 1) = 0
\n\u21d2 m2<\/sup> + 2 = 0 or m2<\/sup> – 1 = 0
\nm2<\/sup> + 2 = 0 is not admissible.
\n\u2234 m2<\/sup> – 1 = 0
\n\u21d2 m = \u00b11
\nFrom y = mx + c we have
\n= mx + \\(\\frac{a}{m}\\)
\n= \\(\\pm x+\\frac{a}{\\pm 1}\\)
\n= \u00b1 x \u00b1 a
\nAlso y = \u00b1 x \u00b1 a satisfy the focus (-a, 0) of the parabola y2<\/sup> = -4ax.
\nHence the common tangents to the circle (1) and parabola (2) intersect at the focus of the parabola y2<\/sup> = -4ax.<\/p>\n

Question 4.
\nThe sum of the ordinates of two points on y2<\/sup> = 4ax is equal to the sum of the ordinates of two other points on the same curve. Show that the chord joining the first two points is parallel to the chord joining the other two points.
\nSolution:
\nGiven equation of parabola is y2<\/sup> = 4ax …..(1)
\nLet A = \\(\\left(\\mathrm{at}_1{ }^2, 2 \\mathrm{at}_1\\right)\\), B = \\(\\left(\\mathrm{at}_2{ }^2, 2 \\mathrm{at}_2\\right)\\), C = \\(\\left(\\mathrm{at}_3{ }^2, 2 \\mathrm{at}_3\\right)\\) and D = \\(\\left(\\mathrm{at}_4{ }^2, 2 \\mathrm{at}_4\\right)\\) be the four points on the parabola (1).
\nGiven that 2at1<\/sub> + 2at2<\/sub> = 2at3<\/sub> + 2at4<\/sub>
\n\u21d2 t1<\/sub> + t2<\/sub> = t3<\/sub> + t4<\/sub> ………(2)
\nThe equation of the chord \\(\\overline{\\mathrm{AB}}\\) of the parabola y2<\/sup> = 4ax is y(t1<\/sub> + t2<\/sub>) = 2x + 2at1<\/sub>t2<\/sub>
\n\u21d2 2x – (t1<\/sub> + t2<\/sub>) y + 2at1<\/sub>t2<\/sub> = 0 ………(3)
\nLet m1<\/sub> be the slope of line (3) then m1<\/sub> = \\(\\frac{2}{t_1+t_2}\\)
\nThe equation of the chord \\(\\overline{\\mathrm{CD}}\\) of the parabola (1) is y(t3<\/sub> + t4<\/sub>) = 2x + 2at3<\/sub>t4<\/sub> and slope of this chord m2<\/sub> = \\(\\frac{2}{t_3+t_4}\\)
\n\u2234 From (2) we have t1<\/sub> + t2<\/sub> = t3<\/sub> + t4<\/sub>
\nHence m1<\/sub> = m2<\/sub>
\n\u2234 Chord \\(\\overline{\\mathrm{AB}}\\) is parallel to the chord \\(\\overline{\\mathrm{CD}}\\).<\/p>\n

\"TS<\/p>\n

Question 5.
\nIf a normal chord at a point ‘t’ on the parabola y2<\/sup> = 4ax subtends a right angle at the vertex then prove that t = \u00b1\u221a2.
\nSolution:
\nGiven equation of parabola is y2<\/sup> = 4ax ……(1)
\nThe equation of normal at a point P(t) on (1) is xt + y = (2at + at3<\/sup>) …….(2)
\nLet the normal at P(t) meet (1) again at Q(t1<\/sub>) then PQ is the chord of the parabola (1)
\n\"TS
\nEquation of chord \\(\\overline{\\mathrm{PQ}}\\) is y(t1<\/sub> + t) – 2x – 2at1<\/sub>t = 0
\n\u21d2 -2x + (t1<\/sub> + t)y – 2at1<\/sub>t = 0 ………(3)
\nEquations (2) and (3) represent the same straight line.
\n\u2234 Coefficients of like terms are proportional
\n\"TS
\n\"TS<\/p>\n","protected":false},"excerpt":{"rendered":"

Students must practice this TS Intermediate Maths 2B Solutions Chapter 3 Parabola Ex 3(b) to find a better approach to solving the problems. TS Inter 2nd Year Maths 2B Solutions Chapter 3 Parabola Exercise 3(b) I. Question 1. Find the equations of the tangent and normal to the parabola y2 = 6x at the positive … Read more<\/a><\/p>\n","protected":false},"author":3,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[3],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/8034"}],"collection":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/users\/3"}],"replies":[{"embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/comments?post=8034"}],"version-history":[{"count":4,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/8034\/revisions"}],"predecessor-version":[{"id":8268,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/8034\/revisions\/8268"}],"wp:attachment":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/media?parent=8034"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/categories?post=8034"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/tags?post=8034"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}