{"id":35870,"date":"2022-11-25T16:45:25","date_gmt":"2022-11-25T11:15:25","guid":{"rendered":"https:\/\/tsboardsolutions.com\/?p=35870"},"modified":"2022-12-03T14:18:46","modified_gmt":"2022-12-03T08:48:46","slug":"maths-2b-parabola-important-questions-long-answer-type","status":"publish","type":"post","link":"https:\/\/tsboardsolutions.com\/maths-2b-parabola-important-questions-long-answer-type\/","title":{"rendered":"TS Inter Second Year Maths 2B Parabola Important Questions Long Answer Type"},"content":{"rendered":"

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

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

Question 1.
\nShow that the equation of a parabola in the standard form is y2<\/sup> = 4ax. [(TS) Mar. ’20, ’18, ’17, ’16; May ’18; (AP) May ’19, ’15; Mar. ’17, ’15]
\nSolution:
\n\"TS
\nLet S be the focus and l = 0 be the directrix of the parabola.
\nLet \u2018P\u2019 be a point on the parabola.
\nLet M, Z be the projection (foot of the perpendiculars) of P, S on the directrix L = 0 respectively.
\nLet ‘N’ be the projection of P on \u2018SZ\u2019.
\nLet \u2018A\u2019 be the midpoint of SZ.
\nSince SA = AZ
\n\u2018A\u2019 lies on the parabola let AS = a.
\nTake AS, the principal axis of the parabola as X-axis and AY \u22a5r to SZ as Y-axis.
\nThen S = (a, 0) and the parabola is in the standard form.
\nLet P(x1<\/sub>, y1<\/sub>)
\nNow PM = NZ = AN + AZ = x1<\/sub> + a
\n‘P’ lies on the parabola then
\n\\(\\frac{\\mathrm{SP}}{\\mathrm{PM}}\\) = 1
\nSP = PM
\n\\(\\sqrt{\\left(x_1-a\\right)^2+\\left(y_1-0\\right)^2}=x_1+a\\)
\nSquaring on both sides,
\n(x1<\/sub> – a)2<\/sup> + (y1<\/sub> – 0)2<\/sup> = (x1<\/sub> + a)2<\/sup>
\n\u21d2 \\(\\mathrm{y}_1^2\\) = (x1<\/sub> + a)2<\/sup> – (x1<\/sub> – a)2<\/sup>
\n\u21d2 \\(\\mathrm{y}_1^2\\) = 4ax1<\/sub>
\nThe locus of \u2018P\u2019 is y2<\/sup> = 4ax
\n\u2234 The equation to the parabola is y2<\/sup> = 4ax<\/p>\n

\"TS<\/p>\n

Question 2.
\nFind the coordinates of the vertex and focus and the equation of the directrix and axes of the parabola y2<\/sup> – x + 4y + 5 = 0. (Mar. ’05)
\nSolution:
\nGiven the equation of the parabola is
\ny2<\/sup> – x + 4y + 5 = 0
\n\u21d2 y2<\/sup> + 4y = x – 5
\n\u21d2 (y)2<\/sup> + 2 . 2 . y + (2)2<\/sup> – (2)2<\/sup> = x – 5
\n\u21d2 (y + 2)2<\/sup> – 4 = x – 5
\n\u21d2 (y + 2)2<\/sup> = x – 1
\n\u21d2 (y + 2)2<\/sup> = 1(x – 1)
\nComparing with (y – k)2<\/sup> = 4a(x – h), we get
\nh = 1, k = -2, a = \\(\\frac{1}{4}\\)
\n(i) Vertex = (h, k) = (1, -2)
\n(ii) Focus = (h + a, k) = (1 + \\(\\frac{1}{4}\\), -2) = (\\(\\frac{5}{4}\\), -2)
\n(iii) Equation of the directrix is x = h – a
\n\u21d2 x = 1 – \\(\\frac{1}{4}\\)
\n\u21d2 x = \\(\\frac{3}{4}\\)
\n\u21d2 4x – 3 = 0
\n(iv) Equation of the axis is y = k
\n\u21d2 y = -2
\n\u21d2 y + 2 = 0<\/p>\n

Question 3.
\nFind the vertex and focus of 4y2<\/sup> + 12x – 20y + 67 = 0.
\nSolution:
\nGiven equation of the parabola is 4y2<\/sup> + 12x – 20y + 67 = 0
\n4y2<\/sup> – 20y = -12x – 67
\n4(y2<\/sup> – 5y) = -12x – 67
\n\"TS
\n\"TS<\/p>\n

Question 4.
\nFind the coordinates of the vertex and focus, the equation of the directrix, and the axis of the parabola y2<\/sup> + 4x + 4y – 3 = 0.
\nSolution:
\nGiven the equation of the parabola is
\ny2<\/sup> + 4x + 4y – 3 = 0
\n\u21d2 y2<\/sup> + 4y = -4x + 3
\n\u21d2 (y)2<\/sup> + 2 . y(2) + (2)2<\/sup> – (2)2<\/sup> = -4x + 3
\n\u21d2 (y + 2)2<\/sup> – 4 = -4x + 3
\n\u21d2 (y + 2)2<\/sup> = -4x + 7
\n\u21d2 (y + 2)2<\/sup> = -4(x – \\(\\frac{7}{4}\\))
\n[y-(-2)]2<\/sup> = -4(x – \\(\\frac{7}{4}\\))
\nComparing with (y – k)2<\/sup> = -4a(x – h), we get
\nh = \\(\\frac{7}{4}\\), k = -2, 4a = 4 \u21d2 a = 1
\n(i) Vertex = (h, k) = (\\(\\frac{7}{4}\\), -2)
\n(ii) Focus = (h – a, k) = (\\(\\frac{7}{4}\\) – 1, -2) = (\\(\\frac{3}{4}\\), -2)
\n(iii) Equation of the directrix is x = h + a
\n\u21d2 x = \\(\\frac{7}{4}\\) + 1
\n\u21d2 x = \\(\\frac{11}{4}\\)
\n\u21d2 x = 4x – 11 = 0
\n(iv) Equation of the axis is y = k
\n\u21d2 y = -2
\n\u21d2 y + 2 = 0<\/p>\n

\"TS<\/p>\n

Question 5.
\nFind the equations of the axis and directrix of the parabola 4x2<\/sup> + 12x – 20y + 67 = 0.
\nSolution:
\nGiven equation of the parabola is 4x2<\/sup> + 12x – 20y + 67 = 0
\n\u21d2 4x2<\/sup> + 12x = 20y – 67
\n\u21d2 4(x2<\/sup> + 3x) = 20y – 67
\n\"TS
\nComparing with (x – h)2<\/sup> = 4a(y – k) we get
\nh = \\(-\\frac{3}{2}\\), k = \\(\\frac{29}{10}\\),
\n4a = 5 \u21d2 a = \\(\\frac{5}{4}\\)
\n(i) Equation of the axis is x = h
\n\u21d2 x = \\(-\\frac{3}{2}\\)
\n\u21d2 2x + 3 = 0
\n(ii) Equation of the directrix is y = k – a
\n\u21d2 y = \\(\\frac{29}{10}-\\frac{5}{4}\\)
\n\u21d2 y = \\(\\frac{33}{20}\\)
\n\u21d2 20y – 33 = 0<\/p>\n

Question 6.
\nFind the coordinates of the vertex and focus and the equations of the directrix and axes of the parabola 3x2<\/sup> – 9x + 5y – 2 = 0.
\nSolution:
\nGiven equation of the parabola
\n3x2<\/sup> – 9x + 5y – 2 = 0
\n\u21d2 3x2<\/sup> – 9x = -5y + 2
\n\u21d2 3(x2<\/sup> – 3x) = -5y + 2
\n\"TS
\n\"TS<\/p>\n

Question 7.
\nFind the equation of the parabola whose axis is parallel to the X-axis and which passes through die points (-2, 1), (1, 2), and (-1, 3). [(AP) May ’18, ’16, (TS) ’17]
\nSolution:
\nLet, the given points are A(-2, 1), B(1, 2), C(-1, 3)
\nThe equation of the parabola whose axis is parallel to the X-axis is
\nx = ly2<\/sup> + my + n ……….(1)
\nSince, eq. (1) passes through point A(-2, 1) then
\n(-2) = l(1)2<\/sup> + m(1) + n
\n\u21d2 -2 = l + m + n
\n\u21d2 l + m + n = -2
\nSince, (1) passes through point B(1, 2) then
\n(1)2<\/sup> = l(2)2<\/sup> + m(2) + n
\n\u21d2 1 = 4l + 2m + n
\n\u21d2 4l + 2m + n = 1 …….(3)
\nSince, (1) passes through point C(-1, 3), then
\n-1 = l(3)2<\/sup> + m(3) + n
\n9l + 3m + n = -1 ………(4)
\nFrom (2) and (3)
\n\"TS
\nSubstitute the values of l, m in (2)
\n\\(\\frac{-5}{2}+\\frac{21}{2}\\) + n = -2
\n\u21d2 -5 + 21 + 2n = -4
\n\u21d2 16 + 2n = -4
\n\u21d2 2n = -20
\n\u21d2 n = -10
\nSubstitute the values of l, m, n in (1),
\nThe required equation of the parabola is
\n\\(\\mathbf{x}=\\frac{-5}{2} \\mathbf{y}^2+\\frac{21}{2} \\mathbf{y}-10\\)
\n\u21d2 -5y2<\/sup> + 21y – 20 = 2x
\n\u21d2 5y2<\/sup> + 2x – 21y + 20 = 0<\/p>\n

\"TS<\/p>\n

Question 8.
\nFind the equation of the parabola passing through the points (-1, 2), (1, -1), and (2, 1) and having its axis parallel to the X-axis.
\nSolution:
\nLet, the given points are A(-1, 2), B(1, -1), C(2, 1)
\nThe equation of the parabola whose axis is parallel to the X-axis is
\nx = ly2<\/sup> + my + n …….(1)
\nSince, (1) passes through point A(-1, 2) then
\n(-1) = l(2)2<\/sup> + m(2) + n
\n\u21d2 -1 = 4l + 2m + n
\n\u21d2 4l + 2m + n = -1 ……..(2)
\nSince, (1) passes through point B(1, -1) then
\n(1) = l(-1)2<\/sup> + m(-1) + n
\n\u21d2 l – m + n = 1 ……..(3)
\nSince (1) passes through point C(2, 1) then
\n2 = l(1)2<\/sup> + m(1) + n
\n\u21d2 l + m + n = 2 ……(4)
\nFrom (2) and (3)
\n\"TS
\n\"TS<\/p>\n

Question 9.
\nFind the equation of the parabola whose X-axis is parallel to the Y-axis and which passes through the point (4, 5), (-2, 11), (-4, 21). (May ’12)
\nSolution:
\nThe equation of the parabola whose axis is parallel to the Y-axis then
\ny = lx2<\/sup> + mx + n ……..(1)
\nSince eq. (1) passes through the point (4, 5) then
\n5 = l(4)2<\/sup> + m(4) + n
\n\u21d2 16l + 4m + n = 5 ………(2)
\nSince eq. (1) passes through the point (-2, 11) then
\n11 = l(-2)2<\/sup> + m(-2) + n
\n\u21d2 4l – 2m + n = 11 ………(3)
\nSince eq. (1) passes through the point (-4, 21) then
\n21 = l(-4)2<\/sup> + m(-4) + n
\n\u21d2 16l + 4m + n = 21 ……….(4)
\n\"TS
\n\"TS<\/p>\n

Question 10.
\nFind the equation of the parabola whose focus is (-2, 3) and whose directrix is the line 2x + 3y – 4 = 0. Also, find the length of the latus rectum and the equation of the axis of the parabola.
\nSolution:
\n2x + 3y – 4 = 0
\nGiven that, focus, S = (-2, 3)
\n\"TS
\nThe equation of the directrix is 2x + 3y – 4 = 0
\nLet, P(x, y) be a point on the parabola.
\nNow, SP = \\(\\sqrt{(\\mathrm{x}+2)^2+(\\mathrm{y}-3)^2}\\)
\nPM = the \u22a5r distance from P(x, y) to the directrix 2x + 3y – 4 = 0
\n\"TS
\nSquaring on both sides
\n(x + 2)2<\/sup> + (y – 3)2<\/sup> = \\(\\frac{(2 x+3 y-4)^2}{13}\\)
\n13x2<\/sup> + 52x + 52 + 13y2<\/sup> + 117 – 78y = 4x2<\/sup> + 9y2<\/sup> + 12xy – 24y – 16x + 16
\n9x2<\/sup> – 12xy + 4y2<\/sup> + 68x – 54y + 153 = 0
\n\u2234 The equation to the parabola is 9x2<\/sup> – 12xy + 4y2<\/sup> + 68x – 54y + 153 = 0
\nNow, the length of the latus rectum = 4a
\n= 2(2a)
\n= 2(the \u22a5r distance from focus S(-2, 3) to the directrix 2x + 3y – 4 = 0)
\n\"TS
\nThe equation of the axis of the parabola is b(x – x1<\/sub>) – a(y – y1<\/sub>) = 0
\n\u21d2 3(x + 2) – 2(y – 3) = 0
\n\u21d2 3x + 6 + 2y – 6 = 0
\n\u21d2 3x + 2y = 0<\/p>\n

\"TS<\/p>\n

Question 11.
\nFind the locus of the point of trisection of the double ordinate of a parabola y2<\/sup> = 4ax (a > 0).
\nSolution:
\nThe given equation of a parabola is y2<\/sup> = 4ax (a > 0)
\n\"TS
\nLet, the ends double-ordinate the parabola
\ny2<\/sup> = 4ax are P(at2<\/sup>, 2at), Q(at2<\/sup>, -2at)
\nLet R(x1<\/sub>, y1<\/sub>) be any point on the locus trisection ratio = 1 : 2
\nR(x1<\/sub>, y1<\/sub>) is the trisection point, then
\n\"TS<\/p>\n

Question 12.
\nShow that the equation of a common tangent to the circle x2<\/sup> + y2<\/sup> = 2a2<\/sup> and the parabola y2<\/sup> = 8ax are y = \u00b1(x + 2a). [(TS) May ’19, ’16 (AP) ’17]
\nSolution:
\nGiven the equation of the parabola is y2<\/sup> = 8ax
\nThe equation of the tangent to the parabola
\ny2<\/sup> = 4ax is y = mx + \\(\\frac{a}{m}\\)
\nThe equation of the tangent to the parabola
\ny2<\/sup> = 8ax is y = mx + \\(\\frac{2a}{m}\\) (\u2235 a = 2a)
\nGiven the equation of the circle is x2<\/sup> + y2<\/sup> = 2a2<\/sup>
\nCentre C = (0, 0)
\nRadius r = \u221a2a
\nSince eq. (1) is a tangent to the circle x2<\/sup> + y2<\/sup> = 2a2<\/sup> then r = d
\n\"TS
\nSquaring on both sides we get
\n2(1 + m2<\/sup>) = \\(\\frac{4}{\\mathrm{~m}^2}\\)
\n\u21d2 m2<\/sup>(1 + m2<\/sup>) = 2
\n\u21d2 m2<\/sup> + m4<\/sup> – 2 = 0
\n\u21d2 m4<\/sup> + m2<\/sup> – 2 = 0
\n\u21d2 m4<\/sup> + 2m2<\/sup> – m2<\/sup> – 2 = 0
\n\u21d2 m2<\/sup>(m2<\/sup> + 2) – 1(m2<\/sup> + 2) = 0
\n\u21d2 (m2<\/sup> + 2) (m2<\/sup> – 1) = 0
\n\u21d2 m2<\/sup> + 2 = 0 or m2<\/sup> – 1 = 0
\n\u21d2 m2<\/sup> = -2 or m2<\/sup> = 1
\n\u21d2 m = \u00b1\u221a-2 \u2209 R or m = \u00b1 1
\nSubstitute the value of \u2018m\u2019 in eq. (1)
\n\u2234 The equation of the common tangents is
\ny = \\(\\pm x+\\frac{2 a}{\\pm 1}\\)
\n\u21d2 y = \u00b1(x + 2a)<\/p>\n

Question 13.
\nShow that the common tangent to the parabola y2<\/sup> = 4ax and x2<\/sup> = 4by is \\(x a^{1 \/ 3}+y b^{1 \/ 3}+a^{2 \/ 3} \\cdot b^{2 \/ 3}=0\\). [(AP) Mar. ’16]
\nSolution:
\nGiven equations of the parabola are
\ny2<\/sup> = 4ax …….(1) and x2<\/sup> = 4by …..(2)
\n\"TS
\nEquation of any tangent to (1) is of the form
\ny = mx + \\(\\frac{a}{m}\\) ……..(3)
\nIf line (3) is a tangent to (2) also.
\nThe points of intersection of (2) and (3) coincide.
\nSubstituting the value of y from (3) in (2), we get
\n\u21d2 x2<\/sup> = \\(4 b\\left(m x+\\frac{a}{m}\\right)\\)
\n\u21d2 x2<\/sup> = 4bmx + \\(\\frac{4ab}{m}\\)
\n\u21d2 mx2<\/sup> = 4bm2<\/sup>x + 4ab
\n\u21d2 mx2<\/sup> – 4bm2<\/sup>x – 4ab = 0
\nThis equation has equal roots, then it’s discriminant = 0
\nb2<\/sup> – 4ac = 0
\n\u21d2 (-4bm2<\/sup>)2<\/sup> – 4(m) (-4ab) = 0
\n\u21d2 16b2<\/sup>m4<\/sup> + 16abm = 0
\n\u21d2 b2<\/sup>m4<\/sup> + abm = 0
\n\u21d2 bm4<\/sup> + am = 0
\n\u21d2 m(bm3<\/sup> + a) = 0
\n\u21d2 m = 0 (or) bm3<\/sup> + a = 0
\n\"TS<\/p>\n

Question 14.
\nThe normal at a point \u2018t1<\/sub>\u2019 on y2<\/sup> = 4ax meets the parabola again in the point \u2018t2<\/sub>\u2019 then prove that t1<\/sub>t2<\/sub> + \\(t_1^2\\) + 2 = 0. (May ’13)
\nSolution:
\n\"TS
\nGiven the equation of the parabola is y2<\/sup> = 4ax
\nThe equation of the normal at P(\\(\\mathrm{at}_1{ }^2\\), 2at1<\/sub>) is
\ny + xt1<\/sub> = 2at1<\/sub> + \\(\\mathrm{at}_1{ }^3\\) …….(1)
\nSince eq. (1) meets the parabola again in the Q(\\(\\mathrm{at}_2{ }^2\\), 2at2<\/sub>) then
\n\"TS
\n\"TS<\/p>\n

Question 15.
\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:
\n\"TS
\nGiven the equation of the parabola is y2<\/sup> = 4ax
\nGiven the equation of the normal is
\nlx + my + n = 0 ……..(1)
\nNow, the equation of the normal at P(at2<\/sup>, 2at) is
\ny + xt = 2at + at3<\/sup> ……..(2)
\nNow, (1) and (2) represent the same line then
\n\"TS
\nWhich is the required condition.<\/p>\n

Question 16.
\nIf a normal chord at a point t on the parabola y2<\/sup> = 4ax subtends a right angle at the vertex then show that t = \u00b1\u221a2. (May ’14)
\nSolution:
\nGiven the equation of the parabola is y2<\/sup> = 4ax
\n\"TS
\n\"TS<\/p>\n

Question 17.
\nShow that the locus of the point of intersection of perpendicular tangents to the parabola y2<\/sup> = 4ax is the directrix x + a = 0.
\nSolution:
\nGiven, the equation of the parabola is y2<\/sup> = 4ax.
\nLet P(x1<\/sub>, y1<\/sub>) be the point of intersection of perpendicular tangents of y2<\/sup> = 4ax.
\nThe equation to the pair of tangents drawn from P(x1<\/sub>, y1<\/sub>) is \\(\\mathrm{S}_1{ }^2\\) = S.S11<\/sub>
\n\u21d2 [yy1<\/sub> – 2a(x + x1<\/sub>)]2<\/sup> = (y2<\/sup> – 4ax) (\\(\\mathbf{y}_1{ }^2\\) – 4ax1<\/sub>)
\nSince the tangents are at right angles, then
\ncoefficient of x2<\/sup> + coefficient of y2<\/sup> = 0
\n4a2<\/sup> + \\(\\mathbf{y}_1{ }^2\\) – (\\(\\mathbf{y}_1{ }^2\\) – 4ax1<\/sub>) = 0
\n\u21d2 4a2<\/sup> + \\(\\mathbf{y}_1{ }^2\\) – \\(\\mathbf{y}_1{ }^2\\) + 4ax1<\/sub> = 0
\n\u21d2 4a2<\/sup> + 4ax1<\/sub> = 0
\n\u21d2 a + x1<\/sub> = 0
\n\u21d2 x1<\/sub> + a = 0
\n\u2234 The equation to the locus of P(x1<\/sub>, y1<\/sub>) is x + a = 0.<\/p>\n

\"TS<\/p>\n

Question 18.
\nShow that the feet of the perpendicular from focus to the tangent of the parabola y2<\/sup> = 4ax lie on the tangent at the vertex.
\nSolution:
\nGiven equation of the parabola is y2<\/sup> = 4ax.
\n\"TS
\nEquation of a tangent to the parabola y2<\/sup> = 4ax is
\ny = mx + \\(\\frac{a}{m}\\)
\n\u21d2 y = \\(\\frac{m^2 x+a}{m}\\)
\n\u21d2 m2<\/sup>x – my + a = 0 ……..(1)
\nEquation of a line passing through the focus S(a, 0) and perpendicular to the line (1) is
\ny – y1<\/sub> = \\(\\frac{-1}{m}\\)(x – x1<\/sub>)
\n\u21d2 y – 0 = \\(\\frac{-1}{m}\\)(x – a)
\n\u21d2 y = \\(\\frac{-1}{m}\\)(x – a)
\n\u21d2 my = -x + a
\n\u21d2 x + my – a = 0 …….(2)
\nSolve (1) and (2)
\n(1) + (2) \u21d2 m2<\/sup>x – my + a + x + my – a = 0
\n\u21d2 x(m2<\/sup> + 1) = 0
\n\u21d2 x = 0 (\u2235 m2<\/sup> \u2260 1)
\n\u2234 The point of intersections of lines (1) and (2) lies on x = 0.
\nWhich is the tangent at the vertex.<\/p>\n

Question 19.
\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 lie on a horizontal line. [Mar. ’19 (TS)]
\nSolution:
\nGiven the equation of the parabola is y2<\/sup> = 4ax
\n\"TS
\nLet P(x1<\/sub>, y1<\/sub>) be any point on the required locus.
\n\u2234 The equation of any tangent to the parabola y2<\/sup> = 4ax is y = mx + \\(\\frac{a}{m}\\)
\nIf this line passes through P then
\ny1<\/sub> = mx1<\/sub> + \\(\\frac{a}{m}\\)
\n\u21d2 y1<\/sub> = \\(\\frac{m^2 x_1+a}{m}\\)
\n\u21d2 my1<\/sub> = m2<\/sup>x1<\/sub> + a
\n\u21d2 x1<\/sub>m2<\/sup> – y1<\/sub>m + a = 0 ………(1)
\nWhich is a quadratic equation in m.
\nIf m1<\/sub>, m2<\/sub> are the slopes of the tangents drawn from P to the parabola then m1<\/sub>, m2<\/sub> are the roots of (1)
\nSum of the slopes = \\(\\frac{-b}{a}\\)
\nm1<\/sub> + m2<\/sub> = \\(\\frac{-\\left(-y_1\\right)}{x_1}=\\frac{y_1}{x_1}\\)
\ntan \u03b81<\/sub> + tan \u03b82<\/sub> = \\(\\frac{\\mathrm{y}_1}{\\mathrm{x}_1}\\)
\nproduct of the slopes = \\(\\frac{c}{a}\\)
\nm1<\/sub>m2<\/sub> = \\(\\frac{\\mathrm{a}}{\\mathrm{x}_1}\\)
\ntan \u03b81<\/sub>\u00a0tan \u03b82<\/sub> = \\(\\frac{\\mathrm{a}}{\\mathrm{x}_1}\\)
\n(\u2235 The tangents made angles \u03b81<\/sub>, \u03b82<\/sub> with its axis (X -axis) then their slopes m1<\/sub> = tan \u03b81<\/sub> and m2<\/sub> = tan \u03b82<\/sub>)
\nGiven that cot \u03b81<\/sub> + cot \u03b82<\/sub> = d
\n\"TS
\n\u2234 P lies on a horizontal line y = ad.<\/p>\n

Question 20.
\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 tan \u03b81<\/sub> + tan \u03b82<\/sub> is a constant, b. Then show that P lies on the line y = bx. [(AP) Mar. ’20]
\nSolution:
\nGiven the equation of the parabola is y2<\/sup> = 4ax
\n\"TS
\nLet, P(x1<\/sub>, y1<\/sub>) be any point on the required locus.
\n\u2234 The equation of any tangent to the parabola y2<\/sup> = 4ax is y = mx + \\(\\frac{a}{m}\\)
\nIf this line passes through P then
\ny1<\/sub> = mx1<\/sub> + \\(\\frac{a}{m}\\)
\n\u21d2 y1<\/sub> = \\(\\frac{m^2 x_1+a}{m}\\)
\n\u21d2 my1<\/sub> = m2<\/sup>x1<\/sub> + a
\n\u21d2 x1<\/sub>m2<\/sup> – y1<\/sub>m + a = 0 ……(1)
\nwhich is a quadratic equation in m.
\nIf m1<\/sub>, m2<\/sub> are the slopes of the tangents drawn from P to the parabola then m1<\/sub>, m2<\/sub> are the roots of (1).
\nSum of the slopes = \\(\\frac{-b}{a}\\)
\nm1<\/sub> + m2<\/sub> = \\(\\frac{-\\left(-y_1\\right)}{x_1}=\\frac{y_1}{x_1}\\)
\ntan \u03b81<\/sub> + tan \u03b82<\/sub> = \\(\\frac{y_1}{x_1}\\)
\n[\u2235 The tangents made angles \u03b81<\/sub>, \u03b82<\/sub> with its axis (X-axis) then their slopes m1<\/sub> = tan \u03b81<\/sub> and m2<\/sub> = tan \u03b82<\/sub>]
\nGiven that, tan \u03b81<\/sub> + tan \u03b82<\/sub> = b
\n\u21d2 \\(\\frac{y_1}{x_1}\\) = b
\n\u21d2 y1<\/sub> = bx1<\/sub>
\n\u2234 P(x1<\/sub>, y1<\/sub>) lies on the line y = bx.<\/p>\n

\"TS<\/p>\n

Question 21.
\nProve that the two parabolas y2<\/sup> = 4ax and x2<\/sup> = 4by intersect (other than the origin) at an angle of \\(\\tan ^{-1}\\left[\\frac{3 a^{\\frac{1}{3}} b^{\\frac{1}{3}}}{2\\left(a^{\\frac{2}{3}}+b^{\\frac{2}{3}}\\right)}\\right]\\). (Mar. ’14)
\nSolution:
\nWithout loss of generality, we can assume that a > 0 and b > 0
\n\"TS
\nLet, P(x, y) be the point of intersection of the parabolas other than the origin.
\nGiven equations of the parabolas are y2<\/sup> = 4ax, x2<\/sup> = b4y
\n\u21d2 y2<\/sup> = 4ax
\nSquaring on both sides
\ny4<\/sup> = 16a2<\/sup>x2<\/sup>
\n\u21d2 y4<\/sup> = 16a2<\/sup>(4by)
\n\u21d2 y4<\/sup> = 64a2<\/sup>by
\n\u21d2 y4<\/sup> – 64a2<\/sup>by = 0
\n\u21d2 y(y3<\/sup> – 64a2<\/sup>b) = 0
\n\u21d2 y = 0 or y3<\/sup> – 64a2<\/sup>b = 0
\n\u21d2 y3<\/sup> = 64a2<\/sup>b
\n\u21d2 y = \\(4 \\mathrm{a}^{2 \/ 3} \\mathrm{~b}^{1 \/ 3}\\)
\nFrom y2<\/sup> = 4ax, we get
\n\"TS
\n\"TS
\n\"TS<\/p>\n

Question 22.
\nShow that the straight line 7x + 6y = 13 is a tangent to the parabola y2<\/sup> – 7x – 8y + 14 = 0 and find the point of contact.
\nSolution:
\nGiven the equation of the parabola is
\ny2<\/sup> – 7x – 8y + 14 = 0 ………(1)
\nGiven the equation of the straight line is
\n7x + 6y = 13
\n\u21d2 7x = 13 – 6y
\n\u21d2 x = \\(\\frac{13-6 y}{7}\\) ………(2)
\nFrom (1) and (2) by eliminating x we get the ordinates of the points of intersection of the line and parabola.
\ny2<\/sup> – 7(\\(\\frac{13-6 y}{7}\\)) – 8y + 14 = 0
\n\u21d2 y2<\/sup> – 13 + 6y – 8y + 14 = 0
\n\u21d2 y2<\/sup> – 2y + 1 = 0
\n\u21d2 (y – 1)2<\/sup> = 0
\n\u21d2 y = 1, 1
\n\u2234 The given line is tangent to the given parabola substitute the value of y = 1 in (2)
\nx = \\(\\frac{13-6}{7}=\\frac{7}{7}=1\\)
\n\u2234 Point of contact = (1, 1)<\/p>\n

Question 23.
\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 the equation of the parabola is y2<\/sup> = 4ax
\nThe equation of the tangent to the parabola y2<\/sup> = 4ax is
\ny = mx + \\(\\frac{a}{m}\\) ……..(1)
\nGiven the equation of the circle is
\n2x2<\/sup> + 2y2<\/sup> = a2<\/sup>
\n\u21d2 x2<\/sup> + y2<\/sup> = \\(\\frac{\\mathrm{a}^2}{2}\\)
\n\u2234 Centre, C = (0, 0)
\nRadius = \\(\\frac{a}{\\sqrt{2}}\\)
\nSince (1) is a tangent to the circle 2x2<\/sup> + 2y2<\/sup> = a2<\/sup> then r = d
\n\"TS
\n\u21d2 m2<\/sup> + m4<\/sup> = 2
\n\u21d2 m4<\/sup> + m2<\/sup> – 2 = 0
\n\u21d2 m4<\/sup> + 2m2<\/sup> – m2<\/sup> – 2 = 0
\n\u21d2 m2<\/sup>(m2<\/sup> + 2) – 1(m2<\/sup> + 2<\/sup>) = 0
\n\u21d2 (m2<\/sup> + 2)(m2<\/sup> – 1) = 0
\n\u21d2 m2<\/sup> + 2 = 0 or m2<\/sup> – 1 = 0
\n\u21d2 m2<\/sup> = -2 or m2<\/sup> = 1
\n\u21d2 m = \u00b1\u221a-2 \u2209 R or m = \u00b11
\nSubstitute the values of m in (1)
\n\u2234 The equations of the common tangents are
\ny = \\(\\pm 1 \\cdot x+\\frac{a}{\\pm 1}\\)
\n\u21d2 y = \u00b1x \u00b1 a
\n\u21d2 y = \u00b1(x \u00b1 a) …….(2)
\nThe focus of the parabola y2<\/sup> = -4ax is S = (-a, 0)
\nNow, (2) intersects at the focus of the parabola y2<\/sup> = -4ax then
\n(2) passes through focus, S(-a, 0)
\n0 = \u00b1(-a + a)
\n\u2234 0 = 0
\n\u2234 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.<\/p>\n

\"TS<\/p>\n

Question 24.
\nShow that the condition that the line y = mx + c may be a tangent to the parabola y2<\/sup> = 4ax is c = \\(\\frac{a}{m}\\). (Mar. ’99, ’94; May ’98)
\nSolution:
\nSuppose y = mx + c ………(1)
\nis a tangent to the parabola y2<\/sup> = 4ax
\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 yy1<\/sub> – 2a(x + x1<\/sub>) = 0
\n\u21d2 yy1<\/sub> – 2ax – 2ax1<\/sub> = 0
\n\u21d2 2ax – yy1<\/sub> + 2ax1<\/sub> = 0 ………(2)
\nNow, (1) & (2) represent the same line
\n\"TS
\nSince \u2018P\u2019 lies on the line
\n\"TS<\/p>\n

Question 25.
\nFind the condition for the line y = mx + c to be a tangent to the parabola x2<\/sup> = 4ay. (Mar. ’12; May ’03)
\nSolution:
\nGiven the equation of the parabola is x2<\/sup> = 4ay
\nLet the line y = mx + c ………(1)
\nbe a tangent to the parabola x2<\/sup> = 4ay.
\n\"TS
\nThe equation of the tangent at P(x1<\/sub>, y1<\/sub>) is S1<\/sub> = 0
\nxx1<\/sub> – 2a(y + y1<\/sub>) = 0
\n\u21d2 xx1<\/sub> – 2ay – 2ay1<\/sub> = 0 ……….(2)
\nNow equations (1) & (2) represent the same line then
\n\"TS
\nSince P(x1<\/sub>, y1<\/sub>) lies on the line y = mx + c then y1<\/sub> = mx1<\/sub> + c
\n\u21d2 -c = m(2am) + c
\n\u21d2 2am2<\/sup> + 2c = 0
\n\u21d2 am2<\/sup> + c = 0
\nWhich is the required condition.<\/p>\n

Question 26.
\nProve that the tangents at the extremities of a focal chord of a parabola intersect at right angles on the directrix.
\nSolution:
\nLet the equation of the parabola is y2<\/sup> = 4ax.
\nLet P(\\(\\mathrm{at}_1^2\\), 2at1<\/sub>), Q(\\(\\mathrm{at}_2^2\\), 2at2<\/sub>) are two points on the parabola.
\n\"TS
\n\"TS
\n\u2234 The tangents are drawn at P, Q are perpendicular
\nThe coordinates of R = [-a, a(t1<\/sub> + t2<\/sub>)]
\nThe equation of the directrix of a parabola y = 4ax is x + a = 0
\nNow, substitute the point R in the directrix x + a = 0
\n\u21d2 -a + a = 0
\n\u21d2 0 = 0
\n\u2234 The tangents at the extremities of a focal chord of a parabola intersect at right angles on the directrix.<\/p>\n

\"TS<\/p>\n

Question 27.
\nFind the equation of the parabola whose focus is S(3, 5) and the vertex is A(1, 3).
\nSolution:
\n\"TS
\n\"TS
\n\"TS<\/p>\n

Question 28.
\nFind the equations of the tangents to the parabola y2<\/sup> = 16x, which are parallel and perpendicular respectively to the line 2x – y + 5 = 0. Find the coordinates of their points of contact also.
\nSolution:
\nGiven parabola is y2<\/sup> = 16x
\nComparing with y2<\/sup> = 4ax
\nwe get 4a = 16 \u21d2 a = 4
\nGiven line is 2x – y + 5 = 0
\n\u21d2 y = 2x + 5
\nComparing with y = mx + c we get
\nm = 2, c = 5
\n(i) Equation of the tangent with slope ‘m’ is
\n\"TS
\n(ii) Slope of the perpendicular given line is m = \\(\\frac{-1}{2}\\)
\nequation of tangent with slope ‘m’ is
\n\"TS<\/p>\n

\"TS<\/p>\n

Question 29.
\nIf L and L’ are the ends of the latus rectum of the parabola x2<\/sup> = 6y. Find the equations of OL and OL’ where \u2018O\u2019 is the origin. Also, find the angle between them.
\nSolution:
\n\"TS
\n\"TS
\n\"TS<\/p>\n

Question 30.
\nTwo parabolas have the same vertex and equal length of latus rectum such that their axes are at right angles. Prove that the common tangent touches each at the end of the latus rectum.
\nSolution:
\nEquations of the parabolas can be taken as y2<\/sup> = 4ax and x2<\/sup> = 4ay
\n\"TS
\nEquation of the tangent at P(2at, at2<\/sup>) to the parabola x2<\/sup> = 4ay is S1<\/sub> = 0
\n\u21d2 xx1<\/sub> – 2a(y + y1<\/sub>) = 0
\n\u21d2 x(2at) – 2a(y + at2<\/sup>) = 0
\n\u21d2 xt – y – at2<\/sup> = 0
\n\u21d2 tx – y – at2<\/sup> = 0 ………(1)
\n\u21d2 y = tx – at2<\/sup>
\nThis is a tangent to y2<\/sup> = 4ax, then
\nc = \\(\\frac{a}{m}\\)
\n\u21d2 -at2<\/sup> = \\(\\frac{a}{t}\\)
\n\u21d2 -t3<\/sup> = 1
\n\u21d2 t3<\/sup> = -1
\n\u21d2 t = -1
\nSubstitute the value of t = -1 in (1)
\n-x – y – a(1) = 0
\n\u21d2 x + y + a = 0
\nEquation of the tangent at L'(a, -2a) to the parabola y2<\/sup> = 4ax is S1<\/sub> = 0
\n\u21d2 yy1<\/sub> – 2a(x + x1<\/sub>) = 0
\n\u21d2 y(-2a) – 2a(x + a) = 0
\n\u21d2 y + x + a = 0
\n\u21d2 x + y + a = 0
\nCommon tangent to the parabolas touches the parabola y2<\/sup> = 4ax at L'(a, -2a).
\nEquation of the tangent at L'(-2a, a) to the parabola x2<\/sup> = 4ay is S1<\/sub> = 0
\n\u21d2 xx1<\/sub> – 2a(y + y1<\/sub>) = 0
\n\u21d2 x(-2a) – 2a(y + a) = 0
\n\u21d2 x + y + a = 0
\n\u2234 Common tangent to the parabolas touches the parabola x2<\/sup> = 4ay at L'(-2a, a).<\/p>\n

Question 31.
\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, the equation of the parabola is y2<\/sup> = 4ax
\n\"TS
\nLet P(\\(\\mathrm{at}_1^2\\), 2at1<\/sub>), Q(\\(\\mathrm{at}_2^2\\), 2at2<\/sub>) be the two ends of a focal chord of the parabola y2<\/sup> = 4ax, then
\nt1<\/sub>t2<\/sub> = -1
\n\u21d2 -t2<\/sub> = \\(\\frac{1}{t_1}\\)
\nLet \\(\\frac{1}{t_1}\\) = m1<\/sub>
\nSlope of the normal at Q(\\(\\mathrm{at}_2^2\\), 2at2<\/sub>) is
\nm2<\/sub> = -t2<\/sub> = \\(\\frac{1}{t_1}\\) = m1<\/sub>
\n\u2234 m1<\/sub> = m2<\/sub>, then the tangent at P and the normal at Q are parallel.<\/p>\n

\"TS<\/p>\n

Question 32.
\nProve that the normal chord at the point other than the origin whose ordinate is equal to its abscissa subtends a right angle at the focus.
\nSolution:
\nLet, the equation of the parabola is y2<\/sup> = 4ax ……..(1)
\n\"TS
\nLet P(at2<\/sup>, 2at) be any point on the parabola given that, whose ordinate is equal to its abscissa, then
\n2at = at2<\/sup>
\n\u21d2 t2<\/sup> = 2t
\n\u21d2 t2<\/sup> – 2t = 0
\n\u21d2 t(t – 2) = 0
\n\u21d2 t = 0, t = 2
\nBut t \u2260 0, then P(4a, 4a)
\nThe equation of the normal at P(4a, 4a) is
\ny + xt = 2at + at3<\/sup>
\n\u21d2 y + x(2) = 2a(2) + a(2)3<\/sup>
\n\u21d2 y + 2x = 4a + 8a
\n\u21d2 y + 2x = 12a
\n\u21d2 y = 12a – 2x …….(2)
\nSubstituting the value of y = 12a – 2x in (1) we get
\n(12a – 2x)2<\/sup> = 4ax
\n\u21d2 4(6a – x)2<\/sup> = 4ax
\n\u21d2 (6a – x)2<\/sup> = ax
\n\u21d2 36a2<\/sup> + x2<\/sup> – 12ax – ax = 0
\n\u21d2 36a2<\/sup> + x2<\/sup> – 13ax = 0
\n\u21d2 x2<\/sup> – 9ax – 4ax + 36a2<\/sup> = 0
\n\u21d2 x(x – 9a) – 4a(x – 9a) = 0
\n\u21d2 (x – 9a) (x – 4a) = 0
\n\u21d2 x – 9a = 0 (or) x – 4a = 0
\n\u21d2 x = 9a (or) x = 4a
\n\u21d2 x = 4a, 9a
\nIf x = 4a, then y = 12a – 8a = 4a
\n\u2234 P = (4a, 4a)
\nIf x = 9a, then y = 12a – 18a = -6a
\n\u2234 Q = (9a, -6a)
\n\u2234 P = (4a, 4a), Q = (9a, -6a)
\nFocus S = (a, 0)
\n\"TS
\nSince m1<\/sub>m2<\/sub> = -1, then
\n\\(\\overline{\\mathrm{SP}}\\) is perpendicular to \\(\\overline{\\mathrm{SQ}}\\)
\n\u2234 The normal chord subtends a right angle at the focus.<\/p>\n

Question 33.
\n(i) If the coordinates of the ends of a focal chord of the parabola y2<\/sup> = 4ax are (x1<\/sub>, y1<\/sub>) and (x2<\/sub>, y2<\/sub>) then prove that x1<\/sub>x2<\/sub> = a2<\/sup>, y1<\/sub>y2<\/sub> = -4a2<\/sup>.
\n(ii) For a focal chord PQ of the parabola y2<\/sup> = 4ax, if SP = l and SQ = l’ then prove that \\(\\frac{1}{l}+\\frac{1}{l^{\\prime}}=\\frac{1}{a}\\).
\nSolution:
\nGiven the equation of the parabola is y2<\/sup> = 4ax
\nLet P(x1<\/sub>, y1<\/sub>) = (\\(a \\mathrm{t}_1^2\\), 2at1<\/sub>) and Q(x2<\/sub>, y2<\/sub>) = (\\(a \\mathrm{t}_2^2\\), 2at2<\/sub>) be two endpoints of a focal chord.
\n\"TS
\n\"TS
\n\"TS<\/p>\n

Question 34.
\nProve that the area of the triangle inscribed in the parabola y2<\/sup> = 4ax is \\(\\frac{1}{8a}\\) |(y1<\/sub> – y2<\/sub>)(y2<\/sub> – y3<\/sub>)(y3<\/sub> – y1<\/sub>)| sq. units where y1<\/sub>, y2<\/sub>, y3<\/sub> are the ordinates of its vertices. [(TS) May. ’15]
\nSolution:
\n\"TS
\n\"TS<\/p>\n

Question 35.
\nProve that the area of the triangle formed by the tangents at (x1<\/sub>, y1<\/sub>), (x2<\/sub>, y2<\/sub>), and (x3<\/sub>, y3<\/sub>) to the parabola y2<\/sup> = 4ax (a > 0) is \\(\\frac{1}{16a}\\) |(y1<\/sub> – y2<\/sub>)(y2<\/sub> – y3<\/sub>)(y3<\/sub> – y1<\/sub>)| sq.units. [(AP) Mar. ’18, (TS) ’15]
\nSolution:
\nGiven parabola y2<\/sup> = 4ax
\n\"TS
\n\"TS
\n\"TS<\/p>\n

\"TS<\/p>\n

Question 36.
\nProve that the orthocentre of the triangle formed by any three tangents to a parabola lies on the directrix of the parabola.
\nSolution:
\nLet the equation of the parabola is y2<\/sup> = 4ax
\nLet A, B, C be the triangle formed by the tangents to the parabola at P(\\(\\mathrm{at}_1^2\\), 2at1<\/sub>), Q(\\(\\mathrm{at}_2^2\\), 2at2<\/sub>) and R(\\(\\mathrm{at}_3^2\\), 2at3<\/sub>) as shown in the figure.
\n\"TS
\nA = Point of intersection of the tangents at P, Q = [at1<\/sub>t2<\/sub>, a(t1<\/sub> + t2<\/sub>)]
\nB = Point of intersection of the tangents at P, R = [at1<\/sub>t3<\/sub>, a(t1<\/sub> + t3<\/sub>)]
\nC = Point of Intersection of the tangents at Q, R = [at2<\/sub>t3<\/sub>, a(t2<\/sub> + t3<\/sub>)]
\nLet, AD and CE be the two altitudes of \u2206ABC.
\n\\(\\overline{\\mathrm{BC}}\\) is the tangent at R, then the equation of \\(\\overline{\\mathrm{BC}}\\) is yt3<\/sub> = x + \\(\\mathrm{at}_3^2\\).
\n\"TS
\n\"TS
\n(1) \u21d2 t3<\/sub>(-a) + y – at1<\/sub> – at2<\/sub> – at1<\/sub>t2<\/sub>t3<\/sub> = 0
\ny = a(t1<\/sub> + t2<\/sub> + t3<\/sub> + t1<\/sub>t2<\/sub>t3<\/sub>)
\n\u2234 Orthocentre H = [-a, a(t1<\/sub> + t2<\/sub> + t3<\/sub> + t1<\/sub>t2<\/sub>t3<\/sub>)]
\nThe equation of the directrix of the parabola y2<\/sup> = 4ax is x + a = 0
\nNow substitute H in the directrix
\n-a + a = 0
\n\u21d2 0 = 0
\n\u2234 H lies on the directrix x + a = 0.<\/p>\n","protected":false},"excerpt":{"rendered":"

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