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>\nTS 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
\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<\/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>\nQuestion 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
\n<\/p>\nQuestion 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<\/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
\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>\nQuestion 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
\n<\/p>\nQuestion 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
\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<\/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
\n<\/p>\nQuestion 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
\n<\/p>\nQuestion 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
\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
\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
\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<\/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
\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<\/p>\nQuestion 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
\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>\nQuestion 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
\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<\/p>\nQuestion 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
\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
\n<\/p>\nQuestion 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
\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
\nWhich is the required condition.<\/p>\nQuestion 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
\n<\/p>\nQuestion 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<\/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
\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>\nQuestion 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
\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
\n\u2234 P lies on a horizontal line y = ad.<\/p>\nQuestion 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 \u03b8