{"id":22406,"date":"2024-01-17T15:35:03","date_gmt":"2024-01-17T10:05:03","guid":{"rendered":"https:\/\/apboardsolutions.in\/?p=22406"},"modified":"2024-01-18T15:50:18","modified_gmt":"2024-01-18T10:20:18","slug":"inter-2nd-year-maths-2b-parabola-solutions-ex-3b","status":"publish","type":"post","link":"https:\/\/apboardsolutions.in\/inter-2nd-year-maths-2b-parabola-solutions-ex-3b\/","title":{"rendered":"Inter 2nd Year Maths 2B Parabola Solutions Ex 3(b)"},"content":{"rendered":"

Practicing the Intermediate 2nd Year Maths 2B Textbook Solutions<\/a> Inter 2nd Year Maths 2B Parabola Solutions Exercise 3(b) will help students to clear their doubts quickly.<\/p>\n

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

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

Question 1.
\nFind the equation of the tangent and normal to the parabola y\u00b2 = 6x at the positive end of the latus rectum.
\nSolution:
\n(a, 2a) Here 4a = 6 \u21d2 a = \\(\\frac{3}{2}\\)
\n(\\(\\frac{3}{2}\\), 3)
\nequation of tangent yy1<\/sub> = 2a (x + x1<\/sub>)
\nyy1<\/sub> = 3(x + x1<\/sub>)
\n3y = 3(x + \\(\\frac{3}{2}\\))
\n2y – 2x – 3 = O’ is the equation of tangent
\nSlope of tangent is 1
\nSlope of normal is -1
\nEquation of normal is y – 3 = -1(x – \\(\\frac{3}{2}\\))
\n2x + 2y – 9 = 0<\/p>\n

Question 2.
\nFind the equation of the tangent and normal to the parabola x\u00b2 – 4x – 8y + 12 = 0 at (4, \\(\\frac{3}{2}\\))
\nSolution:
\n(x – 2)\u00b2 – 4 – 8y + 12 = 0
\n\u21d2 (x – 2)\u00b2 – 8y + 8 = 0
\n\u21d2 (x – 2)\u00b2 = 8(y – 1); 4a = 8 \u21d2 a = 2
\nEquation of tangents at (x1<\/sub>, y1<\/sub>) is
\n(x – 2) (x1<\/sub> – 2) = 2a (y- 1 + y1<\/sub> – 1)
\n\u21d2 (x – 2) (4 – 2) = 2(2) (y – 1 + \\(\\frac{3}{2}\\) – 1)
\n\u21d2 2(x – 2) = 4(\\(\\frac{2y-1}{2}\\)
\nx – 2y – 1 = 0
\nEquation of normal will be
\ny – y1<\/sub> = m(x – x1<\/sub>)
\nm – Slope of normal
\nSlope of tangent is \\(\\frac{1}{2}\\)
\nSlope of normal is – 2
\ny – \\(\\frac{3}{2}\\) = -2(x – 4)
\n2y – 3 = -4x + 16
\n4x + 2y – 19 = 0<\/p>\n

\"Inter<\/p>\n

Question 3.
\nFind the value of k if the line 2y = 5x + k is a tangent to the parabola y\u00b2 = 6x.
\nSolution:
\nGiven line is 2y = 5x + k
\n\u21d2 y = (\\(\\frac{5}{2}\\))x + (\\(\\frac{k}{2}\\))
\nComparing y = (\\(\\frac{5}{2}\\))x + (\\(\\frac{k}{2}\\)) with y = mx + c
\nWe get m = \\(\\frac{5}{2}\\), c = \\(\\frac{k}{2}\\)
\ny = (\\(\\frac{5}{2}\\))x + (\\(\\frac{k}{2}\\)) is a tangent to y\u00b2 = 6x
\n\"Inter<\/p>\n

Question 4.
\nFind the equation of the normal to the parabola y\u00b2 = 4x which is parellel to y – 2x + 5 = 0.
\nSolution:
\nEquation of the parabola is y\u00b2 = 4x
\n\u2234 a = 1
\nThe equation of the given line y – 2x + 5 = 0
\nSlope m = 2
\nThe normal is parallel to the line y – 2x +5 = 0
\nSlope of the normal = 2
\nEquation of the normal at ‘t’ is
\ny + tx = 2at + at\u00b3
\n\u2234 Slope = -t = 2
\n\u21d2 t = -2
\nEquation of the normal is
\ny – 2x = 2.1 (-2) + 1(-2)\u00b3
\n= -4 – 8 = – 12
\n2x-y- 12 = 0.<\/p>\n

Question 5.
\nShow that the line 2x – y + 2 = 0isa tangent to the parabola y\u00b2 = 16x. Find the point of contact also. .
\nSolution:
\nGiven line is 2x – y + 2 = 0
\n\u21d2 y = 2x + 2
\nComparing with y = mx + c we get m = 2, c = 2
\nComparing y\u00b2 = 16x with y\u00b2 = 4ax
\nWe get 4a = 16 \u21d2 a = 4
\n\\(\\frac{a}{m}=\\frac{4}{2}\\) = 2 = c
\nPoint of contact = (\\(\\frac{a}{m^{2}},\\frac{2a}{m}\\)) = (\\(\\frac{4}{2^{2}},\\frac{2(4)}{2}\\))
\n= (1.4)<\/p>\n

Question 6.
\nFind the equation of tangent to the parabola y2 = 16x inclined at an angle 60\u00b0 with its axis and also find the point of contact.
\nSolution:
\n\u03b8 = 60\u00b0; m = tan 60\u00b0 = \u221a3
\n\"Inter<\/p>\n

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

Question 1.
\nFind the equations of tangents to the parabola y\u00b2 = 16x which are parallel and perpendicular respectively to the line 2x – y + 5 = 0, also find the co-ordinates of the points of contact also.
\nSolution:
\nEquation of the parabola is y\u00b2 = 16x
\nThe tangent is parallel to 2x – y + 5 = 0
\nEquation of the tangent can be taken as y = 2x + c
\nCondition for tangency is c = \\(\\frac{a}{m}=\\frac{4}{2}\\) = 2
\nEquation of the parallel tangent is y = 2x + 2
\n2x – y + 2 = 0
\nPoint of contact is (\\(\\frac{a}{m^{2}},\\frac{2a}{m}\\)) = (\\(\\frac{4}{4},\\frac{8}{2}\\)) = (1, 4)
\nSlope of the perpendicular tangept is
\nm’ = –\\(\\frac{1}{m}=-\\frac{1}{2}\\)
\nEquation of the perpendicular tangent is
\ny = m’x + c’
\n= (-\\(\\frac{1}{2}\\)) x + c’
\nwhere c’ = \\(\\frac{a}{m’}\\) = \\(\\frac{4}{\\left(-\\frac{1}{2}\\right)}\\) = -8
\nEquation of the perpendicular tangent is
\ny = –\\(\\frac{1}{2}\\)x – 8
\n2y = – x – 16
\nx + 2y + 16 = 0
\n\"Inter<\/p>\n

Question 2.
\nIf lx + my + n = 0 is a normal to the parabola y\u00b2 = 4ax, then show that al\u00b3 + 2alm\u00b2 + nm\u00b2 = 0.
\nSolution:
\nEquation of the parabola is y\u00b2 = 4ax
\nEquation of the normal is y + tx = 2at + at\u00b3
\ntx + y – (2at + at\u00b3) = 0 ……….. (1)
\nEquation of the given line is
\nlx + my + n = 0 …………. (2)
\n(1), (2) represent the same line.
\nComparing the co-efficients
\n\"Inter
\nMultiplying with m\u00b3, we get
\n-nm\u00b2 = 2alm\u00b2 + al\u00b3
\n\u21d2 al\u00b3 + 2alm\u00b2 + nm\u00b2 = 0<\/p>\n

\"Inter<\/p>\n

Question 3.
\nShow that the equations of common tangents to the circle x\u00b2 + y\u00b2 = 2a\u00b2 and the parabola y\u00b2 = 8ax are y = \u00b1 (x + 2a).
\nSolution:
\nThe equation of tangent to parabola
\ny\u00b2 = 8ax is y = mx + \\(\\frac{2a}{m}\\) ……….. (i)
\nm\u00b2x – my + 2a = 0
\nIf (i) touches circle x\u00b2 + y\u00b2 = 2a\u00b2, then the length of perpendicular from its centre (0, 0) to (i) must be equal to the radius a\u221a2 of the circle.
\n\\(\\frac{2a}{\\sqrt{m^{2} + m^{4}}}\\) = a\u221a2
\nor 4 = 2 (m4<\/sup> + m\u00b2)
\nm4<\/sup> + m\u00b2 – 2 = 0
\n(m\u00b2 + 2) (m\u00b2 – 1) = 0 or m = \u00b11
\nRequired tangents are
\ny = (1)x + \\(\\frac{2a}{(1)}\\) , y = (-1) x + \\(\\frac{2a}{(-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.
\n\"Inter
\nSolution:
\nEquation of the parabola is y\u00b2 = 4ax
\nEquation of the tangent at Q(1<\/sub>) is
\nt1<\/sub>y = x + at\u00b21<\/sub>
\nEquation of the tangent at R(t2<\/sub>) is t2<\/sub>y = x + at\u00b22<\/sub>
\nSolving, point of intersection is T [at1<\/sub>t2<\/sub>, a(t1<\/sub> + t2<\/sub>)]
\nEquation of the chord QR is
\n(t1<\/sub> + t2<\/sub>)y = 2x + 2at1<\/sub>t2<\/sub><\/p>\n

Question 5.
\nFind the condition for the line y = mx + c to be a tangent to the parabola x\u00b2 = 4ay.
\nSolution:
\nEquation of tangent to x\u00b2 = 4ay in terms of ‘m1<\/sub>‘ [slope of tangent] is
\n\"Inter
\nComparing (i) and (ij), we get
\n\"Inter
\nc = – am\u00b2 is required condition.<\/p>\n

Question 6.
\nThree normals are drawn (k, 0) to the parabola y\u00b2 = 8x one of the normal is the axis and the axis and the remaining two normals are perpendicular to each other, then find the value of k.
\nSolution:
\nEquation of any normal to the parabola is
\ny + xt = 2at + at\u00b3
\nThis normal passes through (k, 0)
\n\u2234 kt = 2at + at\u00b3
\nat\u00b3 + (2a – k)t = 0
\nat\u00b2 + (2a – k) = 0
\nGiven m1<\/sub> = 0, m2<\/sub>m3<\/sub> = -1
\n(-t2<\/sub>) (-t3<\/sub>) = -1 \u21d2 t2<\/sub>t3<\/sub>
\n\\(\\frac{2a-k}{a}\\) = -1
\n2a – k = -a
\nk = 2a + a = 3a
\nEquation of the parabola is
\ny\u00b2 = 8x
\n4a = 8
\n\u21d2 a = 2
\nk = 3a = 3(2) = 6.<\/p>\n

\"Inter<\/p>\n

Question 7.
\nShow that the locus of point of intersection of perpendicular tangents to the parabola y\u00b2 = 4ax is the directrix x + a = 0.
\nSolution:
\nEquation of any tangent to the parabola can be taken as y = mx + \\(\\frac{a}{m}\\)
\nThis tangent passes through P(x1<\/sub>, y1<\/sub>)
\ny1<\/sub> = mx1<\/sub> + \\(\\frac{a}{m}\\)
\nmy1<\/sub> = m\u00b2x1<\/sub> + a
\nm\u00b2x1<\/sub> – my1<\/sub> + a = 0
\n\"Inter
\nThe tangents are perpendicular
\n\u21d2 m1<\/sub>m2<\/sub> = -1
\n\\(\\frac{a}{x_1}\\) = -1
\nx1<\/sub> = -a
\nLocus of P(x1<\/sub>, y1<\/sub>) is x = -a, the directrix of the parabola.<\/p>\n

Question 8.
\nTwo Parabolas have the same vertex and equal length of latus rectum such that their axes are at right angle. Prove that the common tangents touch each at the end of latus rectum.
\nSolution:
\n\"Inter
\nEquations of the parabolas can be taken as
\ny\u00b2 = 4ax
\nand x\u00b2 = 4ay
\nEquation of the tangent at (2at, at\u00b2) to
\nx\u00b2 = 4ay is
\n2atx = 2a(y + at\u00b2)
\ny = tx – at\u00b2
\nThis is a tangent to y\u00b2 = 4ax
\n\u2234 The condition is c = \\(\\frac{a}{m}\\)
\n-at\u00b2 = \\(\\frac{a}{t}\\)
\nt\u00b3 = -1 \u21d2 t = -1
\nEquation of the tangent is y = -x – a
\nx + y + a = 0
\nEquation of the tangent at L’ (a, – 2a) is
\ny(-2a) = 2a(x + a)
\nx + y + a = 0
\n\u2234 Common tangent to the parabolas touches the parabola y\u00b2 = 4ax at L (a, -2a) Equation of the tangent at L (-2a, a) to
\nx\u00b2 = 4ay
\nx(-2a) = 2a(y + a)
\nx + y + a = 0
\nCommon tangent to the parabolas touch the parabola at L’ (-2a, a)<\/p>\n

Question 9.
\nShow that the foot of the perpendicular from focus to the tangent of the parabola y\u00b2 = 4ax lies on the tangent at vertex.
\nSolution:
\nEquation of any tangent to the parabola is
\ny = mx + \\(\\frac{a}{m}\\)
\nQ(x1<\/sub>, y1<\/sub>) is the foot of the perpendicular
\n\u2234 y1<\/sub> = mx1<\/sub> + \\(\\frac{a}{m}\\) ……….. (1)
\nSlope of SQ = \\(\\frac{y_{1}}{x_{1}-a}\\)
\n\"Inter
\nSQ and PQ are perpendicular
\n\"Inter
\nSubstituting in (1) we get
\n\"Inter
\nLocus of Q (x1<\/sub>, y1<\/sub>) is x = 0 i.e., the tangent at the vertex of the parabola.<\/p>\n

\"Inter<\/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:
\n\"Inter
\nP(t1<\/sub>), Q(t2<\/sub>) are the ends of a focai chord.
\nSlope of PS = Slope of PQ
\n\"Inter
\nEquation of the tangent at P(t1<\/sub>) is
\nt1<\/sub>y = x + at\u00b21<\/sub>
\nSlope of the tangent at P = \\(\\frac{1}{t_{1}}\\) ………… (2)
\nEquation of the normal at Q(t2<\/sub>) is
\ny + xt2<\/sub> = 2at2<\/sub> + at\u00b32<\/sub>
\nSlope of the normal at Q = -t2<\/sub> ………… (3)
\nFrorn (1), (2)\u201e (3) we get slope of the tangent at P = slope of normal at Q
\nSlope of the tangent at P is parallel to the normal at Q.<\/p>\n

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

Question 1.
\nIf the normal at the point t, on the parabola y\u00b2 = 4ax meets it again at point t\u00b2 then prove that t1<\/sub>t2<\/sub> + t1<\/sub>\u00b2 + 2 = 0.
\nSolution:
\nEquation of normal is
\n\"Inter
\nEquation of the line (i) again meets parabola at (at\u00b22<\/sub>, 2at2<\/sub>)
\n\u2234 2at2<\/sub> – 2at1<\/sub> = t1<\/sub>(at\u00b22<\/sub> – at\u00b21<\/sub>)
\n–\\(\\frac{2}{t_{1}}\\) = t1<\/sub> + t2<\/sub> \u21d2 -2 = t\u00b21<\/sub> + t1<\/sub>t2<\/sub>
\n\u21d2 t\u00b21<\/sub> + t1<\/sub>t2<\/sub> + 2 = 0<\/p>\n

Question 2.
\nFrom an external point P tangents are drawn to the parabola y\u00b2 = 4ax and these tangents make angles \u03b81<\/sub>, \u03b82<\/sub> with its axis such that cot \u03b81<\/sub> + cot \u03b82<\/sub> is a constant ‘a’ show that P lies on a horizontal line.
\nSolution:
\nEquation of any tangent to the parabola is
\ny = mx + \\(\\frac{a}{m}\\)
\nThis tangent passes through P(x1<\/sub>, y1<\/sub>)
\ny1<\/sub> = mx1<\/sub> + \\(\\frac{a}{m}\\) \u21d2 my1<\/sub> = m\u00b2x1<\/sub> + a
\nm\u00b2x1<\/sub> – my1<\/sub> + a = 0
\nSuppose m1<\/sub>, m2<\/sub> are the roots of the equation
\nm1<\/sub> + m2<\/sub> = \\(\\frac{y_{1}}{x_{1}}\\), m1<\/sub>m2<\/sub> = \\(\\frac{a}{x_{1}}\\)
\nGiven cot \u03b81<\/sub> + cot \u03b82<\/sub> = a
\n\"Inter
\nFocus of P(x1<\/sub>; y1<\/sub>) is y = a\u00b2 which is a horizontal line.<\/p>\n

\"Inter<\/p>\n

Question 3.
\nShow that the common tangent to the circle 2x\u00b2 + 2y\u00b2 = a\u00b2 and the parabola 4ax intersect at the focus of the parabola y\u00b2 = -4ax.
\nSolution:
\nGiven circle is 2x\u00b2 + 2y\u00b2 = a\u00b2
\nwith centre = (0, 0); Radius = \\(\\frac{a}{\\sqrt{2}}\\)
\nGiven parabola is y\u00b2 = 4ax a
\nLet y = mx + \\(\\frac{a}{m}\\) be required tangent
\nBut it touches 2x\u00b2 + 2y\u00b2 = a\u00b2
\n\u21d2 Perpendicular distance from (0, 0) = radius
\n\"Inter
\n\"Inter
\nHence the focus of tangent of the parabola is y\u00b2 = – 4ax.<\/p>\n

Question 4.
\nThe sum of the ordinates of two points on y\u00b2 = 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:
\n\"Inter
\nEquation of the parabola is y\u00b2 = 4ax
\nEquation of the chord joining P(t1<\/sub>) and Q(t2<\/sub>) is (t1<\/sub> + t2<\/sub>)y = 2x + 2 at1<\/sub>t2<\/sub>
\nSlope of PQ = \\(\\frac{2}{t_{1}+t_{2}}\\) ………… (1)
\nEquation of the chord joining R(t3<\/sub>) and S(t4<\/sub>) is (t3<\/sub> + t4<\/sub>) y = 2x + 2at3<\/sub>t4<\/sub>
\nSlope of RS = \\(\\frac{2}{t_{3}+t_{4}}\\) ………… (2)
\nGiven 2at1<\/sub> + 2at2<\/sub> = 2at3<\/sub> + 2at4<\/sub>
\ni.e., 2a(t1<\/sub> + t2<\/sub>) = 2a(t3<\/sub> + t4<\/sub>)
\nt1<\/sub> + t2<\/sub> = t3<\/sub> + t4<\/sub>
\nFrom (1), (2), (3) we get slope of PQ = Slope of Rs
\ni.e., PQ and RS are parallel.<\/p>\n

\"Inter<\/p>\n

Question 5.
\nIf a normal chord a point ‘t’ on the parabola y\u00b2 = 4ax subtends a right angle at vertex, then prove that t = \u00b1 \u221a2
\nSolution:
\nEquation of the parabola is y\u00b2 = 4ax ………….. (1)
\nEquation of the normal at’t’ is
\ntx + y = 2at + at\u00b3
\n\"Inter
\nHomogenising (1) with the help of (2) combined equation of AQ, AR is
\ny\u00b2 = \\(\\frac{4ax.(tx+y}{a(2t+t^{3})}\\)
\ny\u00b2(2t + t\u00b3) = 4tx\u00b2 + 4xy
\n4tx\u00b2 + 4xy – (2t + t\u00b3)y\u00b2 = 0
\nAQ, AR are perpendicular
\nCo-eff. of x\u00b2 + Co.eff. of y\u00b2 = 0
\n4t – 2t – t\u00b3 = 0
\n2t – t\u00b3 = 0
\n-t(t\u00b2 – 2) = 0
\nt\u00b2 – 2 = 0 \u21d2 t\u00b2 = 2
\nt = \u00b1 \u221a2<\/p>\n","protected":false},"excerpt":{"rendered":"

Practicing the Intermediate 2nd Year Maths 2B Textbook Solutions Inter 2nd Year Maths 2B Parabola Solutions Exercise 3(b) will help students to clear their doubts quickly. Intermediate 2nd Year Maths 2B Parabola Solutions Exercise 3(b) I. Question 1. Find the equation of the tangent and normal to the parabola y\u00b2 = 6x at the positive … Read more<\/a><\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[15],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts\/22406"}],"collection":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/comments?post=22406"}],"version-history":[{"count":1,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts\/22406\/revisions"}],"predecessor-version":[{"id":22407,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/posts\/22406\/revisions\/22407"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/media?parent=22406"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/categories?post=22406"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.in\/wp-json\/wp\/v2\/tags?post=22406"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}