{"id":3955,"date":"2023-12-19T14:55:27","date_gmt":"2023-12-19T09:25:27","guid":{"rendered":"https:\/\/tsboardsolutions.in\/?p=3955"},"modified":"2023-12-20T17:55:25","modified_gmt":"2023-12-20T12:25:25","slug":"ts-inter-1st-year-maths-1b-solutions-chapter-10-ex-10d","status":"publish","type":"post","link":"https:\/\/tsboardsolutions.in\/ts-inter-1st-year-maths-1b-solutions-chapter-10-ex-10d\/","title":{"rendered":"TS Inter 1st Year Maths 1B Solutions Chapter 10 Applications of Derivatives Ex 10(d)"},"content":{"rendered":"

Students must practice this\u00a0TS Inter 1st Year Maths 1B Study Material<\/a> Chapter 10 Applications of Derivatives Ex 10(d) to find a better approach to solving the problems.<\/p>\n

TS Inter 1st Year Maths 1B Applications of Derivatives Solutions Exercise 10(d)<\/h2>\n

I. Find the angle between the curves given below. (March 2014) (E.Q.)<\/p>\n

Question 1.
\nx + y + 2 = 0, x2<\/sup> + y2<\/sup> – 10y = 0
\nAnswer:
\nx + y + 2 = 0 \u21d2 x = – (y + 2)
\nx2<\/sup> + y2<\/sup> – 10y = 0
\n\u21d2 (y + 2)2<\/sup> + y2<\/sup> – 10y = 0
\n\u21d2 y2<\/sup> + 4y + 4 + y2<\/sup> – 10y = 0
\n\u21d2 2y2<\/sup> – 6y + 4 = 0
\n\u21d2 y2<\/sup> – 3y + 2 = 0
\n\u21d2 (y – 2) (y – 1) = 0
\n\u21d2 y = 1 (or) y = 2
\nx = – (y + 2)
\ny = 1 \u21d2 x = -(1 + 2) = -3
\ny = 2 \u21d2 x = -(2 + 2) = -4
\nThe points of intersection at P (- 3, 1) and Q (- 4, 2)
\nEquation of curve is x + y – 10y = 0
\nDifferentiating w.r.to ‘x’
\n\"TS<\/p>\n

\"TS<\/p>\n

Question 2.
\ny2<\/sup> = 4x, x2<\/sup> + y2<\/sup> = 5. (E.Q.) (May 2007)
\nAnswer:
\nThe given equations of curves are y2<\/sup> = 4x and x2<\/sup> + y2<\/sup> = 5
\nEliminating y, we get x2<\/sup> + 4x = 5
\n\u21d2 x2<\/sup> + 4x – 5 = 0
\n\u21d2 (x – 1) (x + 5) = 0
\n\u21d2 x = 1 or – 5
\nFrom y2<\/sup> = 4x and x = 1
\n\u21d2 y2<\/sup> = 4 \u21d2 y = \u00b12
\nBut when x = -5, y is imaginary and not real
\nPoints of Intersection are
\nP(1, 2) and Q(1, -2)
\nFrom the equation y2<\/sup> = 4x
\n\u21d2 2y \\(\\frac{\\mathrm{dy}}{\\mathrm{dx}}\\) = 4
\n\u21d2 \\(\\frac{d y}{d x}=\\frac{z}{y}\\) …….(1)
\nAt P(1, 2), slope of first curve from (1)
\nm1<\/sub> = \\(\\frac{2}{2}\\) = 1
\nand m2<\/sub> = slope of second curve from (2)
\n= \\(\\frac{-1}{2}\\)
\n\"TS<\/p>\n

Question 3.
\nx2<\/sup> + 3y = 3, x2<\/sup> – y2<\/sup> + 25 = 0 (E.Q.)
\nAnswer:
\nx2<\/sup> + 3y = 3 ……….(1),
\nx2<\/sup> – y2<\/sup> + 25 = 0 ………….(2)
\nare the equations of given curves.
\nFrom (1) x2<\/sup> = 3 – 3y
\nFrom (2), 3 – 3y – y2<\/sup> + 25 = 0
\n\u21d2 – y2<\/sup> – 3y + 28 = 0
\n\u21d2 y2<\/sup> + 3y – 28 = 0
\n\u21d2 (y + 7)(y – 4) = 0
\n\u21d2 y = 4 or y = -7
\nIf y = 4 then x2<\/sup> = 3 – 12 = -9
\nValues of x are not real.
\nIf y = – 7 then x2<\/sup> = 3 + 21
\n\u21d2 x2<\/sup> = 24 = \u00b1 2\u221a6
\n\u2234 Points of intersection are
\nP (2\u221a6 , -7) and Q (-2\u221a6 , -7)
\nFrom (1), 2x + 3 \\(\\frac{d y}{d x}\\) = 0
\n\u21d2 \\(\\frac{d y}{d x}=\\frac{-2 x}{3}\\)
\nSlope of the tangent at P (2\u221a6 , -7) to the curve (1) is m1<\/sub> = \\(\\frac{-2(2 \\sqrt{6})}{3}=\\frac{-4 \\sqrt{6}}{3}\\)
\n\"TS<\/p>\n

\"TS<\/p>\n

Question 6.
\ny2<\/sup> = 8x, 4x2<\/sup> + y2<\/sup> = 32
\nAnswer:
\ny = 8x ….(1)
\n4x2<\/sup> + y2<\/sup> = 32
\n\u21d2 4x2<\/sup> + 8x – 32 = 0
\n\u21d2 x2<\/sup> + 2x – 8 = 0
\n\u21d2 (x + 4) (x – 2) = 0
\n\u21d2 x = 2 or x = – 4
\nWhen x = 2, y2<\/sup> = 16 \u21d2 y = \u00b1 4
\n\u2234 Points of intersection are P (2, 4) and Q (2, – 4)
\nFrom (1), 2y \\(\\frac{\\mathrm{dy}}{\\mathrm{dx}}\\) = 8 \u21d2 \\(\\frac{\\mathrm{dy}}{\\mathrm{dx}}=\\frac{4}{\\mathrm{y}}\\)
\n\"TS<\/p>\n

Question 7.
\nx2<\/sup>y = 4, y(x2<\/sup> + 4) = 8.
\nAnswer:
\nx2<\/sup>y = 4 ………………(1)
\ny(x2<\/sup> + 4) = 8 …………(2)
\nFrom (1), x2<\/sup> = \\(\\frac{4}{y}\\)<\/p>\n

Substitute x2<\/sup> value in equation (2)
\n\u2234 y(\\(\\frac{4}{y}\\)+4) = 8 \u21d2 y\\(\\left(\\frac{4+4 y}{y}\\right)\\) = 8
\n\u21d2 4 + 4y = 8 \u21d2 y = 1
\n\u2234 x2<\/sup> = \\(\\frac{4}{y}\\)
\n\u21d2 x2<\/sup> = 4
\n\u21d2 x = \u00b12
\n\u2234 Points of intersection are P(2, 1), Q(-2, 1)
\nFrom (1)
\n\"TS<\/p>\n

\"TS<\/p>\n

Question 8.
\nShow that the curves 6x2<\/sup> – 5x + 2y = 0 and 4x2<\/sup> + 8y2<\/sup> = 3 touch each other at \\(\\left(\\frac{1}{2}, \\frac{1}{2}\\right)\\). (E.Q)
\nAnswer:
\nThe given curves are 6x2<\/sup> – 5x + 2y = 0 ………..(1)
\nand 4x2<\/sup> + 8y2<\/sup> = 3 ………..(2)
\n\"TS
\nSlopes of two curves at \\(\\left(\\frac{1}{2}, \\frac{1}{2}\\right)\\) are equal and hence the given curves touch each other at \\(\\left(\\frac{1}{2}, \\frac{1}{2}\\right)\\).<\/p>\n","protected":false},"excerpt":{"rendered":"

Students must practice this\u00a0TS Inter 1st Year Maths 1B Study Material Chapter 10 Applications of Derivatives Ex 10(d) to find a better approach to solving the problems. TS Inter 1st Year Maths 1B Applications of Derivatives Solutions Exercise 10(d) I. Find the angle between the curves given below. (March 2014) (E.Q.) Question 1. x + … Read more<\/a><\/p>\n","protected":false},"author":2,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[2],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/3955"}],"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\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/comments?post=3955"}],"version-history":[{"count":4,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/3955\/revisions"}],"predecessor-version":[{"id":6126,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/3955\/revisions\/6126"}],"wp:attachment":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/media?parent=3955"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/categories?post=3955"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/tags?post=3955"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}