{"id":8365,"date":"2024-02-12T09:51:43","date_gmt":"2024-02-12T04:21:43","guid":{"rendered":"https:\/\/tsboardsolutions.in\/?p=8365"},"modified":"2024-02-13T17:46:35","modified_gmt":"2024-02-13T12:16:35","slug":"ts-inter-2nd-year-maths-2b-solutions-chapter-4-ex-4b","status":"publish","type":"post","link":"https:\/\/tsboardsolutions.in\/ts-inter-2nd-year-maths-2b-solutions-chapter-4-ex-4b\/","title":{"rendered":"TS Inter 2nd Year Maths 2B Solutions Chapter 4 Ellipse Ex 4(b)"},"content":{"rendered":"

Students must practice this TS Intermediate Maths 2B Solutions<\/a> Chapter 4 Ellipse Ex 4(b) to find a better approach to solving the problems.<\/p>\n

TS Inter 2nd Year Maths 2B Solutions Chapter 4 Ellipse Exercise 4(b)<\/h2>\n

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

Question 1.
\nFind the equation of the tangent and normal to the ellipse x2<\/sup> + 8y2<\/sup> = 33 at (-1, 2).
\nSolution:
\nGiven ellipse is x2<\/sup> + 8y2<\/sup> = 33
\n\u21d2 \\(\\frac{x^2}{33}+\\frac{y^2}{(33 \/ 8)}=1\\) ……..(1)
\nLet P(x1<\/sub>, y1<\/sub>) = (-1, 2) be a point on (1) then
\nthe equation of tangent at (x1<\/sub>, y1<\/sub>) is \\(\\frac{\\mathrm{xx}_1}{33}+\\frac{\\mathrm{yy}_1}{(33 \/ 8)}-1=0\\)
\n\\(\\frac{x(-1)}{33}+\\frac{y(2)}{(33 \/ 8)}=1\\)
\n\u21d2 -x + 16y = -33
\n\u21d2 x – 16y + 33 = 0
\nEquation of normal at P(-1, 2) on the ellipse
\n\"TS<\/p>\n

Question 2.
\nFind the equation of tangent and normal to the ellipse x2<\/sup> + 2y2<\/sup> – 4x + 12y + 14 = 0 at (2, -1).
\nSolution:
\nGiven equation of ellipse is x2<\/sup> + 2y2<\/sup> – 4x + 12y + 14 = 0
\nLet P(x1<\/sub>, y1<\/sub>) = (2, -1)
\nNow differentiating w.r.t ‘x’
\n\"TS
\n\u2234 Slope of the tangent at (2, -1) is \\(\\frac{d y}{d x}(2,-1)=\\frac{2-2}{-2+6}=0\\)
\n\u2234 The slope of the normal at (2, -1) is \u221e.
\n\u2234 Equation of the tangent to the given ellipse at P(2, -1) is y + 1 = 0, and the equation of normal at P(2, -1) is x – 2 = 0.<\/p>\n

\"TS<\/p>\n

Question 3.
\nFind the equation of the tangents to 9x2<\/sup> + 16y2<\/sup> = 144, which makes equal intercepts on the coordinate axis.
\nSolution:
\nGiven the equation of the ellipse is 9x2<\/sup> + 16y2<\/sup> = 144
\n\u21d2 \\(\\frac{x^2}{16}+\\frac{y^2}{9}=1\\)
\nLet S \u2261 \\(\\frac{x^2}{16}+\\frac{y^2}{9}-1=0\\)
\nCompared with the general equation S = 0 we have
\na2<\/sup> = 16, b2<\/sup> = 9
\n\u21d2 a = 4, b = 3
\n\u2234 Equation of the tangent to the ellipse S = 0 having slope is y = mx \u00b1 \\(\\sqrt{a^2 m^2+b^2}\\)
\n\"TS<\/p>\n

Question 4.
\nFind the coordinates of the points on the ellipse x2<\/sup> + 3y2<\/sup> = 37 at which the normal is parallel to the line 6x – 5y = 2.
\nSolution:
\nGiven equation of ellipse is x2<\/sup> + 3y2<\/sup> = 37 …….(1)
\n\u21d2 \\(\\frac{x^2}{37}+\\frac{y^2}{\\left(\\frac{37}{3}\\right)}=1\\)
\nLet P(x1<\/sub>, y1<\/sub>) be any point on the ellipse (1) then \\(\\frac{x_1^2}{37}+\\frac{y_1^2}{\\left(\\frac{37}{3}\\right)}=1\\)
\n\u21d2 \\(x^2+3 y_1^2=37\\)
\nGiven line is 6x – 5y + 2 = 0 …….(2)
\nSlope of the line = \\(\\frac{6}{5}\\)
\n\u2234 Equation of the normal at P(x1<\/sub>, y1<\/sub>) to the ellipse S = 0 is
\n\"TS
\n\"TS
\n\u2234 The coordinates of the points on ellipse x2<\/sup> + 7y2<\/sup> = 37 at which the normal is parallel to the line 6x – 5y = 2 are (5, 2), (-5, -2).<\/p>\n

Question 5.
\nFind the value of k if 4x + y + k = 0 is a tangent to the ellipse x2<\/sup> + 3y2<\/sup> = 3.
\nSolution:
\nGiven ellipse is x2<\/sup> + 3y2<\/sup> = 3
\n\u21d2 \\(\\frac{x^2}{3}+\\frac{y^2}{1}=1\\)
\nHence a2<\/sup> = 3 and b2<\/sup> = 1
\nThe equation of the given line is 4x + y + k = 0.
\n\u21d2 y = -4x – k where m = -4, and c = -k
\nThe condition for tangency is c2<\/sup> = a2<\/sup>m2<\/sup> + b2<\/sup>
\n\u21d2 k2<\/sup> = 3(16) + 1
\n\u21d2 k2<\/sup> = 49
\n\u21d2 k = \u00b17<\/p>\n

\"TS<\/p>\n

Question 6.
\nFind the condition for the line x cos \u03b1 + y sin \u03b2 = p to be a tangent to the ellipse \\(\\frac{x^2}{a^2}+\\frac{y^2}{b^2}=1\\).
\nSolution:
\nGiven equation of ellipse is \\(\\frac{x^2}{a^2}+\\frac{y^2}{b^2}=1\\) ……..(1)
\nand equation of given line is x cos \u03b1 + y sin \u03b1 = p
\n\u21d2 y sin \u03b1 = p – x cos \u03b1
\n\u21d2 y = -x cot \u03b1 + p cosec \u03b1 …….(2)
\nwhich is of the form y = mx + c where m = -cot \u03b1 and c = p cosec \u03b1
\nCondition for tangency is c2<\/sup> = a2<\/sup>m2<\/sup> + b2<\/sup>
\n\u21d2 p2<\/sup> cosec2<\/sup>\u03b1 = a2<\/sup> cot2<\/sup>\u03b1 + b2<\/sup>
\n\u21d2 p2<\/sup> = \\(a^2 \\frac{\\cot ^2 \\alpha}{{cosec}^2 \\alpha}+\\frac{b^2}{{cosec}^2 \\alpha}\\)
\n\u21d2 p2<\/sup> = a2<\/sup> cos2<\/sup>\u03b1 + b2<\/sup> sin2<\/sup>\u03b1 which is the required condition.<\/p>\n

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

Question 1.
\nFind the equations of tangent and normal to the ellipse 2x2<\/sup> + 3y2<\/sup> = 11 at the point whose ordinate is 1.
\nSolution:
\nLet P(x1<\/sub>, y1<\/sub>) be a given point and given y1<\/sub> = 1 and (x1<\/sub>, y1<\/sub>) lies on 2x2<\/sup> + 3y2<\/sup> = 11.
\n\u21d2 \\(2 \\mathrm{x}_1^2+3 \\mathrm{y}_1{ }^2=11\\)
\n\u21d2 \\(2 \\mathrm{x}_1^2+3=11\\)
\n\u21d2 2\\(\\mathrm{x}_1^2\\) = 8
\n\u21d2 \\(\\mathrm{x}_1^2\\) = 4
\n\u21d2 x1<\/sub> = \u00b12
\nThe required points are (2, 1) and (-2, 1) at which the equations of tangent and normal are to be determined.
\nThe equation of tangent at (x1<\/sub>, y1<\/sub>) to the ellipse 2x2<\/sup> + 3y2<\/sup> = 11
\n\u21d2 \\(\\frac{x^2}{\\left(\\frac{11}{2}\\right)}+\\frac{y^2}{\\left(\\frac{11}{3}\\right)}=1\\) is \\(\\frac{\\mathrm{xx}}{\\left(\\frac{11}{2}\\right)}+\\frac{\\mathrm{yy}}{\\left(\\frac{11}{3}\\right)}=1\\)
\n\"TS
\n\u21d2 -3x – 4y = 2
\n\u21d2 3x + 4y + 2 = 0 ……..(4)
\n\u2234 Tangents are 4x + 3y – 11 = 0 and 4x – 3y – 11 = 0
\nAlso the normals are 3x – 4y – 2 = 0 and 3x + 4y + 2 = 0.<\/p>\n

Question 2.
\nFind the equations to the tangents to the ellipse x2<\/sup> + 2y2<\/sup> = 3 drawn from the point (1, 2) and also find the angle between these tangents.
\nSolution:
\nGiven the equation of the ellipse is x2<\/sup> + 2y2<\/sup> = 3
\n\u21d2 \\(\\frac{x^2}{3}+\\frac{y^2}{\\left(\\frac{3}{2}\\right)}=1\\) which is of the form \\(\\frac{x^2}{a^2}+\\frac{y^2}{b^2}=1\\) where a2<\/sup> = 3 and b2<\/sup> = \\(\\frac{3}{2}\\)
\nEquation of any tangent to the ellipse is y = mx \u00b1 \\(\\sqrt{a^2 m^2+b^2}\\)
\nIf the tangents are drawn from (1, 2) then
\n2 = m \u00b1 \\(\\sqrt{3 m^2+3 \/ 2}\\)
\n\u21d2 2 – m = \\(\\pm \\sqrt{3 m^2+3 \/ 2}\\)
\n\u21d2 m2<\/sup> – 4m + 4 = 3m2<\/sup> + \\(\\frac{3}{2}\\)
\n\u21d2 2m2<\/sup> – 8m + 8 = 6m2<\/sup> + 3
\n\u21d2 4m2<\/sup> + 8m – 5 = 0
\n\u21d2 4m2<\/sup> + 10m – 2m – 5 = 0
\n\u21d2 2m(2m + 5) – 1(2m + 5) = 0
\n\u21d2 2m – 1 = 0 (or) 2m + 5 = 0
\n\u21d2 m = \\(\\frac{1}{2}\\) (or) m = \\(-\\frac{5}{2}\\)
\nThe equation of tangents when m = \\(\\frac{1}{2}\\) is
\n\"TS
\n\u2234 Equations of tangents drawn from the point (1, 2) to the ellipse S = 0 are given by x – 2y + 3 = 0 and 5x + 2y – 9 = 0.
\nAlso, the angle between them is tan-1<\/sup>(12).<\/p>\n

\"TS<\/p>\n

Question 3.
\nFind the equation of the tangents to the ellipse 2x2<\/sup> + y2<\/sup> = 8 which are
\n(i) parallel to x – 2y – 4 = 0
\n(ii) perpendicular to x + y + 2 = 0
\n(iii) which makes an angle \\(\\frac{\\pi}{4}\\) with x-axis.
\nSolution:
\nGiven the equation of the ellipse is 2x2<\/sup> + y2<\/sup> = 8
\n\u21d2 \\(\\frac{x^2}{4}+\\frac{y^2}{8}=1\\)
\nLet S \u2261 \\(\\frac{x^2}{4}+\\frac{y^2}{8}-1=0\\) ……..(1)
\nand compare with general equation a2<\/sup> = 4, and b2<\/sup> = 8
\n\u21d2 a = 2, b = 2\u221a2
\n(i) Parallel to x – 2y – 4 = 0:
\nGiven line is x – 2y – 4 = 0 ……..(2)
\nEquation of any line parallel to x – 2y – 4 = 0 is x – 2y + k = 0 ………(3)
\n\u21d2 2y = x + k
\n\u21d2 y = \\(\\frac{\\mathrm{x}}{2}+\\frac{\\mathrm{k}}{2}\\) where m = \\(\\frac{1}{2}\\) and c = \\(\\frac{k}{2}\\)
\nIf (3) is a tangent to (1) them c2<\/sup> = a2<\/sup>m2<\/sup> + b2<\/sup>
\n\u21d2 \\(\\frac{\\mathrm{k}^2}{4}=4\\left(\\frac{1}{4}\\right)+8\\)
\n\u21d2 \\(\\frac{\\mathrm{k}^2}{4}\\) = 1 + 8
\n\u21d2 k2<\/sup> = 36
\n\u21d2 k = \u00b16
\n\u2234 The equation of the required tangent from (3) is x – 2y \u00b1 6 = 0.<\/p>\n

(ii) Perpendicular to x + y – 2 = 0:
\nGiven line is x + y – 2 = 0 ………(4)
\nEquation of any line perpendicular to (4) is x – y + k = 0 ……….(5)
\n\u2234 y = x + k and m = 1, c = k
\nBy the condition for tangency c2<\/sup> = a2<\/sup>m2<\/sup> + b2<\/sup>
\n\u21d2 k2<\/sup> = 4(1) + 8
\n\u21d2 k2<\/sup> = 12
\n\u21d2 k = \u00b12\u221a 3
\n\u2234 Equation of the required line from (5) is x – y \u00b1 2\u221a3 = 0 ………..(6)<\/p>\n

(iii) Which makes an angle \\(\\frac{\\pi}{4}\\) with x-axis:
\nIf the line makes \\(\\frac{\\pi}{4}\\) with x-axis then m = tan\\(\\frac{\\pi}{4}\\) = 1.
\n\u2234 Equation of the line is y = x + c ………..(7)
\nIf this is a tangent to (1) then c2<\/sup> = a2<\/sup>m2<\/sup> + b2<\/sup>
\n\u21d2 c2<\/sup> = 4(1) + 8
\n\u21d2 c2<\/sup> = 12
\n\u21d2 c = \u00b12\u221a3
\n\u2234 From (7), the equation of the required line is y = x \u00b1 2\u221a3.<\/p>\n

\"TS<\/p>\n

Question 4.
\nA circle of radius 4, is concentric with the ellipse 3x2<\/sup> + 13y2<\/sup> = 78. Prove that a common tangent is inclined to the major axis at an angle \\(\\frac{\\pi}{4}\\).
\nSolution:
\nGiven ellipse is 3x2<\/sup> + 13y2<\/sup> = 78
\n\u21d2 \\(\\frac{x^2}{26}+\\frac{y^2}{6}=1\\)
\nComparing with \\(\\frac{x^2}{a^2}+\\frac{y^2}{b^2}=1\\) we have a2<\/sup> = 26, b2<\/sup> = 6,
\ncentre of the ellipse = (0, 0).
\nThe equation of a circle with centre (0, 0) and radius 4 is x2<\/sup> + y2<\/sup> = 16 ……….(2)
\nEquation of any tangent to the ellipse having slope ‘m’ is y = mx \u00b1 \\(\\sqrt{a^2 m^2+b^2}\\)
\n\u21d2 mx – y + \\(\\sqrt{a^2 m^2+b^2}\\) = 0 (Taking one tangent as common)
\n\u21d2 mx – y + \\(\\sqrt{26 m^2+6}\\) = 0 ………(3)
\nIf (3) is a tangent to (2) then the perpendicular distance from (0, 0) to (3) = Radius of the circle (2)
\n\u2234 \\(\\frac{\\sqrt{26 m^2+6}}{\\sqrt{m^2+1}}\\)
\n\u21d2 26m2<\/sup> + 6 = 16(m2<\/sup> + 1)
\n\u21d2 10m2<\/sup> – 10 = 0
\n\u21d2 m2<\/sup> = 1
\n\u21d2 m = \u00b11
\nIf \u03b8 is the angle made by the common tangent with the major axis of the ellipse then tan \u03b8 = \u00b11
\n\u21d2 \u03b8 = \u00b1\\(\\frac{\\pi}{4}\\)
\nHence common tangent makes an angle \\(\\frac{\\pi}{4}\\) with the major axis of the ellipse.<\/p>\n

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

Question 1.
\nShow that the foot of the perpendicular drawn from the centre on any tangent to the ellipse lies on the curve (x2<\/sup> + y2<\/sup>)2<\/sup> = a2<\/sup>x2<\/sup> + by2<\/sup>.
\nSolution:
\nLet S \u2261 \\(\\frac{x^2}{a^2}+\\frac{y^2}{b^2}-1=0\\) be the equation of ellipse.
\nLet P(x1<\/sub>, y1<\/sub>) be any point on the ellipse.
\nThe equation of the tangent to the ellipse S = 0 having slope m is y = mx \u00b1 \\(\\sqrt{a^2 m^2+b^2}\\)
\n\u21d2 y – mx = \u00b1\\(\\sqrt{a^2 m^2+b^2}\\) ……..(1)
\nThe equation to the perpendicular from centre on this tangent (1) is y – 0 = \\(\\frac{-1}{m}\\)(x – 0)
\n\u21d2 my + x = 0 ……….(2)
\nNow P(x1<\/sub>, y1<\/sub>) is the point of intersection of (1) and (2)
\n\u2234 y1<\/sub> – mx1<\/sub> = \u00b1\\(\\sqrt{a^2 m^2+b^2}\\) ……(3) and my1<\/sub> + x1<\/sub> = 0 ………(4)
\nSquaring (3) and (4) and adding
\n\"TS<\/p>\n

Question 2.
\nShow that the locus of the feet of the perpendiculars drawn from foci to any tangent of the ellipse is the auxiliary circle.
\nSolution:
\nLet S \u2261 \\(\\frac{x^2}{a^2}+\\frac{y^2}{b^2}-1=0\\) be the equation of ellipse.
\n\"TS
\nLet P(x1<\/sub>, y1<\/sub>) be any point on the locus.
\nThe equation of tangent to the ellipse S = 0 having slope ‘m’ is y = mx \u00b1 \\(\\sqrt{a^2 m^2+b^2}\\)
\n\u21d2 y – mx = \u00b1\\(\\sqrt{a^2 m^2+b^2}\\) ……..(1)
\nThe equation to the perpendicular from either focus (\u00b1ae, 0) on the tangent (1) is
\ny – 0 = \\(\\frac{-1}{m}\\)(x \u00b1 ae)
\n\u21d2 my + x = \u00b1ae ………(2)
\nSince P(x1<\/sub>, y1<\/sub>) is a point of intersection of (1) and (2) we have
\ny1<\/sub> – mx1<\/sub> = \u00b1\\(\\sqrt{a^2 m^2+b^2}\\) ……..(3)
\nand my1<\/sub> + x1<\/sub> = \u00b1ae ……….(4)
\nEliminating in from the equation by squaring and adding (3) and (4)
\n(y1<\/sub> – mx1<\/sub>)2<\/sup> + (my1<\/sub> + x1<\/sub>)2<\/sup> = a2<\/sup>m2<\/sup> + b2<\/sup> + a2<\/sup>e2<\/sup>
\n= a2<\/sup>m2<\/sup> + a2<\/sup> (1 – e2<\/sup>) + a2<\/sup>e2<\/sup>
\n= a2<\/sup>m2<\/sup> + a2<\/sup>
\n= a2<\/sup>(m2<\/sup> + 1)
\n\u2234 \\(\\mathrm{y}_1^2\\left(1+\\mathrm{m}^2\\right)+\\mathrm{x}_1^2\\left(1+\\mathrm{m}^2\\right)=\\mathrm{a}^2\\left(1+\\mathrm{m}^2\\right)\\)
\n\u21d2 \\(x_1^2+y_1{ }^2=a^2\\)
\n\u2234 The Locus of P(x1<\/sub>, y1<\/sub>) is x2<\/sup> + y2<\/sup> = a2<\/sup> which is the equation of the auxiliary circle of the ellipse.<\/p>\n

\"TS<\/p>\n

Question 3.
\nThe tangent and normal to the ellipse x2<\/sup> + 4y2<\/sup> = 4 at a point P(\u03b8) on it meets the major axis in Q and R respectively. If 0 < \u03b8 < \\(\\frac{\\pi}{2}\\) and QR = 2 then show that \u03b8 = \\(\\cos ^{-1}\\left(\\frac{2}{3}\\right)\\). (March 2012)
\nSolution:
\nGiven ellipse is x2<\/sup> + 4y2<\/sup> = 4
\n\u21d2 \\(\\frac{x^2}{4}+\\frac{y^2}{1}=1\\)
\n\"TS
\nLet S \u2261 \\(\\frac{x^2}{4}+\\frac{y^2}{1}-1=0\\) be the given ellipse.
\nComparing this with \\(\\frac{x^2}{a^2}+\\frac{y^2}{b^2}-1=0\\) we get a2<\/sup> = 4, b2<\/sup> = 1
\n\u21d2 a = 2, b = 1
\nThe equation of the tangent at P(\u03b8) on the ellipse S = 0 is \\(\\frac{x}{a} \\cos \\theta+\\frac{y}{b} \\sin \\theta=1\\)
\n\u21d2 \\(\\frac{\\mathrm{x}}{2} \\cos \\theta+\\frac{\\mathrm{y}}{1} \\sin \\theta=1\\) ……..(1)
\nThe tangent (1) meets the major axis at the Q
\n\u2234 y-coordinate of Q = 0; Put y = 0 in (1)
\n\"TS
\n\"TS
\n\u21d2 -3 cos2<\/sup>\u03b8 + 4 = 4 cos\u03b8
\n\u21d2 3 cos2<\/sup>\u03b8 + 4 cos \u03b8 – 4 = 0
\n\u21d2 3 cos2<\/sup>\u03b8 + 6 cos \u03b8 – 2 cos \u03b8 – 4 = 0
\n\u21d2 3 cos \u03b8 (cos \u03b8 + 2) – 2(cos \u03b8 + 2) = 0
\n\u21d2 (cos \u03b8 + 2) (3 cos \u03b8 – 2) = 0
\n\u21d2 cos \u03b8 + 2 = 0; solution is not admissive (\u2235 -1 \u2264 cos \u03b8 \u2264 1)
\nand 3 cos \u03b8 – 2 = 0
\n\u21d2 cos \u03b8 = \\(\\frac{2}{3}\\)
\n\u21d2 \u03b8 = \\(\\cos ^{-1}\\left(\\frac{2}{3}\\right)\\)<\/p>\n","protected":false},"excerpt":{"rendered":"

Students must practice this TS Intermediate Maths 2B Solutions Chapter 4 Ellipse Ex 4(b) to find a better approach to solving the problems. TS Inter 2nd Year Maths 2B Solutions Chapter 4 Ellipse Exercise 4(b) I. Question 1. Find the equation of the tangent and normal to the ellipse x2 + 8y2 = 33 at … 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\/8365"}],"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=8365"}],"version-history":[{"count":5,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/8365\/revisions"}],"predecessor-version":[{"id":8414,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/8365\/revisions\/8414"}],"wp:attachment":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/media?parent=8365"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/categories?post=8365"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/tags?post=8365"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}