{"id":14057,"date":"2024-03-23T11:09:10","date_gmt":"2024-03-23T05:39:10","guid":{"rendered":"https:\/\/tsboardsolutions.in\/?p=14057"},"modified":"2024-03-25T17:12:21","modified_gmt":"2024-03-25T11:42:21","slug":"ts-inter-2nd-year-maths-2b-ellipse-important-questions","status":"publish","type":"post","link":"https:\/\/tsboardsolutions.in\/ts-inter-2nd-year-maths-2b-ellipse-important-questions\/","title":{"rendered":"TS Inter 2nd Year Maths 2B Ellipse Important Questions"},"content":{"rendered":"

Students must practice these\u00a0TS Inter 2nd Year Maths 2B Important Questions<\/a> Chapter 4 Ellipse to help strengthen their preparations for exams.<\/p>\n

TS Inter 2nd Year Maths 2B Ellipse Important Questions<\/h2>\n

Very Short Answer Type Questions<\/span><\/p>\n

Question 1.
\nIf the length of the latus rectum is equal to half of its minor axis of an ellipse in the standard form, then find the eccentricity of the ellipse.
\nSolution:
\nLet \\(\\frac{x^2}{a^2}+\\frac{y^2}{b^2}\\) + 1(a > b) be the ellipse in its standard form.
\nGiven that the length of the latus rectum = \\(\\frac{1}{2}\\) (minor axis)
\n\"TS<\/p>\n

\"TS<\/p>\n

Question 2.
\nThe orbit of the Earth is an ellipse with eccentricity with the \\(\\frac{1}{60}\\) Sun at one of its foci, the major axis being approximately 186 x 106<\/sup> miles in length. Find the shortest and longest distance of the Earth from the Sun.
\nSolution:
\nLet the earths orbit be an ellipse given by
\n\\(\\frac{x^2}{a^2}+\\frac{y^2}{b^2}\\) + 1(a > b)
\nSince the major axis is 186 x 106<\/sup> miles
\nwe have 2a = 186 x 106<\/sup> \u21d2 a = 93 x 106<\/sup> miles
\nIf e is the eccentricity of ellipse then e = \\(\\frac{1}{60}\\)
\nThe longest and shortest distances of the earth from the sun are respectively a + ae and a – ae.
\nHere the longest distance of earth from the sun = a +ae \\(\\left(1+\\frac{1}{60}\\right)\\)
\n= 9445 x 104 <\/sup>miles and shortest distance of earth from the sun = a – ae
\n= 93 x 104\u00a0<\/sup> \\(\\left(1-\\frac{1}{60}\\right)\\)
\n= 9145 x 104\u00a0<\/sup> miles<\/p>\n

Short Answer Type Questions<\/span><\/p>\n

Question 1.
\nFind the eccentricity, coordinates of foci, length of latus rectum and equations of directrices of the following ellipse
\n9x2<\/sup> + 16y2<\/sup>– 36x + 32y – 92 = 0
\nSolution:
\nGiven equation of ellipse is
\n9x2<\/sup> + 16y2<\/sup>– 36x + 32y – 92 = 0
\nwhich can be written as
\n\"TS<\/p>\n

\"TS<\/p>\n

(ii) 3x+ y2<\/sup> – 6x -2y -5 = 0
\nSolution:
\nGiven equation can be written as
\n3x2<\/sup> – 6x + y2<\/sup> – 2y = 5
\n\u21d2 3(x2<\/sup>– 2x) + (y2<\/sup>– 2y) = 5
\n\u21d2 3(x2 <\/sup>-2x + 1)+(y2<\/sup>-2y+1) = 5 + 4
\n\u21d2 3(x – 1)2<\/sup> + (y – 1)2 <\/sup>= 9
\n\u21d2 \\(\\frac{(x-1)^2}{3}+\\frac{(y-1)^2}{9}=1\\)
\nwhich is of the form
\n\\(\\frac{(\\mathrm{x}-\\mathrm{h})^2}{\\mathrm{a}^2}+\\frac{(\\mathrm{y}-\\mathrm{k})^2}{\\mathrm{~b}^2}=1\\)
\nwhich a2<\/sup> = 3 and b2<\/sup> = 9 and a < b.
\nAlso (h, k) = (1, 1); eccentrIcity
\n\"TS<\/p>\n

Question 2.
\nFind the equation of the ellipse referred to its major and minor axes as the coordinate axes X, Y – respectively with latus rectum of length 4 and distance between foci \\(4 \\sqrt{2}\\).
\nSolution:
\nLet the equation of ellipse be
\n\"TS<\/p>\n

\"TS<\/p>\n

Question 3.
\nC is the centre, A A\u2019 and B B\u2019 are major and minor axis of ellipse \\(\\frac{x^2}{a^2}+\\frac{y^2}{b^2}=1\\).If
\nPN is the ordinate of a point P on the ellipse then show that \\(\\frac{(\\mathrm{PN})^2}{(\\mathrm{AN})(\\mathrm{AN})}+\\frac{(\\mathrm{BC})^2}{(\\mathrm{CA})^2} \\)
\nSolution:
\nLet P(\u03b8) = (a cos \u03b8, b sin \u03b8) be any point on the ellipse \\(\\frac{x^2}{a^2}+\\frac{y^2}{b^2}=1\\)
\n\"TS<\/p>\n

Question 4.
\nS and T are the foci of an ellipse and B is one end of the minor axis. If STB is an equilateral triangle, then find the
\neccentricity of the ellipse.
\nSolution:
\nLet \\(\\frac{x^2}{a^2}+\\frac{y^2}{b^2}\\) = 1 :(a>b) be an ellipse whose foci are S and T. B is the end of minor axis such that STB is an equilateral triangle.
\nthan SB = ST = SB. Also S = (ae, 0).
\nT = (- ae. 0) and B = (0, b).
\n\"TS<\/p>\n

Question 5.
\nShow that among the points on the ellipse \\(\\frac{x^2}{a^2}+\\frac{y^2}{b^2}\\)=1,(a>b) ;(-a, 0) is the farthest point and (a, 0) is the nearest point form the focus (ae, 0).
\nSolution:
\nLet P(x, y) be any point on the ellipse so that < x < a and S = (ae, 0) is the focus. Since (x. y) is on the ellipse.
\n\"TS
\nMaximum value of SP is a + ae when P(-a.0)
\nand Minimum value of SP is a – ae when P (a. 0).
\nThe nearest point is (a, 0) and the farthest point is (-a, 0).<\/p>\n

\"TS<\/p>\n

Question 6.
\nFind the equation of tangent and normal to the ellipse 9x2<\/sup> + 16y2<\/sup> = 144 at the end of latus rectum in the first quadrant.
\nSolution:
\nGiven equation of ellipse is 9x2<\/sup> + 16y2<\/sup> = 144
\n\"TS
\n\"TS<\/p>\n

Question 7.
\nIf a tangent to the ellipse = \\(\\frac{x^2}{a^2}+\\frac{y^2}{b^2}=1,(a>b)\\) meets its major axis and minor axis at M and N respectively then prove that \\(\\frac{a^2}{(\\mathrm{CM})^2}+\\frac{b^2}{(\\mathrm{CN})^2}=1\\) where C is the centre of the ellipse.
\nSolution:
\n\"TS<\/p>\n

\"TS<\/p>\n

Question 8.
\nFind the condition for the line,
\n(i) lx + my + n = 0 to be a tangent to the ellipse \\(\\frac{x^2}{a^2}+\\frac{y^2}{b^2}=1\\)
\n(ii) lx + my + n = 0 to be a normal to the ellipse \\(\\frac{x^2}{a^2}+\\frac{y^2}{b^2}=1\\)
\nSolution:
\nLet lx + my + n = 0 be a tangent at
\nP (\u03b8) (a cos \u03b8 . b sin \u03b8) on the ellipse \\(\\frac{x^2}{a^2}+\\frac{y^2}{b^2}=1\\)
\n\"TS<\/p>\n

(ii) Let lx + myn = 0 be a normal to the ………………\u00a0 (2)
\nEllipse at the point P (\u03b8). Then equation of normal at ‘\u03b8\u2019 is
\n\"TS<\/p>\n

\"TS<\/p>\n

Question 9.
\nIf PN is the ordinate of a point P on the ellipse \\(\\frac{x^2}{a^2}+\\frac{y^2}{b^2}=1\\) and the tangent at P
\nmeets the x-axis at T then show that (CN) (CT) = a2<\/sup> where C is the centre of ellipse.
\nSolution:
\nLet P (\u03b8) = P (a cos \u03b8, b sin \u03b8) be a point on the ellipse \\(\\frac{x^2}{a^2}+\\frac{y^2}{b^2}=1\\) .
\nThen the equation of the tangent at p (\u03b8) is.
\n\"TS<\/p>\n

Question 10.
\nShow that the points of intersection of the perpendicular tangents to any ellipse lie on the circle.
\nSolution:
\nLet the equation of ellispe be \\(\\frac{x^2}{a^2}+\\frac{y^2}{b^2}=1\\) (a>b)
\nAny tangent to the above ellipse is of the form y = mx \u00b1 \\(\\pm \\sqrt{a^2 m^2+b^2}\\)
\nLet the perpendicular tangents intersect at
\n\"TS
\nThis being a quadratic is \u2018rn has two roots m1<\/sub> and m2<\/sub> which corresponds to the slopes of tangents drawn from P to ellipse then
\n\"TS
\n(\u2235 Product of slopes = – 1 for perpendicular tangents)
\n\u21d2 x1<\/sub>2<\/sup>+y1<\/sub>2 <\/sup>= a2<\/sup> +b2<\/sup>
\n\u2234 Locus of (x1<\/sub>, y1<\/sub>) is x2<\/sup> + y2<\/sup> = a2<\/sup>+ b2<\/sup> which is a circle.<\/p>\n

\"TS<\/p>\n

Long Answer Type Questions<\/span><\/p>\n

Question 1.
\nIf \u03b81<\/sub>, \u03b82<\/sub> are the eccentric angles of the extremeties of a focal chord (other than the verticles) of the ellipse \\(\\frac{x^2}{a^2}+\\frac{y^2}{b^2}=1\\) (a> b) and e is its eccentricity. Then show that
\n\"TS
\nSolution:
\n\"TS<\/p>\n

Let P(\u03b81<\/sub>), Q(\u03b82<\/sub>) be the two extremeties of a focal chord of the ellipse
\n\\(\\frac{x^2}{a^2}+\\frac{y^2}{b^2}=1,(a>b)\\)
\n\u2234 P = (acos \u03b81<\/sub>, b sin \u03b81<\/sub>),(\u03b81<\/sub> \u2260 0)
\nQ = (a cos \u03b82<\/sub>, b sin \u03b82<\/sub>), (\u03b82<\/sub> \u2260 \u03c0)
\nand focus S = (ae, 0). Now PQ is a focal chord and hence P, S. Q are collinear.
\n\u2234 Slope of \\(\\overline{\\mathrm{PS}}\\) = slope of \\(\\overline{\\mathrm{SQ}}\\)
\n\\(\\frac{b \\sin \\theta_1}{a\\left(\\cos \\theta_1-\\mathrm{e}\\right)}=\\frac{\\mathrm{b} \\sin \\theta_2}{\\mathrm{a}\\left(\\cos \\theta_2-\\mathrm{e}\\right)}\\)
\n\"TS<\/p>\n

Question 2.
\nIf the normal at one end of a latus rectum of the ellipse \\(\\frac{x^2}{a^2}+\\frac{y^2}{b^2}=1\\) passes through one end of the minor axis, then show that e4<\/sup> + e2<\/sup> = 1 (e is the eccentricity of the ellipse)
\nSolution:
\nLet L be the one end of the latus rectum of \\(\\frac{x^2}{a^2}+\\frac{y^2}{b^2}=1\\). Then the coordinates of
\n\\(L=\\left(a e, \\frac{b^2}{a}\\right)\\)
\n\u2234Equation of normal at L is
\n\"TS<\/p>\n

\"TS<\/p>\n

Question 3.
\nIf a circle is concentric with the ellipse, find the inclination of their common tangent to the major axis of the ellipse.
\nSolution:
\nLet the circle x2<\/sup> + y2<\/sup> = r2<\/sup> and the ellipse be \\(\\frac{x^2}{a^2}+\\frac{y^2}{b^2}=1\\) with a> b.
\nThe major axis of ellipse is X – axis.
\nIf r< b <a, then the circle lies completely in the ellipse making no common tangents.
\nIf b < a < r (ellipse lies completely in circle) no common tangent is passive.<\/p>\n

Case (i) : If b <r <a
\n\"TS
\nLet one of the common tangent make angle \u03b8\u00a0with positive X- axis and suppose the equation of tangent to the circle be x Cos \u03b1 + y sin \u03b1 = r where a is the angle made by the radius of circle with positive X – axis.
\n\u2234 \\(\\theta=\\frac{\\pi}{2}+\\alpha \\text { (or) } \\theta=\\alpha-\\frac{\\pi}{2}\\)
\nSince x cos \u03b1 + y sina r touches the ellipse also, we have a2<\/sup> cos2<\/sup>a + b2<\/sup> sin2<\/sup> = r2<\/sup><\/p>\n

\"TS<\/p>\n

Case (ii): When r = a the circle touches the ellipse at the ends of major axis of the ellipse so that the common tangents are x = \u00b1 a and \u03b8 = \\(\\frac{\\pi}{2}\\)<\/p>\n

\"TS<\/p>\n

Case (iii): When r = b, the circle touches the ellipse at the ends of minor axis of ellipse so that common tangents
\ny = \u00b1 b making \u03b8 = 0.<\/p>\n","protected":false},"excerpt":{"rendered":"

Students must practice these\u00a0TS Inter 2nd Year Maths 2B Important Questions Chapter 4 Ellipse to help strengthen their preparations for exams. TS Inter 2nd Year Maths 2B Ellipse Important Questions Very Short Answer Type Questions Question 1. If the length of the latus rectum is equal to half of its minor axis of an ellipse … 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":[3],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/14057"}],"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=14057"}],"version-history":[{"count":6,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/14057\/revisions"}],"predecessor-version":[{"id":14230,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/14057\/revisions\/14230"}],"wp:attachment":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/media?parent=14057"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/categories?post=14057"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/tags?post=14057"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}