{"id":35520,"date":"2022-09-19T17:20:22","date_gmt":"2022-09-19T11:50:22","guid":{"rendered":"https:\/\/apboardsolutions.guru\/?p=35520"},"modified":"2022-09-19T17:20:22","modified_gmt":"2022-09-19T11:50:22","slug":"inter-2nd-year-maths-2b-ellipse-solutions-ex-4a","status":"publish","type":"post","link":"https:\/\/apboardsolutions.guru\/inter-2nd-year-maths-2b-ellipse-solutions-ex-4a\/","title":{"rendered":"Inter 2nd Year Maths 2B Ellipse Solutions Ex 4(a)"},"content":{"rendered":"

Practicing the Intermediate 2nd Year Maths 2B Textbook Solutions<\/a> Inter 2nd Year Maths 2B Ellipse Solutions Exercise 4(a) will help students to clear their doubts quickly.<\/p>\n

Intermediate 2nd Year Maths 2B Ellipse Solutions Exercise 4(a)<\/h2>\n

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

Question 1.
\nFind the equation of the ellipse with focus at (1, -1) e = \\(\\frac{2}{3}\\) and directrix as x + y + 2 = 0.
\nSolution:
\nP(x1<\/sub>, y1<\/sub>) is any point on the ellipse. Equation of the directrix is
\nx + y + 2 = 0
\nDraw PM perpendicular to ZM, Join SP
\nBy Definition of ellipse SP = e . PM
\nSP\u00b2 = e\u00b2. PM\u00b2
\n\"Inter
\n9[(x1<\/sub> – 1)\u00b2 + (y1<\/sub> + 1)\u00b2] = 2[x1<\/sub> + y1<\/sub> + 2]\u00b2
\n9[x\u00b21<\/sub> – 2x1<\/sub> + 1 + y\u00b21<\/sub> + 2y1<\/sub> + 1] = 2[x\u00b21<\/sub> + y\u00b21<\/sub> + 4 + 2x1<\/sub>y1<\/sub> + 4x1<\/sub> + 4y1<\/sub>]
\n9x\u00b21<\/sub> +9y\u00b21<\/sub> – 18x1<\/sub> 18y1<\/sub> +18 = 2x\u00b21<\/sub> +2y\u00b21<\/sub> + 4x1<\/sub>y1<\/sub> + 8x1<\/sub> + 8y1<\/sub> + 8
\n7x\u00b21<\/sub> – 4x1<\/sub>y1<\/sub> + 7y\u00b21<\/sub> – 26x1<\/sub> + 10y1<\/sub> +10 = 0 focus of P (x1<\/sub>, y1<\/sub>) is
\n7x\u00b2 – 4xy + 7y\u00b2 – 26x + 10y + 10 = 0
\nThis is the equation of the required Ellipse.
\n\"Inter<\/p>\n

Question 2.
\nFind the equation of the ellipse in the standard form whose distance between foci is 2 and the length of latus rectum is \\(\\frac{15}{2}\\).
\nSolution:
\nLatus rectum = \\(\\frac{15}{2}\\) ;
\ndistance between foci = 2
\n\\(\\frac{2b^2}{2}\\) = \\(\\frac{15}{2}\\) ; 2ae = 2
\nae = 1
\n\u21d2 b\u00b2 = a\u00b2 – a\u00b2e\u00b2
\n\u21d2 b\u00b2 = a\u00b2 – 1
\n\u21d2 \\(\\frac{15}{2}\\)a = a\u00b2 – 1
\n\u21d2 4a\u00b2 – 15a – 4 = 0
\na = 4 or a = –\\(\\frac{1}{4}\\)
\nb\u00b2 = a\u00b2 – 1
\n= 16 – 1
\nEquation of the ellipse is \\(\\frac{x^{2}}{16}+\\frac{y^{2}}{15}\\) = 1<\/p>\n

\"Inter<\/p>\n

Question 3.
\nFind the equation of the ellipse in the standard form such that distance between foci is 8 and distance between directrices is 32.
\nSolution:
\nDistance between foci = 8.
\nDistance between directrices = 32.
\n2ae = 8, \\(\\frac{2a}{e}\\) = 32
\nae = 4, \\(\\frac{a}{e}\\) = 16
\n(ae)\\(\\frac{a}{e}\\) = 64
\na\u00b2 = 64
\nb\u00b2 = a\u00b2 – a\u00b2 e\u00b2
\n= 64 – 16 = 48
\nEquation of the ellipse is
\n\u2234 \\(\\frac{x^{2}}{64}+\\frac{y^{2}}{48}\\) = 1<\/p>\n

Question 4.
\nFind the eccentricity of the ellipse, (in standard form), if its length Solution:
\nLatus rectum = \\(\\frac{2b^2}{a}\\)
\nMajor axis = 2a
\nGiven \\(\\frac{2b^2}{a}\\) = \\(\\frac{1}{2}\\) .2a.
\n2b\u00b2 = a\u00b2
\nBut b\u00b2 = a\u00b2 (1 – e\u00b2)
\n2a\u00b2(1 – e\u00b2) = a\u00b2
\n1 – e\u00b2 = \\(\\frac{1}{2}\\)
\ne\u00b2 = \\(\\frac{1}{2}\\) \u21d2 e = \\(\\frac{1}{\\sqrt{2}}\\)<\/p>\n

\"Inter<\/p>\n

Question 5.
\nThe distance of a point on the ellipse x\u00b2 + 3y\u00b2 = 6 from its centre is equal to 2. Find the eccentric angles.
\nSolution:
\nEquation of the ellipse is x\u00b2 + 3y\u00b2 = 6
\n\\(\\frac{x^{2}}{6}+\\frac{y^{2}}{2}\\)
\na = \u221a6 , b = \u221a2
\nAny point on the ellipse is
\nP (\u221a6 cos \u03b8, \u221a2 sin \u03b8)
\nGiven CP = 2 \u21d2 CP\u00b2 = 4
\n6 cos\u00b2 \u03b8 + 2 sin\u00b2 \u03b8 = 4
\n6(1 – sin\u00b2 \u03b8 + 2 sin\u00b2 \u03b8 = 4
\n6 – 6 sin\u00b2 \u03b8 + 2 sin\u00b2 \u03b8 = 4
\n\"Inter<\/p>\n

Question 6.
\nFind the equation of ellipse in the standard form, if it passes through the points (-2, 2) and (3, -1).
\nSolution:
\nEquation of the ellipse in standard form is
\n\\(\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}\\) = 1
\npasses through-(-2, 2); (3, -1)
\n\\(\\frac{4}{a^{2}}+\\frac{4}{b^{2}}\\) = 1 ……… (i)
\n\\(\\frac{9}{a^{2}}+\\frac{1}{b^{2}}\\) = 1 ……….(ii)
\nSolving (i) and (ii) we get
\n\"Inter
\n3x\u00b2 + 5y\u00b2 = 32.<\/p>\n

Question 7.
\nIf the ends of major axis of an ellipse are (5, 0) and (-5, 0). Find the equation of the ellipse in the standard form if its focus lies on the line 3x – 5y – 9 = 0.
\nSolution:
\n(a, 0): (5, 0), (-a, 0) : (-5, 0)
\na = 5, b\u00b2 = a\u00b2 (1 – e\u00b2)
\nFocus lies on the line 3x – 5y – 9 = 0
\n3(ae) – 5 (0) – 9 = 0
\n3(5e) – 9 = 0
\n\"Inter
\nEquation of the ellipse is
\n\\(\\frac{x^{2}}{25}+\\frac{y^{2}}{16}\\) \u21d2 16x\u00b2 + 25y\u00b2 = 400<\/p>\n

\"Inter<\/p>\n

Question 8.
\nIf the length of the major axis of an ellipse is three times the length of its minor axis then find the eccentricity of the ellipse.
\nSolution:
\nMajor axis = 3 minor axes
\n2a = 3(2b) \u21d2 a = 3b
\na\u00b2 = 9b\u00b2 \u21d2 a\u00b2 = 9a\u00b2 (1 – e\u00b2)
\n(1 – e\u00b2) = \\(\\frac{1}{9}\\) \u21d2 e\u00b2 = 1 – \\(\\frac{1}{9}\\) = \\(\\frac{8}{9}\\)
\ne = \\(\\frac{2\\sqrt{2}}{3}\\)
\nEccentricity of the ellipse = \\(\\frac{2\\sqrt{2}}{3}\\)<\/p>\n

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

Question 1.
\nFind the length of major axis, minor axis, latus rectum, eccentricity, co-ordinates of centre, foci and the equations of directrices of the following ellipse.
\ni) 9x\u00b2 + 16y\u00b2 = 144
\nii) 4x\u00b2 + y\u00b2 – 8x + 2y + 1 =0
\niii) x\u00b2 + 2y\u00b2 – 4x + 12y + 14 = 0
\nSolution:
\nGiven equation is 9x\u00b2 + 16y\u00b2 = 144
\n\\(\\frac{x^{2}}{16}+\\frac{y^{2}}{9}\\) =1
\n\u2234 a = 4, b = 3
\nLength of major axis = 2a = 2. 4 = 8
\nLength of minor axis = 2b = 2. 3 = 6
\nLength of latus rectum = \\(\\frac{2b^{2}}{a}=\\frac{2.9}{4}=\\frac{9}{2}\\)
\nEccentricity = \\(\\sqrt{\\frac{a^2-b^2}{a^2}}=\\sqrt{\\frac{16-9}{16}}=\\frac{\\sqrt{7}}{4}\\)
\nCentre is C (0,0)
\nFoci are (\u00b1 ae, 0) = (\u00b1\u221a7, 0)
\nEquations of the directrices are x = \u00b1 \\(\\frac{a}{e}\\)
\nx = \u00b1 4.\\(\\frac{4}{\\sqrt{7}}\\) = \u00b1\\(\\frac{16}{\\sqrt{7}}\\)
\n\u221a7x = +16<\/p>\n

(ii) Given equation is 4x\u00b2 + y\u00b2 – 8x + 2y + 1 =0
\n4(x\u00b2 – 2x) + (y\u00b2 + 2y) = -1
\n4(x – 1)\u00b2 + (y + 1)\u00b2 = 4 + 1 – 1 = 4
\n\\(\\frac{(x-1)^{2}}{2}+\\frac{(y+1)^{2}}{2}\\)= 1
\nHence a < b \u21d2 y – axis is major axis
\na = 1, b = 2
\nLength of major axis = 2b = 4
\nLength of minor axis = 2a = 2
\nLength of latus rectum \\(\\frac{2a^{2}}{a}=\\frac{2}{2}=1\\)
\nEccentricity = \\(\\sqrt{\\frac{b^2-a^2}{b^2}}=\\sqrt{\\frac{4-1}{4}}=\\frac{\\sqrt{3}}{2}\\)
\nCentre is C (-1, -1)
\nbe = 2.\\(\\frac{\\sqrt{3}}{2}\\) = \u221a3
\nFoci are (-1, 1 \u00b1\u221a3)
\nEquations of the directrices are y + 1 = \u00b1 \\(\\frac{b}{e}\\)
\n= \u00b1 \\(\\frac{4}{\\sqrt{3}}\\)
\n\u221a3y + \u221a3 = \u00b14
\n\u221a3y + \u221a3 \u00b1 4 = o<\/p>\n

(iii) Given equation is x\u00b2 + 2y\u00b2 – 4x+ 12y + 14 = 0
\nx\u00b2 – 4x + 2 (y\u00b2 + 6y) = – 14
\n\u21d2 (x\u00b2 – 4x + 4) + 2(y\u00b2 + 6y + 9) = 4 + 18 – 14
\n\u21d2 (x – 2)\u00b2 + 2(y + 3)\u00b2 = 8
\n\"Inter
\na = 2\u221a2, b = 2, h = 2, k =-3
\nLength of major axis = 2a = 2(2 \u221a2) = 4 \u221a2
\nLength of minor axis = 2b = 2(2) = 4
\nLength of latus rectum
\n\"Inter
\nCentre = (h, k) = (2, – 3)
\nFoci = (h \u00b1 ae, k) = (2 \u00b1 2, -3)
\n= (4, -3), (0, – 3)
\nEquations of the directrices are x- h = \u00b1 \\(\\frac{a}{e}\\)
\nx – 2 = \u00b1\\(\\frac{2 \\sqrt{2}}{\\left(\\frac{1}{\\sqrt{2}}\\right)}\\) x – 2 = \u00b14
\ni.e., x = 6, x = -2.<\/p>\n

\"Inter<\/p>\n

Question 2.
\nFind the equation of the ellipse in the form \\(\\frac{(x-h)^2}{a^2}+\\frac{(y-k)^2}{b^2}\\) = 1, given the following data.
\ni) Centre (2, -1), one end of major axis (2, – 5), e = \\(\\frac{1}{3}\\).
\nSolution:
\nCentre (2, -1) \u21d2 h = 2, k = -1
\nend of major axis (2, -5)
\n\"Inter
\nEquation of the ellipse is
\n\"Inter
\n9(x – 2)\u00b2 + 8(y + 1)\u00b2 = 128
\ni.e., 9 (x – 2)\u00b2 + 8 (y + 1)\u00b2 = 128.<\/p>\n

ii) Centre (4, -1), one end of major axis is (-1, -1) and passing through (8, 0).
\nSolution:
\na = \\(\\sqrt{(4+1)^2+(-1+1)^2}\\)
\na = 5
\npasses through (8, 0)
\n\"Inter
\nRequired equation be
\n\\(\\frac{(x-4)^2}{25}\\) + \\(\\frac{9}{25}\\)(y + 1)\u00b2 = 1
\n\u21d2 (x – 4)\u00b2 + 9(y + 1)\u00b2 = 25<\/p>\n

iii) Centre (0, -3), e = \\(\\frac{2}{3}\\), semi-minor axis = 5.
\nSolution:
\nb = 5
\n\u21d2 b\u00b2 = a\u00b2 – a\u00b2e\u00b2
\n\u21d2 25 = a\u00b2 – a\u00b2. \\(\\frac{4}{9}\\) = a\u00b2\\(\\frac{5}{9}\\)
\n\u21d2 45 = a\u00b2
\n\"Inter<\/p>\n

iv) Centre (2, -1); e = \\(\\frac{1}{2}\\); length of latus rectum 4.
\nSolution:
\nEquation of the ellipse is
\n\\(\\frac{9(x-2)^2}{64}+\\frac{3(y+1)^2}{16}\\) = 1
\n9(x – 2)\u00b2 + 12(y + 1)\u00b2 = 64.<\/p>\n

Question 3.
\nFind the radius of the circle passing through the foci of an ellipse 9x\u00b2 + 16y\u00b2 = 144 and having least radius.
\nSolution:
\nEquation of the ellipse 9x\u00b2 + 16y\u00b2 = 144.
\n\"Inter
\n\\(\\frac{x^2}{16}+\\frac{y^2}{9}\\) = 1
\na\u00b2 = 16, b\u00b2 = 9
\na = 4, b = 3
\nThe circle passes through S and S\u00b9 and has least radius.
\nIf S S\u00b9 is diameter.
\na\u00b2e\u00b2 = a\u00b2 – a\u00b2 (1 – e\u00b2) = a\u00b2 – b\u00b2 = 16 – 9 = 7
\nEquation of the required circle is x\u00b2 + y\u00b2 = 7.<\/p>\n

\"Inter<\/p>\n

Question 4.
\nA man running on a race course notices that the sum of the distances of the two flag posts from him is always 10m and the distance between the flag posts is 8m. Find the equation of the race course traced by the man.
\nSolution:
\nS and S\u00b9 are the flags and P is the position of the man.
\n\"Inter
\nGiven SP + S\u00b9P = 10 and SS\u00b9 = 8
\nThe path traced by the man is an ellipse whose foci are S and S\u00b9.
\n2a = 10 \u21d2 a = 5
\nSS\u00b9 = 8 \u21d2 2ae = 8 \u21d2 ae = 4
\nb\u00b2 = a\u00b2 (1 – e\u00b2) = 25 (1 – \\(\\frac{16}{25}\\)) = 9
\nEquation of the ellipse is \\(\\frac{x^2}{a^2}+\\frac{y^2}{b^2}\\) = 1
\n\\(\\frac{x^2}{25}+\\frac{y^2}{9}\\) = 1.<\/p>\n

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

Question 1.
\nA line of fixed length (a + b) moves so that its ends are always on two perpendicular straight lines fixed. Prove that a marked point on the line, which divides this line into portions of lengths ‘a’ and ‘b’ describes an ellipse and also find the eccentricity of the ellipse when a = 8, b = 12.
\nSolution:
\nTake the perpendicular lines as co-ordinates axes. AB is the fixed line. Let OA = \u03b1, and OB = \u03b2 so that equation of AB is \\(\\frac{x}{\\alpha}+\\frac{y}{\\beta}\\) = 1
\nWhere \u03b1\u00b2 + \u03b2\u00b2 = (a + b)\u00b2 ………….. (1)
\nP(x, y) divides AB in the ratio a = b
\nCo-ordinates of P are (\\(\\frac{b\\alpha}{a+b}+\\frac{a\\beta}{a+b}\\)) = (x, y)
\n\"Inter
\nSubstituing the values of \u03b1, \u03b2 in (1) we get,
\n\"Inter
\nP describes an ellipse.
\nGiven a = 8, b = 12, so that b > a
\n\"Inter<\/p>\n

\"Inter<\/p>\n

Question 2.
\nProve that the equation of the chord joining the points \u03b1, \u03b2 on the ellipse \\(\\frac{x^2}{a^2}+\\frac{y^2}{b^2}\\)
\n\"Inter
\nSolution:
\nThe given points on the ellipse are
\nP(a cos \u03b1, b sin \u03b1) and Q (a cos \u03b2, b sin \u03b2).
\n\"Inter
\nEquation of the chord PQ is
\n\"Inter<\/p>\n","protected":false},"excerpt":{"rendered":"

Practicing the Intermediate 2nd Year Maths 2B Textbook Solutions Inter 2nd Year Maths 2B Ellipse Solutions Exercise 4(a) will help students to clear their doubts quickly. Intermediate 2nd Year Maths 2B Ellipse Solutions Exercise 4(a) I. Question 1. Find the equation of the ellipse with focus at (1, -1) e = and directrix as x …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[14],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/35520"}],"collection":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/comments?post=35520"}],"version-history":[{"count":1,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/35520\/revisions"}],"predecessor-version":[{"id":35521,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/35520\/revisions\/35521"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/media?parent=35520"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/categories?post=35520"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/tags?post=35520"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}