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Q.Find the eccentricity, coordinates of foci, length of latus rectum and equations of directrices of the following ellipse : 9x2+16y2−36x+32y−92=09x^2 + 16y^2 - 36x + 32y - 92 = 0.

Telangana TsbieTelangana Board of Intermediate Education 2023Subjective· 4mImportance★★★★★
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In standard form the ellipse is (x−2)216+(y+1)29=1\dfrac{(x-2)^2}{16}+\dfrac{(y+1)^2}{9}=1, giving e=74e=\tfrac{\sqrt7}{4}, foci (2±7,−1)(2\pm\sqrt7,-1), latus rectum 92\tfrac92, directrices x=2±167x=2\pm\tfrac{16}{\sqrt7}.

Group and complete the squares:

9(x2−4x)+16(y2+2y)=929(x^2 - 4x) + 16(y^2 + 2y) = 92

9(x−2)2−36+16(y+1)2−16=929(x-2)^2 - 36 + 16(y+1)^2 - 16 = 92

9(x−2)2+16(y+1)2=1449(x-2)^2 + 16(y+1)^2 = 144.

Divide by 144144:

(x−2)216+(y+1)29=1\dfrac{(x-2)^2}{16} + \dfrac{(y+1)^2}{9} = 1.

So a2=16,b2=9a^2 = 16, b^2 = 9, i.e. a=4,b=3a = 4, b = 3, centre (2,−1)(2, -1).

Eccentricity: e=1−b2a2=1−916=74e = \sqrt{1 - \dfrac{b^2}{a^2}} = \sqrt{1 - \dfrac{9}{16}} = \dfrac{\sqrt7}{4}.

Foci: (2±ae,−1)=(2±4⋅74,−1)=(2±7,−1)(2 \pm ae, -1) = \left(2 \pm 4\cdot\tfrac{\sqrt7}{4}, -1\right) = (2 \pm \sqrt7, -1).

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