Question #4f402

1 Answer
Jul 19, 2016

Reqd. Eqn. of the Ellipse is : x^2/12+y^2/16=1:x212+y216=1. graph{x^2/12+y^2/16=1 [-10, 10, -5, 5]}

Explanation:

Let us suppose that the reqd. eqn. of ellipse is

S : x^2/a^2+y^2/b^2=1S:x2a2+y2b2=1.

Given that pt.(-3,2) in SrArr 9/a^2+4/b^2=1............(1).

We take a note, that the Major Axis of the Ellipse is Y-axis.

:. b>a, and, a^2=b^2(1-e^2)..................(2) (e<1).

Length l of Latus Retum=2a^2/b

Centre is C(0,0) and Focii are S(0,be) and S(0-be), so that,

CS, or, CS'=be.

By what is given, l=3CSrArr 2a^2/b=3berArr 2a^2=3b^2e

By (2), then, 2cancelb^2(1-e^2)=3cancelb^2e

:. 2e^2+3e-2=0rArr (e+2)(2e-1)=0, giving,

e=-2, which is impossible, or, e=1/2(<1, as desired.)

Then, from (2), a^2=3/4b^2, or, a^2/3=b^2/4, i.e., 4/b^2=3/a^2

Using this in (1), 9/a^2+3/a^2=1, i.e., 12/a^2=1, or, a^2=12

Finally, since, b^2=4/3a^2=4/3*12=16

Thus, with a^2=12, and, b^2=16 we get the reqd. eqn. of Ellipse,

S : x^2/12+y^2/16=1.

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