Question #4f402

1 Answer
Jul 19, 2016

Reqd. Eqn. of the Ellipse is # : x^2/12+y^2/16=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=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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