If e1 and e2 are the eccentricities of the ellipse, x218+y24=1 and the hyperbola, x29-y24=1 respectively and e1,e2 is a point on the ellipse, 15x2+3y2=k, then k is equal to:
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a
17
b
16
c
15
d
14
answer is B.
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Detailed Solution
We have e1=1-418=79 and e2=1+49=139 Given e1,e2 lie on 15x2+3y2=k⇒15e12+3e22=k⇒k=1579+3139=16