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If the ellipse (x^2)/(a^2-7)+(y^2)/(13-5...

If the ellipse `(x^2)/(a^2-7)+(y^2)/(13-5a)=1` is inscribed in a square of side length `sqrt(2)a` , then `a` is equal to (a)`6/5` (b) `(-oo,-sqrt(7))uu(sqrt(7),(13)/5)` (c) `(-oo,-sqrt(7))uu((13)/5,sqrt(7),)` (d)no such a exists

A

`6//5`

B

`11//10`

C

`13//10`

D

no such a exists

Text Solution

Verified by Experts

Since sides of square are tangent to ellipse and perpendicular to each other, vertices lie on the director circle.
Equation of director circle is `x^(2)+y^(2)=(a^(2)-7)+(13+-5a)`
`or x^(2)+y^(2)=a^(2)-5a+6`
Thus, radius of the circle is a`:. a^(2)-5a+6=a^(2)`
or a=6/5
But for an ellipse, we have must `a^(2)-7gt 0 and 13-5agt0`
`:. ane=(6)/(5)`
Hence, no such a exists.
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