Home
Class 12
MATHS
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 `6/5` `(-oo,-sqrt(7))uu(sqrt(7),(13)/5)` `(-oo,-sqrt(7))uu((13)/5,sqrt(7),)` 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.
Promotional Banner

Similar Questions

Explore conceptually related problems

If area of the ellipse (x^(2))/(16)+(y^(2))/(b^(2))=1 inscribed in a square of side length 5sqrt(2) is A, then (A)/(pi) equals to :

(sqrt(7)+2sqrt(3))(sqrt(7)-2sqrt(3))

Simplify (5+sqrt(7))(5+sqrt(2))

sqrt(x+5)+sqrt(20-x)=7

Simplify: (i)\ ((sqrt(2))/5)^8\ -:((sqrt(2))/5)^(13)

int_2^5 (sqrt(x))/(sqrt(x)+sqrt(7-x)) dx

{:(sqrt(5)x - sqrt(7)y = 0),(sqrt(7)x - sqrt(3)y = 0):}

The set of all real numbers x for which x^2-|x+2|+x >0 is (-oo,-2) b. (-oo,-sqrt(2))uu(sqrt(2),oo) c. (-oo,-1)uu(1,oo) d. (sqrt(2),oo)

Write in ascending order: 6 sqrt(5) , 7 sqrt(3) and 8 sqrt(2)

Show that : (1)/(3-2sqrt(2))- (1)/(2sqrt(2)-sqrt(7)) + (1)/(sqrt(7)-sqrt(6))-(1)/(sqrt(6)-sqrt(5))+(1)/(sqrt(5)-2)=5 .