Question

If tangents `P Qa n dP R` are drawn from a point on the circle `x^2+y^2=25` to the ellipse `(x^2)/4+(y^2)/(b^2)=1,(b<4),` so that the fourth vertex `S` of parallelogram `P Q R S` lies on the circumcircle of triangle `P Q R` , then the eccentricity of the ellipse is `(sqrt(5))/4` (b) `(sqrt(7))/4` (c) `(sqrt(7))/4` (d) `(sqrt(5))/3`<br>A. `sqrt(5)//4`
B. `sqrt(7)//3`
C. `sqrt(7)//4`
D. `sqrt(5)//3`

Answer 1

A cyclic parallelogram will be a reactangel or a squar.
So, `angleQPR=90^(@)`
Therefore, , P lines on the director circle of the ellipse `(x^(2))/(16)+(y^(2))/(b^(2))=1`
Hence `x^(2),y^(2)=25` is the director of `(x^(2))/(16)+(y^(2))/(b^(2))=1` then, `16+b^(2)=25`
or `b^(2)=9`
Now, `a^(2)(1-e^(2))=9`
or `1-e^(2)=(9)/(16)`
or `e^(2)=(7)/(16)`
or `e=(sqrt(7))/(4)`