Question

If tangents pq and pr are drawn from a point on the circle x^2 + y^2 = 25 to the ellipse x^2/16+y^2/b^2=1, (b<4) so that the fourth vertex s of the parallelogram pqrslies on the circumcircleof triangle pqr, then find the eccentricity of the ellipse.

Answer 1

The eccentricity of the ellipse is e =  √7 / 4

Step-by-step explanation:

S = X^2+y^2 = 25

E = x^2 / 4 + y^2 / b^2 = 1 (b<4)

Ptp is on therefore S = 0

PQRS is 11 g

S = X^2+y^2 = 25 can be the director circle of E = 0 (PQ ⊥ PR)

DC:  x^2 + y^2 = a^2 + b^2

       x^2 + y^2 = 16 + b^2

i.e. 16 + b^2 = 25

b^2 = 9

b = 3

e = √ 1 - b^2 / a^2 = √ 1- 9 /16  = √7 / 4

Thus the eccentricity of the ellipse is e =  √7 / 4