Q.
Given ellipse 16x2+7y2=1 and the hyperbola 144x2−81y2=251, if the ordinate of one of the points of intersection is produced to cut an asymptote at P, then which of the following is true?
16x2+7y2=1…(1) ⇒a2=16,b2=7
i.e. a=4,b=7 ∴e2=a2a2−b2⇒e=43 ∴ foci ≡(±ae,0)=(±3,0) (144/25)x2−(81/25)y2=1…..(2) ⇒a2=25144,b2=2581
i.e. a=512,b=59 ∴e2=a2a2+b2⇒e=45 foci ≡(±ae,0)=(±3,0)
solving (1) and (2) we get y2=2563⇒y=±537⇒x=±516
one of the point of intersection is (516,537)
The equation of the asymptote is 144x2−81y2=0
The abscissa of P is 516
Its ordinate is given by 81y2=25×14416×16 ∴y=±512 ∴P≡(516,512)⇒(516)2+(512)2=16
Equation of the auxiliary circle formed on major axis of ellipse x2+y2=16 P lies on it.