Question

x2
y2
If a hyperbola passes through the focus of the ellipse +
= 1 and its transverse and
25 16
conjugate axes coincide with the major and minor axes of the ellipse, and the product of
eccentricities is 1, then​

Answer 1

Step-by-step explanation:

Let the equation of the hyperbola be a2x2−by2=1   ...(1)

Since it passes through the focus (±25−16,0)=(±3,0)

of the ellipse, so a29=1⇒a2=9   ...(2)

Also product of this eccentricities =1

⇒1−2516×1+a2b2=1

⇒1+9b2=925⇒b2=16

Therefore equation of hyperbola will be 9x2−16y2=1

Its focus =(±a2+b2,0)=(±5,0).