Let a hyperbola passes through the focus of the ellipse `(x^(2))/(25)-(y^(2))/(16)=1`. The transverse and conjugate axes of this hyperbola coincide with the major and minor axes of the given ellipse, also the product of eccentricities of given ellipse and hyperbola is 1, then

Options

A. `((x^(2))/(9)-(y^(2))/(16))^(2)=(x^(2)+y^(2))/(25)`
B. `((x^(2))/(9)-(y^(2))/(16))=((x^(2)+y^(2))/(25))^(2)`
C. `((x^(2))/(9)-(y^(2))/(16))=(x^(2)+y^(2))/(25)`
D. `((x^(2))/(9)-(y^(2))/(16))^(2)=((x^(2)+y^(2))/(25))^(2)`

Explanation

For the given ellipse, `(x^(2))/(25)+(y^(2))/(16)=1, e=sqrt(1-(16)/(25))=(3)/(5)`. So, eccentricity of hyperbola `=(5)/(3)`.
Let the hyperbola be, `(x^(2))/(A^(2))-(y^(2))/(B^(2))=1 " " ...(1)`
Then, `B^(2)=A^(2)((25)/(9)-1)=(16)/(9)A^(2)`. Also, foci of ellipse are `(+-3,0)`.
As, hyperbola passes through `(+-3,0)`. So, `(9)/(A^(2))=1rArrA^(2)=9, B^(2)=16`
`rArr` Equation of hyperbola is `(x^(2))/(9)-(y^(2))/(16)=1`
(i) Vertices of hyperbola are `(+-3,0)rArr`(A) is correct.
Focal length of hyperbola `=10rArr ` (B) is incorrect.
(ii) Any point of hyperbola is P `(3 sec theta, 4 tan theta)`.
Equation of auxiliary circle of ellipse is `x^(2)+y^(2)=25`.
`:.` Equation of chord of contact to the circle `x^(2)+y^(2)=25`, with respect to `P(3 sec theta, 4 tan theta)`, is
`3x sex theta + 4 y tan theta = 25 ...(1)`
If (h, k) is the mid point of chord of contact, then its equation is
`hx+ky-25=h^(2)+k^(2)-25 rArr hx + ky=h^(2)=k^(2) " " ...(2)`
As, equations (1) and (2) represent the same straight line, so on comparing , we get
`(3 sec theta)/(h)= (4 tan theta)/(k)=(25)/(h^(2)+k^(2))`
`:. sec theta = ((25)/(h^(2)+k^(2))).(h)/(3), tan theta =((25)/(h^(2)+k^(2))).(k)/(4)`
`:. ` Eliminating ` theta`, we get, `((25)/(h^(2)+k^(2)))^(2)((h^(2))/(9)-(k^(2))/(16))=1`. (As, `sec^(2)theta-tan ^(2)theta=1`)
`:. ` Locus of (h, k) is `((x^(2))/(9)-(y^(2))/(16))((x^(2)+y^(2))/(25))^(2)`