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  4. Let a hyperbola passes through the focus of the ellipse 16 x2+25 y2=400. The transverse and conjugate axes of this hyperbola coincide with the major and minor axes of the given ellipse. The eccentricity of the hyperbola is reciprocal of that the ellipse. Tangents are drawn from any point on the hyperbola to the auxiliary circle of the ellipse, then the locus of mid-point of chord of contact is

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Q. Let a hyperbola passes through the focus of the ellipse $16 x^2+25 y^2=400$. The transverse and conjugate axes of this hyperbola coincide with the major and minor axes of the given ellipse. The eccentricity of the hyperbola is reciprocal of that the ellipse.
Tangents are drawn from any point on the hyperbola to the auxiliary circle of the ellipse, then the locus of mid-point of chord of contact is

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Solution:

For the given ellipse, $\frac{ x ^2}{25}+\frac{ y ^2}{16}=1, e =\sqrt{1-\frac{16}{25}}=\frac{3}{5}$.
So, eccentricity of hyperbola $=\frac{5}{3}$.
Let the hyperbola be, $\frac{ x ^2}{ A ^2}-\frac{ y ^2}{ B ^2}=1 \ldots$ (1)
Then, $B^2=A^2\left(\frac{25}{9}-1\right)=\frac{16}{9} A^2$. Also, foci of ellipse are $( \pm 3,0)$.
As, hyperbola passes through $( \pm 3,0)$. So, $\frac{9}{A^2}=1 \Rightarrow A^2=9, B^2=16$
$\Rightarrow$ Equation of hyperbola is $\frac{x^2}{9}-\frac{y^2}{16}=1$
Any point of hyperbola is $P (3 \sec \theta, 4 \tan \theta)$.
Equation of auxiliary circle of ellipse is $x^2+y^2=25$.
$\therefore $ Equation of chord of contact to the circle $x ^2+ y ^2=25$, with respect to $P (3 \sec \theta, 4 \tan \theta)$, is $3 x \sec \theta+4 y \tan \theta=25 \ldots$ (1)
If $(h, k)$ is the mid point of chord of contact, then its equation is
$hx + ky -25= h ^2+ k ^2-25 \Rightarrow hx + ky = h ^2+ k ^2$......(2)
As, equations (1) and (2) represent the same straight line, so on comparing, we get
$ \frac{3 \sec \theta}{ h }=\frac{4 \tan \theta}{ k }=\frac{25}{ h ^2+ k ^2} $
$\therefore \sec \theta=\left(\frac{25}{ h ^2+ k ^2}\right) \cdot \frac{ h }{3}, \tan \theta=\left(\frac{25}{ h ^2+ k ^2}\right) \cdot \frac{ k }{4}$
$\therefore \text { Eliminating } \theta, \text { we get, }\left(\frac{25}{ h ^2+ k ^2}\right)^2\left(\frac{ h ^2}{9}-\frac{ k ^2}{16}\right)=1 .\left(\text { As, } \sec ^2 \theta-\tan ^2 \theta=1\right) $
$\therefore \text { Locus of }( h , k ) \text { is }\left(\frac{ x ^2}{9}-\frac{ y ^2}{16}\right)\left(\frac{ x ^2+ y ^2}{25}\right)^2$