Question

Tangents are drawn at two points of a hyperbola $xy=1$. Let tangent at one point passes through the foot of ordinate of the other point. If the locus of the point of intersection of the two tangents is the hyperbola $xy=a$, then the value of $9a$ is
('Foot of ordinate of a point' is the foot of its perpendicular from the point to the $x-$axis)

Answer 1

Let two points on the hyperbola $xy=1$ be $A(t_1,\frac{1}{t_1}),B(t_2,\frac{1}{t_2})$
Equation of tangent at $(ct,\frac{c}{t})$ on a rectagular hyperbola $xy=c^2$ is,
$\frac{x}{t}+yt=2c$
Tangent at $A$,
$\frac{x}{t_1}+y{t}_1=2\cdots \left(1\right)$
Tangent at $B$,
$\frac{x}{t_2}+y{t}_2=2\cdots \left(2\right)$

As tangent at $A$ passes through the foot of ordinate of point $B$, so putting $(t_2,0)$ in equation of tangent at $A$,
$\frac{t_2}{t_1}=2\Rightarrow t_2=2t_1$

Let the point of intersection of both tangents be $(h,k)$
Using equation $\left(1\right)$ and $\left(2\right)$,
$\frac{h}{t_1}+k{t}_1=2\cdots \left(3\right)$
$\frac{h}{2t_1}+2k{t}_1=2\cdots \left(4\right)$

From equation $\left(3\right)$ and $\left(4\right)$,
$\frac{2h}{t_1}-\frac{h}{2t_1}=4-2\Rightarrow \frac{3h}{4}=t_1$
Putting in equation $\left(3\right)$,
$\frac{h}{\frac{3h}{4}}+k\times \frac{3h}{4}=2\Rightarrow hk=\frac{8}{9}$

Therefore, the locus of the point of intersection is, $xy=\frac{8}{9}=a$
$\therefore 9a=8$