Question

Chords of contact are drawn from the points on a tangent of a hyperbola $x^2-y^2=16$ to a parabola $y^2=16x$. If all the chords of contact pass through a fixed point $Q$, then the locus of the point $Q$ for different tangents on hyperbola is an ellipse, whose length of latus rectum is .

Answer 1

The given hyperbola is,
$x^2-y^2=16$
General point on the hyperbola is given by $(4\sec \theta ,4\tan \theta )$
Tangent to the hyperbola at $(4\sec \theta ,4\tan \theta )$ is
$\left(4\sec \theta \right)x-\left(4\tan \theta \right)y=16\Rightarrow x\sec \theta -y\tan \theta =4$
Any point on the tangent will be given by
$(\lambda ,\frac{\left(\sec \theta \right)\lambda -4}{\tan \theta })$
Equation of the chord of contact to the parabola $y^2=16x$ is
$T=0\Rightarrow y\times \left(\frac{\left(\sec \theta \right)\lambda -4}{\tan \theta }\right)=8(x+\lambda )\Rightarrow y\times (\lambda -4\cos \theta )=8\sin \theta (x+\lambda )\Rightarrow (y-8\sin \theta )\lambda -4y\cos \theta -8x\sin \theta =0\Rightarrow y=8\sin \theta ,x=-\frac{4y\cos \theta }{8\sin \theta }=-4\cos \theta$
Let the coordinates of $Q$ be $(h,k)$
$h=-4\cos \theta ,k=8\sin \theta \Rightarrow {\left(\frac{h}{-4}\right)}^2+{\left(\frac{k}{8}\right)}^2=1\Rightarrow \frac{h^2}{16}+\frac{k^2}{64}=1$
Length of the latus rectum will be
$L=\frac{2a^2}{b}=\frac{2\times 16}{8}=4$