Question

Find the locus of the point of intersection of tangents drawn at the extremities of a normal chord of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1.$ Deduce the corresponding result if the hyperbola be rectangular i.e. $x^2-y^2=a^2.$

Answer 1

Let (h, k) be the pole of the normal chord or point of intersection of
tangents at its extremities so that its equation is $\frac{hx}{a^2}-\frac{ky}{b^2}=1$ ....(1)
i.e, polar of (h, k) or chord of contact of the point (h, k).
Since (1) is a normal chord its equation is of the form
$ax\cos \theta +by\cot \theta =a^2+b^2$ ...(2)
Comparing (1) and (2), we get $\frac{h}{a^3\cos \theta }=\frac{-k}{b^3\cot \theta }=\frac{1}{a^2+b^2}$
$\therefore (a^2+b^2)\sec \theta =\frac{a^3}{h}$
and $(a^2+b^2)\tan \theta =-\frac{b^3}{k}$
Squaring and subtracting thereby eliminating $\theta ,$ we get
$(a^2+b^2)\cdot 1=\frac{a^6}{h^2}-\frac{b^6}{k^2}$
Generalising, the locus of the point (h, k) is
$\frac{a^6}{x^2}-\frac{b^6}{y^2}={(a^2+b^2)}^2$
Note
: In case the hyperbola is $x^2-y^2=a^2$ i.e.
rectangular then putting $b=a$ in [3], we get
$a^5\frac{(y^2-x^2)}{y^2{x}^2}=4a^4$
$\therefore a^2(y^2-x^2)=4x^2{y}^2$
or $\frac{1}{x^2}-\frac{1}{y^2}=\frac{4}{a^2}$