Question

If the sum of the slopes of the normal from a point ${}^{\prime }{P}^{\prime }$ to the hyperbola $xy=c^2$ is equal to $\lambda (\lambda \in R^{+})$, then locus of point ${}^{\prime }{P}^{\prime }$ is:

Answer 1

Equation of normal at any point $(ct,\frac{c}{t})$ is $c{t}^4-x{t}^3+ty-c=0$

$\Rightarrow$ Slope of normal $=t^2$

Let $p(h,k)$ be the point through which the normal is passing.

Then $c{t}^4-h{t}^3+tk-c=0$

$\Rightarrow \sum t_i=\frac{h}{c}$ and $\sum t_i{t}_j=0$

Hence, sum of the slopes of the normal

$=\sum {t_i}^2={(\sum t_i)}^2=h^2=c^2\lambda$

Therefore, requires locus is $x^2=\lambda c^2$