Question

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

• A. $x^2=\lambda c^2$
• B. $y^2=\lambda c^2$
• C. $xy=\lambda c^2$
• D. $x^2-y^2=\lambda c^2$

Answer 1

The correct option is A $x^2=\lambda c^2$

Let $(ct,\frac{c}{t})$ be any point on the hyperbola $xy=c^2$
Normal at this point
$x{y}^{\prime }+y=0\Rightarrow y^{\prime }=\frac{-y}{x}$
Slope of normal $=\frac{x}{y}{|}_{(ct,\frac{c}{t})}$
$=\frac{ct}{\frac{c}{t}}=t^2...\left(1\right)$
The equation of normal
$y-\frac{c}{t}=t^2(x-ct)$
Let the coordinate of $P$ be $(h,k)$ and the normal passes through point $P$.
$k-\frac{c}{t}=t^2(h-ct)$
$kt-c=t^3(h-ct)$
$\Rightarrow c{t}^4-t^{3}h+kt-c=0$
Given that sum of slopes of normal =$\lambda$
$t_1^2+t_2^2+t_3^2+t_4^2=\lambda$ (from equation (1))
$\Rightarrow (t_1+t_2+t_3+t_4{)}^2-2\sum t_1{t}_2=\lambda$
$\Rightarrow {\left(\frac{h}{c}\right)}^2-0=\lambda$
$\Rightarrow h^2=\lambda c^2$
$\Rightarrow x^2=\lambda c^2$