Question

The tangent to the hyperbola $xy=c^2$ at the point $P$ intersects the $x$-axis at $T$ and the $y$-axis at $T^{\prime }$. The normal to the hyperbola at $P$ intersects the $x$-axis at $N$ and the y-axis at $N^{\prime }$. The areas of the triangles PNT and $P{N}^{\prime }{T}^{\prime }$ are $\Delta$ and ${\Delta }^{\prime }$ respectively, then $\frac{1}{\Delta }+\frac{1}{{\Delta }^{\prime }}$ is -

• A. equal to 1
• B. depends on $t$
• C. depends on $c$
• D. equal to 2

Answer 1

Answer: (C)

Let point $P$ be $\left(ct,\frac{c}{t}\right)$.
Equation of tangent at $P$ is
$x+y{t}^2=2ct$
$\therefore T$ is $(2ct,0)\&T^{\prime }$ is $\left(0,\frac{2c}{t}\right)$
Now equation of normal at $P$ is
$t^{2}x-y=c{t}^3-\frac{c}{t}$
$\therefore N\left(ct-\frac{c}{t^3},0\right)\&N^{\prime }\left(0,\frac{c}{t}-c{t}^3\right)$
$\Delta =\frac{1}{2}\frac{c}{t}\left(2ct-ct+\frac{c}{t^3}\right)=\frac{1}{2}\frac{c^2}{t^4}\left(t^4+1\right)$
${\Delta }^{\prime }=\frac{1}{2}ct\left(\frac{c}{t}+c{t}^3\right)=\frac{1}{2}{c}^2\left(1+t^4\right)$
$\therefore \frac{1}{\Delta }+\frac{1}{{\Delta }^{\prime }}=\frac{2}{c^2}$.