Question

The tangent to the hyperbola $x y=c^2$ 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 PNT 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)

Tangent: $\frac{x}{c t}+\frac{y t}{c}=2$
put $y=0 ; x=2 c t(T) $
$x=0 ; y=\frac{2 c}{t}\left(T^{\prime}\right)$

||| ly $\text { normal is } y-\frac{c}{t}=t^2(x-c t)$
$\text { put } y=0 ; x=c t-\frac{c}{t^3}(N)$
$ x=0 ; \frac{c}{t}-c t^3\left(N^{\prime}\right)$
$\text { Area of } \Delta \text { PNT }=\frac{c}{2 t}\left(c t+\frac{c}{t^3}\right) \Rightarrow \Delta=\frac{c^2\left(1+t^4\right)}{2 t^4} $
$\text { area of } \Delta \text { PNT }=c t\left(\frac{c}{t}+c^3\right) \Rightarrow \Delta^{\prime}=\frac{c^2\left(1+t^4\right)}{2} $
$\therefore \frac{1}{\Delta}+\frac{1}{\Delta^{\prime}}=\frac{2 t^4}{c^2\left(1+t^4\right)}+\frac{2}{c^2\left(1+t^4\right)}=\frac{2}{c^2\left(1+t^4\right)}\left(t^4+1\right)=\frac{2}{c^2}$
which is independent of $t$.