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

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