Question

The straight line $x+y=\sqrt{2}p$ touches the hyperbola $4x^2-9y^2=36$ , then find value of $2p^2$ .

Answer 1

Answer: 5

The given hyperbola is $4x^2-9y^2=36$
$\Rightarrow \frac{x^2}{9}-\frac{y^2}{4}=1$
On comparing it with the standard hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
We get $a^2=9,b^2=4$
And for the line $x+y=\sqrt{2}p,$
$\Rightarrow y=-x+\sqrt{2}p$ ,
which is of the form $y=mx+c,$
Hence, we have $m=-1$ and $c=\sqrt{2}p$
If a line $y=mx+c$ touches a hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1,$ then
$c^2=a^2{m}^2-b^2$
$\Rightarrow {\left(\sqrt{2}p\right)}^2=9{\left(-1\right)}^2-4$
$\Rightarrow 2p^2=5$ .