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

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
$c^{2}=a^{2}m^{2}-b^{2}$
$\Rightarrow \left(\sqrt{2} p\right)^{2}=9\left(- 1\right)^{2}-4$
$\Rightarrow 2p^{2}=5$ .