Question

Find the value of $p$ for which the curves $x^2=9p(9-y)$ and $x^2=p(y+1)$ cut each other at right angles.

Answer 1

The given curves are $x^2=9p(9-y)$ .... $\left(1\right)$

$x^2=p(y+1)$ ...... $\left(2\right)$

Solving $\left(1\right)$ and $\left(2\right)$ simultaneously, we get

$9p(9-y)=p(y+1)\Rightarrow 81-9y=y+1$

$\Rightarrow 10y=80\Rightarrow y=8$

$x^2=9p(9-8)\Rightarrow x^2=9p\Rightarrow x=\pm 3\sqrt{p}$

Therefore, given curves intersect at points $(\pm 3\sqrt{p},8)$ i.e., at points $P(3\sqrt{p},8)$ and $Q(-3\sqrt{p},8)$

Differentiating $\left(1\right)$ and $\left(2\right)$ w.r.t $x,$ we get

$2x=-9p\frac{dy}{dx}\Rightarrow \frac{dy}{dx}=\frac{-2x}{9p}$ ...... $\left(3\right)$ and

$2x=p\frac{dy}{dx}\Rightarrow \frac{dy}{dx}=\frac{2x}{p}$ ...... $\left(4\right)$

Slope of tangent to the curve $\left(1\right)$ at point $P$ is $m_1=\frac{-2\times 3\sqrt{p}}{9p}=\frac{-2}{3\sqrt{p}}$

Slope of tangent to the curve $\left(2\right)$ at point $P$ is $m_2=\frac{2\times 3\sqrt{p}}{p}=\frac{6}{\sqrt{p}}$

Now given curves $\left(1\right)$ and $\left(2\right)$ cut at right angles,

Therefore, $m_1\times m_2=-1$

$\Rightarrow \frac{-2}{3\sqrt{p}}\times \frac{6}{\sqrt{p}}=-1$

$\Rightarrow \frac{4}{p}=1\Rightarrow p=4$

Similarly, when the curves intersect at right angles at point Q, then $p=4.$