Question

Let $P\equiv (-6,0)$ and $S\equiv x^2-y^2+16=0$.Which of the following is/are true

Answer 1

Given, $S\equiv x^2-y^2+16=0\cdots \left(i\right),P\equiv (-6,0)$
Let $Q(x_1,y_1)$ be the nearest point on $S=0$ from $P$
Differentiating $\left(i\right)$ w.r.t ${}^{\prime }{x}^{\prime },\frac{dy}{dx}=\frac{x}{y}$
Now, Slope of normal at $Q=$ Slope of $PQ$
$\Rightarrow -\frac{y_1}{x_1}=\frac{y_1}{x_1+6}$
$\Rightarrow x_1=-3$ $\{\because y_1\ne 0\}$
$\Rightarrow y_1^2=25$
$\Rightarrow y_1=\pm 5$
$\therefore Q\equiv (-3,5)$ or $(-3,-5)$
$PQ=\sqrt{34}$