Question

If the straight line $y=x+1$ and the circle $(x+2)^2+(y-d)^2=r^2$ touch the parabola $y^2=a x$ at $P$ then find the value of $\left(\frac{ r ^2+ d + a }{9}\right)$.

Answer 1

$y=x+1$ touches the parabola $y^2=a x$
$\therefore 1=\frac{ a }{4 \cdot 1} \Rightarrow a =4\left\{\Theta c =\frac{ a }{ m }\right\}$
Point of contact is $\left(\frac{ a }{ m ^2}, \frac{2 a }{ m }\right) \equiv(1,2)$
Now, normal at ' $p$ ' is $x + y +\lambda=0$
Satisfying $(1,2)$, get $\lambda=-3$
$\therefore x + y -3=0$
Centre of the circle $(-2, d )$ lies on the above normal
$\therefore d =5$
$r =$ distance between $(-2,5)$ and $(1,2)$
$=\sqrt{9+9}=\sqrt{18} $
$\therefore r ^2=18$
$\therefore \frac{ r ^2+ d + a }{9}=\frac{18+5+4}{9}=3$