Question

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

Answer 1

Answer: 3

$y=x+1$ touches the parabola $y^2=ax$
$\therefore 1=\frac{a}{4\cdot 1}\Rightarrow a=4\left\{\Theta c=\frac{a}{m}\right\}$
Point of contact is $\left(\frac{a}{m^2},\frac{2a}{m}\right)\equiv (1,2)$
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 $(1,2)$
$=\sqrt{9+9}=\sqrt{18}$
$\therefore r^2=18$
$\therefore \frac{r^2+d+a}{9}=\frac{18+5+4}{9}=3$