Question

The common tangent between the curves $x^2+y^2=1$ and $y^2=4(x-2)$, passing through $(x_1,y_1)$ on $y^2=4(x-2)$ satisfies $a{x}_1^2-9x_1+b=0$. Then $a+b$ is equal to

Answer 1

$y^2=4(x-2)$
$\Rightarrow \frac{dy}{dx}=\frac{4}{2y}=\frac{2}{y}$
Let the point of tangency be $(x_1,y_1)$
$\Rightarrow$ $(x_1,y_1)$ lies on $y^2=4(x-2)$
$\Rightarrow y_1^2=4(x_1-2)$
$\Rightarrow$ Equation of tangent at $(x_1,y_1)$
$\Rightarrow y-y_1=\frac{2}{y_1}(x-x_1)$
$y{y}_1-y_1^2=2x-2x_1$
$\Rightarrow y{y}_1-2x=y_1^2-2x_1=4x_1-8-2x_1$
$\Rightarrow y{y}_1-2x=2x_1-8\cdots \left(i\right)$
Now $y{y}_1-2x=2x_1-8$ is also tangent to $x^2+y^2=1$
$\Rightarrow$ Using condition of tangency for circle distance of centre from line must be equal to radius.
$\Rightarrow r=1=\frac{|0-0-2x_1+8|}{\sqrt{4+y_1^2}}$
$\Rightarrow 4+y_1^2=(2x_1-8{)}^2$
$\Rightarrow 4+(4x_1-8)=4x_1^2+64-32x_1$
$\Rightarrow 4x_1^2-36x_1+68=0$
$\Rightarrow x_1^2-9x_1+17=0$
$\Rightarrow a+b=18$