Question

If the tangent to the conic, $y–6=x^2$ at $(2,10)$ touches the circle, $x^2+y^2+8x–2y=k$ (for some fixed$\left(k\right)$ at a point $(\alpha ,\beta )$; then $(\alpha ,\beta )$ is :

• A. $(\frac{-7}{17},\frac{6}{17})$
• B. $(\frac{-6}{17},\frac{10}{17})$
• C. $(\frac{-4}{17},\frac{1}{17})$
• D. $(\frac{-8}{17},\frac{2}{17})$

Answer 1

The correct option is D $(\frac{-8}{17},\frac{2}{17})$

$y–6=x^2$
$y^{\prime }=2x$
At $(2,10),y^{\prime }=4$
So, equation of tangent at $(2,10)$ is
$4x–y+2=0$
$(\alpha ,\beta )$ is on the line
So, $\beta =4\alpha +2$
Now slope of tangent of circle at $(\alpha ,\beta )$
$2x+2y{y}^{\prime }+8-2y^{\prime }=0y^{\prime }=\frac{2x+8}{2-2y}=\frac{2\alpha +8}{2-2\beta }=4\Rightarrow \frac{2\alpha +8}{2-2(4\alpha +2)}=4\alpha =-\frac{8}{17}\beta =-4\times \frac{8}{17}+2=\frac{2}{17}$