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 k) at a point $(\alpha ,\beta )$; then $(\alpha ,\beta )$ is;

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

Answer 1

Answer: (D)

$(-\frac{8}{17},\frac{2}{17})$

Given equation of conic is
$y-6=x^2$
$\frac{dy}{dx}=2x$
Slope of tangent at $(2,10)$ is $4.$
Equation of tangent to conic is
$y-10=4(x-2)$
$\Rightarrow y-10=4x-8$
$\Rightarrow y=4x+2$ .....(1)
Given equation of circle is
$x^2+y^2+8x-2y=k$
$\Rightarrow (x+4{)}^2+(y-1{)}^2=k+17$
Given $(\alpha ,\beta )$ is a point of tangency for fixed $k$
Radius = Length of tangent from center $(-4,1)$
$\Rightarrow \sqrt{k+17}=\frac{15}{\sqrt{17}}$
$\Rightarrow k+17=\frac{225}{17}$
So, equation of circle at $(\alpha ,\beta )$ is
$(\alpha +4{)}^2+(\beta -1{)}^2=\frac{225}{17}$
Now, eqn (1) is tangent to the circle for fixed $k$ at $(\alpha ,\beta )$
$\Rightarrow (\alpha +4{)}^2+(4\alpha +1{)}^2=\frac{225}{17}$
$\Rightarrow 17{\alpha }^2+16\alpha +17=\frac{225}{17}$
$\Rightarrow 289{\alpha }^2+272\alpha +64=0$
$\Rightarrow \alpha =\frac{-8}{17}$ (Here $D=0$)
So, by (1), $\beta =\frac{2}{17}$