Question

If $x-2y+k=0$ is a tangent to the parabola $y^2-4x-4y+8=0$, then the slope of the tangent drawn at $(l,k)$ on the given parabola is

• A. $-\frac{5}{2}$
• B. $2$
• C. $-2$
• D. $\frac{2}{5}$

Answer 1

Answer: (D)

We have, $x-2y+k=0$ is tangent to the parabola $y^2-4x-4y+8=0$, then
Now put $x=2y-k$ in equation of parabola, we get
$\therefore y^2-4(2y-k)-4y+8=0$
$\Rightarrow y^2-8y+4k-4y+8=0$
$\Rightarrow y^2-12y+4k+8=0$
Since line is tangent to parabola
$\therefore D=0$
$(-12{)}^2-4(4k+8)=0$
$\Rightarrow 144-16k-32=0$
$\Rightarrow 16k=112$
$\Rightarrow k=7$
Now, we have
$y^2-4x-4y+8=0$
$\therefore 2y\frac{dy}{dx}-4-4\frac{dy}{dx}=0$
$\Rightarrow 2\frac{dy}{dx}(y-2)=4$
$\Rightarrow \frac{dy}{dx}=\frac{2}{y-2}$
$\therefore$ Slope of tangent at
$(l,k)=\frac{2}{k-2}=\frac{2}{7-2}[\because k=7]$
$=\frac{2}{5}$