Question

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