Q95 / 160 · AP EAMCET 2023 (15 May Shift 1) · Mathematics
Let the equation of the tangent at a point $\mathrm{P}$ on the parabola $x^2-4 x-4 y+16=0$ be $2 x-y-5=0$. If the equation of the normal drawn at $\mathrm{P}$ to this parabola is $\mathrm{ax}+\mathrm{y}+\mathrm{c}=0$, then $\mathrm{ac}=$.0
- A-20
- B20
- C5
- Correct answerD-5
Solution
$\because x^2-4 x-4 y+16=0$
$$
\Rightarrow 2 x-4-4 \frac{d y}{d x}=0 \Rightarrow \frac{d y}{d x}=\frac{4-2 x}{-4}=-1+\frac{1}{2} x
$$
$2 x-y-5=0$ is tangent on the parabola.
$$
\begin{aligned}
& \therefore m=2 \\
& \Rightarrow-1+\frac{1}{2} x=2 \because m=\frac{d y}{d x} \\
& \Rightarrow \frac{1}{2} x=3 \Rightarrow x=6
\end{aligned}
$$
So, $2 \times 6-y-5=0 \Rightarrow y=7$.
$$
\therefore P \equiv(6,7)
$$
Now, equation of normal at $P(6,7)$ is
$$
\begin{aligned}
& \Rightarrow(y-7)=-\frac{1}{2}(x-6) \\
& \Rightarrow 2 y-14=-x+6 \Rightarrow x+2 y-20=0 . \\
& \Rightarrow \frac{1}{2} x+y-10=0...(1)
\end{aligned}
$$
The given equation normal is $\Rightarrow a x+y+c=0...(2)$ from (1) and (2) $\Rightarrow a=\frac{1}{2}$ and $c=-10$
$$
\therefore a c=\frac{1}{2} \times(-10)=-5
$$
Hence, the correct answer is option: 'D'.