Q95 / 160 · TS EAMCET 2023 (12 May Shift 2) · Mathematics
The equations of common tangents to the parabola $y^2=16 x$ and the circle $x^2+y^2 \div 8$ are
- A$y=x+2, y=x-2$
- B$y=x+1, y=x-2$
- C$y=2 x+4, y=-2 x+4$
- Correct answerD$y=x+4, y=-x-4$
Solution
$$
y^2=16 x
$$
Equation of parabola's tangent
$$
\begin{aligned}
y & =m x+\frac{a}{m} \\
\Rightarrow \quad y & =m x+\frac{4}{m}
\end{aligned}
$$
For circle, equation of tangent,
$$
\begin{aligned}
y & =m x \pm r \sqrt{1+m^2} \\
\Rightarrow \quad y & =m x \pm 2 \sqrt{2} \sqrt{1+m^2}
\end{aligned}
$$
$\because$ (i) and (ii) represents same line,
$$
\begin{aligned}
& \frac{4}{m}= \pm 2 \sqrt{2} \sqrt{\left(1+m^2\right)} \\
\Rightarrow & \frac{16}{m^2}=8\left(1+m^2\right) \\
\Rightarrow & m^2\left(1+m^2\right)=2 \\
\therefore \quad & m= \pm 1 \\
\therefore \quad & \text { Equation of tangents, } y=x+4, y=-x-4 .
\end{aligned}
$$
Hence, the correct answer is option: 'D'.