Question

The equation of one of the common tangents to the parabola $y^2 = 8x$ and $x^2 + y^2 - 12x + 4 = 0$ is

• A. $y = - x + 2$
• B. $y = x - 2$
• C. $y = x + 2$
• D. None of these

Answer 1

Answer: (C)

$y = x + 2$

Answer 2

Any tangent to parabola $y^2 = 8x$ is $y = mx + \frac{2}{m}$ ......(i) It touches the circle $x^2 + y^2 - 12x + 4 = 0$ . if the length of perpendicular from the centre $(6, 0)$ is equal to radius $\sqrt{32}$ $\therefore \, \, \frac{6m+ \frac{2}{m}}{\sqrt{m^{2} + 1 }} = \pm\sqrt{32}$ $\Rightarrow \left(3m + \frac{1}{m}\right)^{2} = 8\left(m^{2} + 1\right)$ $\Rightarrow \left(3m^{2 } + 1\right)^{2} = 8 \left(m^{4} + m^{2}\right)$ $\Rightarrow m^{4} - 2m^{2} + 1 = 0\Rightarrow m = \pm1$ Hence, the required tangents are $y = x + 2$ and $y = -x - 2$.