Question

The equation(s) of the common tangent(s) to the parabola ${{y}^{2}}=2x$ and the circle ${{x}^{2}}+{{y}^{2}}+4x=0$ is / are: (This question has multiple correct options)
(a) $2\sqrt{6}x+y=12$
(b) $x+2\sqrt{6}y+12=0$
(c) $x-2\sqrt{6}y+12=0$
(d) $2\sqrt{6}x-y=12$

Answer 1

Hint:Let y = mx + c be the equation of common tangent to the parabola ${{y}^{2}}=2x$ and the circle ${{x}^{2}}+{{y}^{2}}+4x=0$. Use the formula $c=\dfrac{a}{m}$ to get one equation. Then use the fact that for the line y = mx + c to become tangent to the circle, the perpendicular distance from the centre of the circle to the line should be equal to radius of the circle to find another equation. Solve these to get values of c and m. put those values in the equation of the line y = mx + c to get the final answer.

Complete step-by-step answer:
In this question, we need to find the equation(s) of the common tangent(s) to the parabola ${{y}^{2}}=2x$ and the circle ${{x}^{2}}+{{y}^{2}}+4x=0$.
Let y = mx + c be the equation of common tangent to the parabola ${{y}^{2}}=2x$ and the circle ${{x}^{2}}+{{y}^{2}}+4x=0$.
The condition for the line y = mx + c to be the tangent to the parabola ${{y}^{2}}=4ax$ is
$c=\dfrac{a}{m}$.
On comparing the parabola ${{y}^{2}}=2x$ with the parabola ${{y}^{2}}=4ax$, we get to know that
$a=\dfrac{1}{2}$.
Substituting this to the above expression, we will get the following:
$c=\dfrac{1}{2m}$ … (1)
Now, for the line y = mx + c to become tangent to the circle, the perpendicular distance from the centre of the circle to the line should be equal to radius of the circle.
From the equation of the circle, we will find that the radius of the circle given by ${{x}^{2}}+{{y}^{2}}+4x=0$ is 2 units and its centre is (-2, 0).
We know that the distance of a point (p, q) from a line ax + by +c = 0 is given by:
$\dfrac{|ap+bq+c|}{\sqrt{{{a}^{2}}+{{b}^{2}}}}$
Using this formula for the line y = mx + c and the point (-2, 0) and equating that to the radius of the circle, which is 2 units, we will get the following:
$\dfrac{-2m+c}{\sqrt{1+{{m}^{2}}}}=2$ …(2)
Substituting equation (1) in equation (2), and then solving for m, we will get the following:
$m=\pm \dfrac{1}{2\sqrt{6}}$
Putting these in equation (1), we will get the following:
$c=\pm 2\sqrt{6}$
Substituting these values in the equation of the line y = mx + c .
So, the equations for the line y = mx + c are $x+2\sqrt{6}y+12=0$ and $x-2\sqrt{6}y+12=0$
So, the equations of the common tangents to the parabola ${{y}^{2}}=2x$ and the circle ${{x}^{2}}+{{y}^{2}}+4x=0$ are $x+2\sqrt{6}y+12=0$ and $x-2\sqrt{6}y+12=0$.
Hence, options (b) and (c) are correct.

Note: In this question, it is very important to know the following formulae/ facts: The condition for the line y = mx + c to be the tangent to the parabola ${{y}^{2}}=4ax$ is $c=\dfrac{a}{m}$, for the line y = mx + c to become tangent to the circle, the perpendicular distance from the centre of the circle to the line should be equal to radius of the circle and that the distance of a point (p, q) from a line ax + by +c = 0 is given by: $\dfrac{|ap+bq+c|}{\sqrt{{{a}^{2}}+{{b}^{2}}}}$.The equation $x^2+y^2+2gx+2fy+c=0$ represents a circle with centre (–g,–f) and radius ${\sqrt{g^2+f^2-c}}$. This is called the general equation of a circle.From given equation ${{x}^{2}}+{{y}^{2}}+4x=0$ comparing with general equation of circle $2g=4$ and $2f=0$ so centre will be $(-g,-f)$ i.e $(-2 ,0)$ and $radius$ =${\sqrt{2^2+0-0}}=2units$.