Question

Two common tangents to the circle \(x^2 + y^2 = 2a^2\) and parabola \(y^2 = 8ax\) are

• A. \(x = \pm (y + 2a)\)
• B. \(y = \pm (x + 2a)\)
• C. \(x = \pm (y + a)\)
• D. \(y = \pm (x + a)\)

Hint:

For parabola \(y^2 = 4ax\), tangent is \(y = mx + a/m\).

Answer 1

The Correct Option is B

To solve the problem of finding the common tangents to the circle \(x^2 + y^2 = 2a^2\) and the parabola \(y^2 = 8ax\), we need to understand the properties of tangents to conic sections and how they interact with each other.

The equation of the circle is \(x^2 + y^2 = 2a^2\), which is centered at the origin with a radius \(\sqrt{2}a\). 

The equation of the parabola is \(y^2 = 8ax\), which opens to the right with its vertex at the origin.

The condition for common tangents to a circle and a parabola involves the geometric constructions that comply with both conics. We need to find a linear equation that simultaneously satisfies the tangent condition for the given circle and parabola.

First, consider the standard form of the tangent to the parabola \(y^2 = 8ax\). The equation of the tangent can be written as \(y = mx + \frac{2a}{m}\).

Next, for the circle \(x^2 + y^2 = 2a^2\), the standard form of the tangent is \(y = mx \pm \sqrt{2a^2(1+m^2)}\).

To have common tangents, the equations must be equivalent, requiring:

\(\frac{2a}{m} = \pm \sqrt{2a^2(1+m^2)}\)

Squaring both sides and simplifying, this condition will yield the values of \(m\) which satisfy both the equation of the circle's tangents and the parabola's tangents.

After solving the above conditions, the common tangent turns out to be of the form:

\(y = \pm (x + 2a)\)

This means the line equations \(y = x + 2a\) and \(y = -x - 2a\) are the common tangents to both the given circle and the parabola.

Thus, the correct answer is the option \(y = \pm (x + 2a)\).