Question:medium

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

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For parabola \(y^2 = 4ax\), tangent is \(y = mx + a/m\).
Updated On: Jun 17, 2026
  • \(x = \pm (y + 2a)\)
  • \(y = \pm (x + 2a)\)
  • \(x = \pm (y + a)\)
  • \(y = \pm (x + a)\)
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The Correct Option is B

Solution and Explanation

To find the common tangents to the circle \(x^2 + y^2 = 2a^2\) and the parabola \(y^2 = 8ax\), we need to analyze the equations and derive their common tangents.

  1. First, consider the standard equations:
    • The circle is given by \(x^2 + y^2 = 2a^2\), which is centered at the origin with a radius of \(\sqrt{2}a\).
    • The parabola is given by \(y^2 = 8ax\), which opens to the right and has vertex at the origin.
  2. Common tangents to a circle and a parabola can be found using the concept that they both lie at points where their respective distances from any point \((x, y)\) on the tangent line equate.
  3. We systematically derive the tangents:
    • The general tangent to the parabola \(y^2 = 8ax\) can be given by the equation \(y = mx + \frac{2a}{m}\).
    • The general tangent to the circle \(x^2 + y^2 = 2a^2\) can be written in the form: \(y = mx \pm \sqrt{2a^2(1 + m^2)}\).
  4. Equating these for common tangents involves equating coefficients, particularly of \(x\) and the constant terms.
  5. Additionally, substitute and simplify the expressions to find when they represent valid common tangents.
  6. After simplifying, we find the common tangents suitable with the structure \(y = \pm(x + 2a)\).

By verifying through balancing coefficients and forming valid tangents, the derived solution is the correct form of the common tangents present in the options given:

The correct answer is: \(y = \pm(x + 2a)\).

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