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The common tangent to the parabola ${y^2} = 4ax$ and ${x^2} = 4ay$ is
  {\text{A}}{\text{. }}x + y + a = 0 \\
  {\text{B}}{\text{. }}x + y - a = 0 \\
  {\text{C}}{\text{. }}x - y + a = 0 \\
  {\text{D}}{\text{. }}x - y - a = 0 \\

Last updated date: 09th Apr 2024
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MVSAT 2024
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Hint: Whenever you get the questions of common tangent you have to write the equation of tangent to a given curve and then apply the condition of tangent for others. Here in case of parabola tangent means that equation has equal apply the condition and get a common tangent.

The equation of any tangent to ${y^2} = 4ax$ is $y = mx + \dfrac{a}{m}$,if it touches ${x^2} = 4ay$.
We know tangent touches at a single point so in case of parabola it is a quadratic equation and touches at single point means it has real and equal roots.
Then the equation ${x^2} = 4a\left( {mx + \dfrac{a}{m}} \right)$ has equal roots or, $m{x^2} - 4a{m^2}x - 4{a^2} = 0$ has equal roots.
We know condition of equal roots $
  D = 0 \\
  {\text{i}}{\text{.e }}{{\text{b}}^2} - 4ac = 0 \\ )$
$ \Rightarrow {b^2} = 16{a^2}{m^4},4ac = - 16{a^2}{m^4}$
$ \Rightarrow 16{a^2}{m^4} = - 16{a^2}{m^4} \Rightarrow m = - 1\left( {\because m \ne 0} \right)$
Putting $m = - 1$ in $y = mx + \dfrac{a}{m}$, we get $y = - x - a$
Or, $x + y + a = 0$
Hence option ${\text{A}}$ is correct.

Note: The key concept of solving questions of common tangent is first select a curve and write any general tangent to it and then apply the condition according to the second curve given in question. If the second curve is a circle then distance from center to tangent will be it’s radius but here in case of parabola equal roots will be conditioned.