Question

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

Answer 1

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 roots.so apply the condition and get a common tangent.

The equation of any tangent to $y^2=4ax$ is $y=mx+{\displaystyle \frac{a}{m}}$,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 $x^2=4a\left(mx+{\displaystyle \frac{a}{m}}\right)$ has equal roots or, $m{x}^2-4a{m}^{2}x-4a^2=0$ has equal roots.
We know condition of equal roots $$\begin{gathered} D=0 \\ \text{i}\text{.e }{\text{b}}^2-4ac=0 \\ ) \end{gathered}$$
$\Rightarrow b^2=16a^2{m}^4,4ac=-16a^2{m}^4$
$\Rightarrow 16a^2{m}^4=-16a^2{m}^4\Rightarrow m=-1\left(\because m\ne 0\right)$
Putting $m=-1$ in $y=mx+{\displaystyle \frac{a}{m}}$, 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.