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Q.
The equation of the common tangent to the parabolas $y^{2}=4ax$ and $x^{2}=4by$ is given by
NTA AbhyasNTA Abhyas 2022
Solution:
Any tangent to the parabola $y^{2}=4ax$ is
$y=mx+\frac{a}{m}\ldots \left(\right.1\left.\right)$
$\text{ meets }x^{2}=4by\text{ where }x^{2}=4b\left(mx + \frac{a}{m}\right)$
$\text{ or }\left(mx\right)^{2}-4\left(bm\right)^{2}x-4ab=0\ldots \left(\right.2\left.\right)$
If the line ( $1$ ) be a tangent to the second parabola, then roots of ( $2$ ) must be equal. The condition for this is
$16b^{2}m^{4}+16abm=0\text{ or }m^{3}=-\frac{a}{b} \\ \therefore m=-\left(\frac{a}{b}\right)^{1 / 3}=-\frac{a^{1 / 3}}{b^{1 / 3}}$
Substituting this value of $m$ in $\left(\right.1\left.\right),$ we $get$
$y=-\frac{a^{1 / 3}}{b^{1 / 3}}x+a\left(- \frac{b^{1 / 3}}{a^{1 / 3}}\right) \\ \text{ i. }e\cdot \left(xa\right)^{1 / 3}+\left(yb\right)^{1 / 3}+a^{2 / 3}b^{2 / 3}=0$
which is the required equation of the common tangent.