If the polar with respect to $y^2=4ax$ touches the ellipse  $\frac{x^2}{{\alpha }^2}+\frac{y^2}{{\beta }^2}=1$, the locus of its pole is

 If  $P(x_1,y_1)$ is the pole , then polar p is

$y{y}_1=2a(x+x_1)\Rightarrow y=\frac{2a}{y_1}x+\frac{2a{x}_1}{y_1}$

Since this line touches $\frac{x^2}{{\alpha }^2}+\frac{y^2}{{\beta }^2}=1$ we get

${\left(\frac{2a{x}_1}{y_1}\right)}^2={\alpha }^2{\left(\frac{2a}{y_1}\right)}^2+{\beta }^2\Rightarrow 4a^2{x}_1^2=4a^2{\alpha }^2+{\beta }^2{y}_1^2$

$\Rightarrow 4a^2{x}_1^2-{\beta }_1^2{y}_1^2=4a^2{\alpha }^2\Rightarrow \frac{x_1^2}{{\alpha }^2}-\frac{{\beta }^2{y}_1^2}{4a^2{\alpha }^2}=1$

$\therefore$Locus of p is  $\frac{x^2}{{\alpha }^2}-\frac{y^2}{4a^2{\alpha }^2/{\beta }^2}=1$