Any ordinate $MP$of an ellipse meets the auxiliary circle in $Q$ Then the locus of the point of intersection of the normal at $P$ and $Q$ is the  . 

Let the equation of ellipse be $\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}+\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1$ then equation of its auxiliary circle will be $x^2+y^2=a^2$

Let Coordinates of $P$ and $Q$ are $(a\cos \varphi ,b\sin \varphi )$ and $(a\cos \varphi ,a\sin \varphi )$ respectively. 

Equation of normal at $P$ to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is $ax\sec \varphi -bycosec\varphi =a^2-b^2$

and equation of normal at $Q$ to the circle $x^2+y^2=a^2$ is $y=x\tan \varphi$

 $\Rightarrow \tan \varphi =\frac{y}{x}$

So, $cosec\varphi =\frac{\sqrt{\left(x^2+y^2\right)}}{y}$ and $\sec \varphi =\frac{\sqrt{\left(x^2+y^2\right)}}{x}$

Substituting the values of $\sec \varphi$ and $cosec\varphi$ in equation of normal we get ,$ax\times \frac{\sqrt{\left(x^2+y^2\right)}}{x}-by\times \frac{\sqrt{\left(x^2+y^2\right)}}{y}=a^2-b^2$

$\Rightarrow (a-b)\sqrt{\left(x^2+y^2\right)}=(a+b)(a-b)$

$\Rightarrow \sqrt{x^2+y^2}=a+b$

Squaring both sides we get, $x^2+y^2=(a+b{)}^2$ which is required locus.

Hence option-$1$ is the correct answer.