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Statement $-1$ : If two tangents are drawn to an ellipse from a single point and if they are perpendicular to each other, then locus of that point is always a circle
Statement $-2$ : For an ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$ , locus of that point from which two perpendicular tangents are drawn, is $x^2 + y^2 = (a + b)^2$ .
Statement $-1$ is true, statement $-2$ is true but statement $-1$ is not the correct explanation for statement $-2$
Statement $-1$ is true, statement $-2$ is false
Statement $-1$ is false, statement $-2$ is true
Both statements are true, and statement $-1$ is the true explanation of statement $-2$
Solution
$y=m x \pm \sqrt{a^{2} m^{2}+b^{2}}$
passing through a point $(\mathrm{h}, \mathrm{k})$
$\mathrm{k}=\mathrm{mh} \pm \sqrt{\mathrm{a}^{2} \mathrm{m}^{2}+\mathrm{b}^{2}}$
$m^{2}\left(a^{2}-h^{2}\right)+2 m h k-\left(k^{2}-b^{2}\right)=0$
$\because$ tangents are $\perp$
$\mathrm{so}, \mathrm{m}_{1} \cdot \mathrm{m}_{2}=-1$
$m_{1} \cdot m_{2}=c / a=\frac{-k^{2}-b^{2}}{a^{2}-h^{2}}=-1$
so, $x^{2}+y^{2}=a^{2}+b^{2}$
