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#### The tangents drawn at those points of the ellipse $\mathrm{\frac{x^2}{a}+\frac{y^2}{b}=(a+b)}$, where it is cut by any tangent to  $\mathrm{\frac{x^2}{a^2}+\frac{y^2}{b^2}=1}$ intersect at.Option: 1 $75^{\circ}$Option: 2 $90^{\circ}$Option: 3 $60^{\circ}$Option: 4 $120^{\circ}$

The given ellipses are
$\mathrm{\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 }$......................(1)
$\mathrm{\frac{x^2}{a(a+b)}+\frac{y^2}{b(a+b)}=1 }$.....................(2)

Chord of contact of $\mathrm{\left(x_1, y_1\right) }$ w.r.t. (2) ellipse is
\mathrm{\begin{aligned} & \frac{x x_1}{a(a+b)}+\frac{y y_1}{b(a+b)}=1 \\ & \text { or } \quad \mathrm{lx}+\mathrm{my}=\mathrm{n} \text { (say) } \\ & \text { If it touches } \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \\ & \end{aligned} }
then $\mathrm{a^2 1^2+b^2 m^2=n^2 \quad from corollary\} }$
$\mathrm{\begin{array}{ll} \Rightarrow \quad & \frac{a^2 x_1^2}{a^2(a+b)^2}+\frac{b^2 y_1^2}{b^2(a+b)^2}=1 \\ \Rightarrow \quad & x_1^2+y_1^2=(a+b)^2 \end{array}}$
locus of T $\mathrm{\left(x_1, y_1\right)}$ is
$\mathrm{x^2+y^2=(a+b)^2}$