The locus of the point of intersection of the perpendicular tangents to the ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$ is
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The locus of the point of intersection of the perpendicular tangents to the ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$ is

The locus of the point of intersection of the perpendicular tangents to the ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$ is
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The locus of the point of intersection of the perpendicular tangents to the ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$ is

(a) $x^2+y^2=9$

(b) $x^2+y^2=4$

(c) $x^2+y^2=13$

(d) $x^2+y^2=5$

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SOLUTION — Here, $a^2=9, b^2=4$

$\therefore \quad$ Locus is $x^2+y^2=a^2+b^2$

$\Rightarrow x^2+y^2=9+4 \Rightarrow x^2+y^2=13$

So, The correct option of this question will be (C).

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