The point of intersection of two perpendicular tangents drawn to the ellipse $x^{2} + 2 y^{2} = 10$ will lie on

x^{2} + y^{2} = 5

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x^{2} + 2 y^{2} = 10

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x^{2} - 2 y^{2} = 10

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x^{2} + y^{2} = 15

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Solution

The correct option is A $x^{2} + y^{2} = 15$
point of intersection of two perendicular tangents is director circle of that ellipse whose equation is $x^{2} + y^{2} = a^{2} + b^{2}$
Ellipse $= x^{2} + 2 y^{2} = 10$
$\frac{x^{2}}{10} + \frac{y^{2}}{5} = 1$
director circle (locus) $= x^{2} + y^{2} = 10 + 5$ $x^{2} + y^{2} = 15$
D is correct.

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The point of intersection of two perpendicular tangents drawn to the ellipse x2+2y2=10 will lie on

The correct option is A x2+y2=15point of intersection of two perendicular tangents is director circle of that ellipse whose equation is x2+y2=a2+b2 Ellipse =x2+2y2=10 x210+y25=1 director circle (locus) =x2+y2=10+5 x2+y2=15 D is correct.

The point of intersection of two perpendicular tangents drawn to the ellipse $x^{2} + 2 y^{2} = 10$ will lie on