How many tangents to the circle $x^{2} + y^{2} = 3$ are there which are normal to the ellipse $\frac{x^{2}}{9} + \frac{y^{2}}{4} = 1$

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Solution

The correct option is D \N
Equation of normal at $p (3 c o s θ, 2 s i n θ)$ is $3 x s e c θ - 2 y c o s e c θ = 5$
$\frac{5}{\sqrt{9 s e c^{2} θ + 4 c o s e c^{2} θ}} = \sqrt{3} B u t m i n i m u m v a l u e o f 9 s e c^{2} θ + 4 c o s e c^{2} θ = 25 ∴ n o s u c h θ^{- 1} e x i s t s$
So the number of tangents to the circle $x^{2} + y^{2} = 3$ which are normal to the ellipse $\frac{x^{2}}{9} + \frac{y^{2}}{4} = 1$ is 0.

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Similar questions

x^{2} + y^{2} = 3

\frac{x^{2}}{9} + \frac{y^{2}}{4}

\frac{x^{2}}{9} + \frac{y^{2}}{2} = 1

x^{2} + y^{2} = 3

x^{2} + y^{2} = 25

\frac{x^{2}}{16} + \frac{y^{2}}{9} = 1,

(4, \sqrt{3})

\frac{x^{2}}{16} + \frac{y^{2}}{9} = 1

(4, \sqrt{3})

\frac{x^{2}}{16} + \frac{y^{2}}{9} = 1.

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