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The number of normals that can be drawn from a point to an ellipse $\mathrm{\frac{x^2}{a^2}+\frac{y^2}{b^2}=1}$ is

Option: 1
$4$

Option: 2
$3$

Option: 3
$2$

Option: 4
$1$

Answers (1)

best_answer

The equation of normal at $\mathrm{P(a \cos \theta, b \sin \theta)}$ on

$\mathrm{\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,}$ is

$\mathrm{a x \sec \theta-b y \operatorname{cosec} \theta=a^2-b^2}$

Let it be drawn from $\mathrm{M\left(x_1, y_1\right)}$ . The condition is

$\mathrm{a x_1 \sec \theta-b y_1 \operatorname{cosec} \theta=a^2-b^2}$

Put $\mathrm{ \tan \frac{\theta}{2}=t.}$

$\mathrm{ \Rightarrow \quad \sin \theta=\frac{2 t}{1+t^2} \: and \: \cos \theta=\frac{1-t^2}{1+t^2}}$

Condition (1) becomes

$\mathrm{ a x_1\left(\frac{1+t^2}{1-t^2}\right)-b y_1 \frac{1+t^2}{2 t}=a^2-b^2 }$

$\mathrm{ Or \quad b y_1 t^4+2 t^3\left(a x_1+a^2 e^2\right)+2 t\left(a x_1-a^2 e^2\right)-b y_1=0 }$ ........(2)

Let its roots be $\mathrm{ t_1, t_2, t_3, t_4. }$

Then, the number of points (like P ) where normals pass through $\mathrm{ M\left(x_1, y_1\right) }$ is four.

Hence option 1 is correct.

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