Let F₁ & F₂ be foci of an ellipse frac{{{x^2}}}{4} + frac{{{y^2}}}{9} = 1 such that a

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10-2. Parabola, Ellipse, Hyperbola

normal

Let $F_1$ & $F_2$ be the foci of an ellipse $\frac{{{x^2}}}{4} + \frac{{{y^2}}}{9} = 1$ such that a ray from $F_1$ strikes the elliptical mirror at the point $P$ and get reflected. Then equation of angle bisector of the angle between incident ray and reflected ray can be

$y = x + \frac{5}{{\sqrt {13} }}$

$y = 2x - \frac{5}{{\sqrt {13} }}$

$x + y -5 = 0$

$3x -4y -5 = 0$

angle bisectors are either tangents or normals to

the ellipse $\mathrm{y}=\mathrm{mx} \pm \sqrt{\mathrm{a}^{2} \mathrm{m}^{2}+\mathrm{b}^{2}}$

or $\mathrm{y}=\mathrm{mx} \pm \frac{\left(\mathrm{a}^{2}-\mathrm{b}^{2}\right) \mathrm{m}}{\sqrt{\mathrm{a}^{2}+\mathrm{m}^{2} \mathrm{b}^{2}}}$

Now put $\mathrm{m}=1$

$y=x \pm \frac{5}{\sqrt{13}}$

Standard 11

Mathematics

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easy

normal

medium

hard