Let F_1 and F_2 be the foci of an ellipse fra | vedclass

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(A) For an ellipse,the reflected ray from one focus passes through the other focus.Therefore,the incident ray is $PF_1$ and the reflected ray is $PF_2

Vedclass · Accepted Answer

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

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(A) For an ellipse,the reflected ray from one focus passes through the other focus.Therefore,the incident ray is $PF_1$ and the reflected ray is $PF_2$.The angle bisector of the angle between the incident ray $PF_1$ and the reflected ray $PF_2$ is the normal to the ellipse at point $P$.The equation of the ellipse is $\frac{x^2}{4} + \frac{y^2}{9} = 1$,where $a^2 = 4$ and $b^2 = 9$.Since $b^2 > a^2$,the major axis is along the $y$-axis.The equation of the normal to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with slope $m$ is $y = mx \pm \frac{(a^2 - b^2)m}{\sqrt{a^2 + b^2m^2}}$.Substituting $a^2 = 4$,$b^2 = 9$,and $m = 1$ (assuming the bisector has slope $1$ as per options):$y = x \pm \frac{(4 - 9)(1)}{\sqrt{4 + 9(1)^2}} = x \pm \frac{-5}{\sqrt{13}}$.Thus,the equation is $y = x \pm \frac{5}{\sqrt{13}}$.

Let $F_1$ and $F_2$ be the foci of an ellipse $\frac{x^2}{4} + \frac{y^2}{9} = 1$. $A$ ray from $F_1$ strikes the elliptical mirror at point $P$ and is reflected. What is the equation of the angle bisector of the angle between the incident ray and the reflected ray?

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