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  4. Slope of normal to the ellipse at a point P is (3/4) and eccentricity of ellipse is (1/3) . If this normal makes acute angle 'β ' with its focal chord through P , then sinβ is

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Q. Slope of normal to the ellipse at a point $P$ is $\frac{3}{4}$ and eccentricity of ellipse is $\frac{1}{3}$ . If this normal makes acute angle $'\beta '$ with its focal chord through $P$ , then $sinβ$ is

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Solution:

Solution
$sinα=\frac{PM}{PN}$
$sinβ=\frac{PS}{PN}$
$\frac{sinβ}{sinα}=\frac{PS}{PM}=\frac{1}{3}$
$sinβ=\frac{1}{3}\cdot sinα=\frac{1}{3}\cdot \frac{3}{5}=\frac{1}{5}$