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 sinÎ² is

Question

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 sinÎ² is (A) (1/4) (B) (3/10) (C) (1/5) (D) (1/2)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/m-d9dwgkrsjhcezo0t.jpg" /> sinÎ±=(PM/PN) sinÎ²=(PS/PN) (sinÎ²/sinÎ±)=(PS/PM)=(1/3) sinÎ²=(1/3)⋅ sinÎ±=(1/3)⋅ (3/5)=(1/5)

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 $^{'} β^{'}$ with its focal chord through $P$ , then $s in β$ is

$\frac{1}{4}$

$\frac{3}{10}$

$\frac{1}{5}$

$\frac{1}{2}$

$s in α = \frac{PM}{PN}$
$s in β = \frac{PS}{PN}$
$\frac{s in β}{s in α} = \frac{PS}{PM} = \frac{1}{3}$
$s in β = \frac{1}{3} \cdot s in α = \frac{1}{3} \cdot \frac{3}{5} = \frac{1}{5}$

Q. Slope of normal to the ellipse at a point P is 43​ and eccentricity of ellipse is 31​ . If this normal makes acute angle ′β′ with its focal chord through P , then sinβ is

41​

103​

51​

21​