The eccentric angle of the point where the line, 5 x-3 y=8 √2 is a normal to the ellipse (x2/25)+(y2/9)=1 is

Question

The eccentric angle of the point where the line, 5 x-3 y=8 √2 is a normal to the ellipse (x2/25)+(y2/9)=1 is (A) (3 π/4) (B) (π/4) (C) (π/6) (D)  tan -1 2

Tardigrade Admin · Accepted Answer

Normal to the ellipse (x2/25)+(y2/9)=1 at the point ' θ prime is (5 x/ cos θ)-(3 y/ sin θ)=16 Comparing with, 5 x-3 y=8 √2 cos θ= sin θ=(1/√2) ⇒ θ=(π/4)

Q. The eccentric angle of the point where the line, $5 x - 3 y = 82$ is a normal to the ellipse $\frac{x ^{2}}{25} + \frac{y ^{2}}{9} = 1$ is

$\frac{3 π}{4}$

$\frac{π}{4}$

$\frac{π}{6}$

$tan^{- 1} 2$

Normal to the ellipse $\frac{x ^{2}}{25} + \frac{y ^{2}}{9} = 1$ at the point ' $θ^{'}$ is $\frac{5 x}{c o s θ} - \frac{3 y}{s i n θ} = 16$
Comparing with, $5 x - 3 y = 82 cos θ = sin θ = \frac{1}{2} \Rightarrow θ = \frac{π}{4}$

Q. The eccentric angle of the point where the line, 5x−3y=82​ is a normal to the ellipse 25x2​+9y2​=1 is

43π​

4π​

6π​

tan−12