If √3 b x+a y=2 a b touches the ellipse (x2/a2)+(y2/b2)=1 then the eccentric angle θ of the point of contact is

Question

If √3 b x+a y=2 a b touches the ellipse (x2/a2)+(y2/b2)=1 then the eccentric angle θ of the point of contact is (A) (π/6) (B) (π/4) (C) (π/3) (D) (π/2)

Tardigrade Admin · Accepted Answer

Let point of contact be ( a cos θ, b sin θ) ∴ equation of tangent be (x/a) cos θ+(y/b) sin θ=1 equation of given line is (x/a) ⋅ (√3/2)+(y/b) ⋅ (1/2)=1 ∴ cos θ=(√3/2) and sin θ=(1/2) ⇒ θ=(π/6)

Q. If $3 b x + a y = 2 ab$ touches the ellipse $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$ then the eccentric angle $θ$ of the point of contact is

$\frac{π}{6}$

$\frac{π}{4}$

$\frac{π}{3}$

$\frac{π}{2}$

Let point of contact be $(a cos θ, b sin θ)$
$∴$ equation of tangent be $\frac{x}{a} cos θ + \frac{y}{b} sin θ = 1$
equation of given line is $\frac{x}{a} \cdot \frac{3}{2} + \frac{y}{b} \cdot \frac{1}{2} = 1$
$∴ cos θ = \frac{3}{2}$ and $sin θ = \frac{1}{2} \Rightarrow θ = \frac{π}{6}$

Q. If 3​bx+ay=2ab touches the ellipse a2x2​+b2y2​=1 then the eccentric angle θ of the point of contact is

6π​

4π​

3π​

2π​