The line 2 x + y = 3 cuts the ellipse 4 x 2 + y 2 = 5 at points P and Q . If θ is the acute angle between the normals at P and Q , then θ is equal to

Step-by-step solution

Solving y = 3 - 2 x and 4 x 2 + y 2 = 5 , we get the intersection points P 1 2 , 2 and Q 1,1 The equations of the tangents at these points are 4 x 1 2 + 2 y = 5 and 4 x 1 + y 1 = 5 ⇒ The slopes of the tangents at P & Q are - 1 & - 4 respectively ⇒ The slopes of the normals at P & Q are 1 & 1 4 respectively ⇒ tan ⁡ θ = 1 - 1 4 1 + 1 1 4 = 3 5 ⇒ θ = tan - 1 ⁡ 3 5 ⇒ cos ⁡ θ = 5 34 , sin ⁡ θ = 3 34 ⇒ θ = sin - 1 ⁡ 3 34