If the normal to the ellipse 3 x 2 + 4 y 2 = 12 at a point P on it is parallel to the line, 2 x + y = 4 and the tangent to the ellipse at P passes through Q ( 4,4 ) then P Q is equal to:

Step-by-step solution

Given the equation of an ellipse is x 2 4 + y 2 3 = 1 ,where a = 2 ,   b = 3 . As we know equation of normal to the ellipse x 2 a 2 + y 2 b 2 = 1 can be taken as a x sec ⁡ θ - b y cosec ⁡ θ = a 2 - b 2 ⇒ 2 x sec ⁡ θ - 3 y cosec ⁡ θ = 1 for the given ellipse. Now, this normal is parallel to 2 x + y = 5 . ⇒ - 2 sec ⁡ θ 3 cosec ⁡ θ = - 2 ⇒ tan ⁡ θ = - 3 ⇒ θ = 2 π 3 , - π 3 Hence point P a cos &#8289