Question

If the normal to the ellipse 3x² + 4y² = 12 at a point P on it is parallel to the line, 2x + y = 4 and the tangent to the ellipse at P passes through Q(4, 4) then PQ is equal to :
(A) √157/2
(B) 5√5/2
(C) √221/2
(D) √61/2

Answer 1

Answer:

The given ellipse is x2 + 3y2 = 6

Or x2/6 + y2/2 = 1 ...... (1)

If the coordinates of the required point on the ellipse (1) be (√6 cos Φ, √2 sin Φ) then the tangent at the point is x/√6 cos Φ + y/√2 sin Φ = 1 ...... (2)

Slope of (2) = (-cos Φ)/√6 ×√2/(sin Φ ) = (-√2)/√6 cot Φ

As the tangents are equally inclined to the axes so we have

-1/√3 cot Φ = + tan 45o = + 1

Hence, tan Φ = + 1/√3

The coordinates of the required points are

(±√6 × √3/2, ±√2 × 1/2) and (±√6 × √3/2, ±√2 × 1/2)

= (± (3√2)/2, ±1/√2) and (± (3√2)/2, ± 1/√2)

Again the length of perpendicular from (0, 0) and (2),

= (√6.√2)/√(2 cos2Φ + 6 sin2Φ)

= (2√3)/√((2.3/4) + (6.1/4) )

= (2√3)/√3

= 2.