If the normal at the point P(θ) to the ellipse (x2/14) + (y2/5) = 1 intersects it again at the point Q(2θ), then cos θ equals to

Question

If the normal at the point P(θ) to the ellipse (x2/14) + (y2/5) = 1 intersects it again at the point Q(2θ), then cos θ equals to (A)  -(2/3)  (B)  (2/3)  (C)  (3/2)  (D)  -(3/2)

Tardigrade Admin · Accepted Answer

Any point on the ellipse is (√14 cos θ, √5 sin θ) ∴ Equation of normal at (√14 cos θ, √5 sin θ) is √14 x sec θ, √5y cosec θ = 9 Since, it passes through (√14 cos 2θ, √5 sin 2θ) ∴ √14√14 cos 2θ sec θ - √5 √5 sin 2θ cosec θ = 9 ⇒ 14 (cos 2θ/cos θ) - 5 (sin θ/sin θ) = 9 ⇒ 14(2 cos2 θ -1) - 10 cos2 θ = 9 cos θ ⇒ 18 cos2θ - 9 cos θ - 14 = 0 ⇒ (3 cos θ + 2)(6 cos θ - 7) = 0 ⇒ cos θ = -(2/3), cos θ ≠ (7/6)

Q. If the normal at the point $P (θ)$ to the ellipse $\frac{x ^{2}}{14} + \frac{y ^{2}}{5} = 1$ intersects it again at the point $Q (2 θ),$ then $cos θ$ equals to

$- \frac{2}{3}$

$\frac{2}{3}$

$\frac{3}{2}$

$- \frac{3}{2}$

Q. If the normal at the point P(θ) to the ellipse 14x2​+5y2​=1 intersects it again at the point Q(2θ), then cosθ equals to

−32​

32​

23​

−23​

Q. If the normal at the point $P (θ)$ to the ellipse $\frac{x ^{2}}{14} + \frac{y ^{2}}{5} = 1$ intersects it again at the point $Q (2 θ),$ then $cos θ$ equals to

$- \frac{2}{3}$

$\frac{2}{3}$

$\frac{3}{2}$

$- \frac{3}{2}$