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

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a

-23

b

23

c

-67

d

67

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detailed solution

Correct option is A

Normal at P(14cos⁡ θ,5sin⁡ θ) to the ellipse is14xcos⁡ θ−5ysin⁡ θ=14−5=9Since it passes through Q(2θ)14×14cos⁡ 2θcos⁡ θ−5×5sin⁡ 2θsin⁡ θ=9⇒142cos2⁡ θ−1cos⁡ θ−5×2sin⁡ θcos⁡ θsin⁡ θ=9⇒18cos2⁡ θ−9cos⁡ θ−14=0⇒(3cos⁡ θ+2)(6cos⁡θ−7)=0⇒cos⁡ θ=−2/3.

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