First slide

Ellipse

Question

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

Moderate

A

$\frac{2}{3}$
B

$- \frac{2}{3}$
C

$\frac{3}{2}$
D

$- \frac{3}{2}$

Solution

$The normal at P (a \cos θ, b \sin θ) is$
$\frac{a x}{\cos θ} - \frac{b x}{\sin θ} = a^{2} - b^{2}, where a^{2} = 14, b^{2} = 5$
$It meets the curve again at Q (2 θ), i.e., (a \cos 2 θ, b \sin 2 θ)$
$∴ \frac{a}{\cos θ} (a \cos 2 θ) - \frac{b}{\sin θ} (b \sin 2 θ) = a^{2} - b^{2}$
$\begin{array}{l} \Rightarrow \frac{14}{\cos θ} (\cos 2 θ) - \frac{5}{\sin θ} (\sin 2 θ) = 14 - 5 \\ \Rightarrow 28 \cos^{2} θ - 14 - 10 \cos^{2} θ = 9 \cos θ \\ \Rightarrow 18 \cos^{2} θ - 9 \cos θ - 14 = 0 \\ \Rightarrow (6 \cos θ - 7) (3 \cos θ + 2) = 0 \Rightarrow \cos θ = - \frac{2}{3} \end{array}$

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