Question

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

• A. $2/3$
• B. $-2/3$
• C. $3/4$
• D. None of these

Answer 1

The correct option is B $-2/3$

Normal at the point P(theta) to the ellipse $\frac{x²}{14}+\frac{y²}{5}=1$ intersects it again at the point $Q\left(2\theta \right).$

we know, standard equation of ellipse is

$\frac{x²}{a²}+\frac{y²}{b²}=1$ compare it with given equation

so, $a²=14$ then, $a=\surd 14$

$b²=5$ then, $b=\surd 5$

now equation of normal passing through point $P\left(\theta \right)$ is given by,

$\frac{ax}{cos\theta }-\frac{by}{sin\theta }=a²-b².$

or, $\frac{\sqrt{1}4x}{cos\theta }-\frac{\sqrt{5}y}{sin\theta }=14-5=9$ ....(1)

it again meets the curve at the point $Q\left(2\theta \right)$

so, $Q\left(2\theta \right)=(\surd 14cos2\theta ,\surd 5sin2\theta )$

now, put it in equation (1),

or, $\frac{14cos2\theta }{\cos \theta }-\frac{5sin2\theta }{\sin \theta }=9$

or, $14(2cos²\theta -1)\cos \theta -\frac{10sin\theta cos\theta }{\sin \theta }=9$

or, $28cos\theta -14sec\theta -10cos\theta =9$

or, $18cos\theta -\frac{14}{cos\theta }=9$

or, $18cos²\theta -14-9cos\theta =0$

or, $18cos²\theta -21cos\theta +12cos\theta -14=0$

or, $3cos\theta (6cos\theta -7)+2(cos\theta -7)=0$

or, $(3cos\theta +2)(6cos\theta -7)=0$

or, $cos\theta =\frac{-2}{3}$