JEE Advanced 2013 · Paper 2 · Mathematics Q56

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MathematicsSingle correct+3 / -1Question 56 of 60

Passage

Paragraph for Questions 55 and 56

Let $PQ$ be a focal chord of the parabola $y^2=4ax$ . The tangents to the parabola at $P$ and $Q$ meet at a point lying on the line $y=2x+a$ , $a>0$ .

If chord $PQ$ subtends an angle $\theta$ at the vertex of $y^2=4ax$ , then $\tan\theta=$

\dfrac23\sqrt7

\dfrac{-2}3\sqrt7

\dfrac23\sqrt5

\dfrac{-2}3\sqrt5

Answer and worked solution

Official answer: D

Slopes from the vertex. $OP$ and $OQ$ have slopes $\frac2{t_1}$ and $\frac2{t_2}$ .
$\tan\theta=\frac{\frac2{t_1}-\frac2{t_2}}{1+\frac4{t_1t_2}}=\frac{2(t_2-t_1)/(t_1t_2)}{1-4}=\frac{2(t_2-t_1)}3.$

Sign. $1+\frac4{t_1t_2}=-3<0$ , so the angle at the vertex is obtuse; with $|t_1-t_2|=\sqrt5$ , $\tan\theta=-\frac23\sqrt5$ .

(D)

Topics:Parabola Angle between lines

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JEE Advanced 2013 · Paper 2, Mathematics question 56