JEE PYQ: Parabola Question 30
Question 30 - 2019 (12 Jan Shift 1)
The equation of a tangent to the parabola, $x^2 = 8y$, which makes an angle $\theta$ with the positive direction of $x$-axis, is:
(1) $y = x\tan\theta + 2\cot\theta$
(2) $y = x\tan\theta - 2\cot\theta$
(3) $x = y\cot\theta + 2\tan\theta$
(4) $x = y\cot\theta - 2\tan\theta$
Show Answer
Answer: (3)
Solution
$x^2 = 8y$, $a = 2$. Parametric point $P(2at, at^2) = (4t, 2t^2)$. Tangent: $tx = y + 2t^2$. Slope $= t = \tan\theta$. So $x = \frac{y}{t} + 2t = y\cot\theta + 2\tan\theta$.
BETA
