JEE PYQ: Hyperbola Question 6
Question 6 - 2020 (07 Sep Shift 1)
A line parallel to the straight line $2x - y = 0$ is tangent to the hyperbola $\frac{x^2}{4} - \frac{y^2}{2} = 1$ at the point $(x_1, y_1)$. Then $x_1^2 + 5y_1^2$ is equal to:
(1) 6
(2) 8
(3) 10
(4) 5
Show Answer
Answer: (1)
Solution
Tangent at $(x_1, y_1)$: $\frac{xx_1}{4} - \frac{yy_1}{2} = 1$. Slope $= \frac{x_1}{2y_1} = 2$, so $x_1 = 4y_1$. Since $(x_1, y_1)$ lies on the hyperbola: $\frac{x_1^2}{4} - \frac{y_1^2}{2} = 1$. With $x_1 = 4y_1$: $4y_1^2 - \frac{y_1^2}{2} = 1$, giving $y_1^2 = \frac{2}{7}$ and $x_1^2 = \frac{32}{7}$. So $x_1^2 + 5y_1^2 = \frac{32}{7} + \frac{10}{7} = 6$.
BETA
