JEE PYQ: Hyperbola Question 11
Question 11 - 2019 (08 Apr Shift 2)
If the eccentricity of the standard hyperbola passing through the point $(4, 6)$ is 2, then the equation of the tangent to the hyperbola at $(4, 6)$ is:
(1) $x - 2y + 8 = 0$
(2) $2x - 3y + 10 = 0$
(3) $2x - y - 2 = 0$
(4) $3x - 2y = 0$
Show Answer
Answer: (3)
Solution
$e = 2$, so $b^2 = a^2(e^2 - 1) = 3a^2$. Passes through $(4,6)$: $\frac{16}{a^2} - \frac{36}{3a^2} = 1$, so $\frac{16 - 12}{a^2} = 1$, $a^2 = 4$, $b^2 = 12$. Tangent at $(4,6)$: $\frac{4x}{4} - \frac{6y}{12} = 1$, i.e. $x - \frac{y}{2} = 1$, so $2x - y - 2 = 0$.
BETA
