JEE PYQ: Hyperbola Question 10
Question 10 - 2020 (08 Jan Shift 2)
If a hyperbola passes through the point $P(10, 16)$ and it has vertices at $(\pm 6, 0)$, then the equation of the normal to it at $P$ is:
(1) $3x + 4y = 94$
(2) $2x + 5y = 100$
(3) $x + 2y = 42$
(4) $x + 3y = 58$
Show Answer
Answer: (2)
Solution
$a = 6$. From $\frac{100}{36} - \frac{256}{b^2} = 1$: $\frac{256}{b^2} = \frac{64}{36}$, so $b^2 = 144$. Hyperbola: $\frac{x^2}{36} - \frac{y^2}{144} = 1$. Normal at $P(10,16)$: $\frac{36x}{10} + \frac{144y}{16} = 36 + 144 = 180$, i.e. $\frac{18x}{5} + 9y = 180$, so $18x + 45y = 900$, giving $2x + 5y = 100$.
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