JEE PYQ: Hyperbola Question 12
Question 12 - 2019 (09 Apr Shift 1)
If the line $y = mx + 7\sqrt{3}$ is normal to the hyperbola $\frac{x^2}{24} - \frac{y^2}{18} = 1$, then a value of $m$ is:
(1) $\frac{\sqrt{5}}{2}$
(2) $\frac{\sqrt{15}}{2}$
(3) $\frac{2}{\sqrt{5}}$
(4) $\frac{3}{\sqrt{5}}$
Show Answer
Answer: (3)
Solution
For $lx + my + n = 0$ normal to $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$: $\frac{a^2}{l^2} - \frac{b^2}{m^2} = \frac{(a^2+b^2)^2}{n^2}$. Here $a^2 = 24$, $b^2 = 18$, line: $mx - y + 7\sqrt{3} = 0$. So $\frac{24}{m^2} - \frac{18}{1} = \frac{(42)^2}{(7\sqrt{3})^2} = \frac{1764}{147} = 12$. Thus $\frac{24}{m^2} = 30$, $m^2 = \frac{4}{5}$, $m = \frac{2}{\sqrt{5}}$.
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