Question

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 :

• A. \(\frac{3}{\sqrt{5}}\)
• B. \(\frac{\sqrt{15}}{2}\)
• C. \(\frac{\sqrt{5}}{2}\)
• D. \(\frac{2}{\sqrt{5}}\)

Hint:

Use the condition for the line \(y = mx + c\) to be a normal to \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\): \(c^2 = \frac{m^2(a^2+b^2)^2}{a^2 - b^2m^2}\).

Answer 1

The Correct Option is D

Here \(a^2 = 24,\ b^2 = 18,\ a^2 + b^2 = 42\) and \(c = 7\sqrt{3}\), so \(c^2 = 147\).
Condition for normality: \(c^2 = \frac{m^2(a^2+b^2)^2}{a^2 - b^2m^2}\).
\(147 = \frac{1764\,m^2}{24 - 18m^2}\)
\(147(24 - 18m^2) = 1764\,m^2\)
\(3528 - 2646m^2 = 1764m^2\)
\(4410\,m^2 = 3528\)
\(m^2 = \frac{3528}{4410} = \frac{4}{5}\)
\(m = \pm\frac{2}{\sqrt{5}}\). Among the options only \(\frac{2}{\sqrt{5}}\) appears.
Answer: (4) \(\frac{2}{\sqrt{5}}\)