Hyperbola
Normal Properties
Grade 11
Question:
<p>If the normal to the hyperbola \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\) at any point \(P(a\sec\theta, b\tan\theta)\) meets the transverse and conjugate axes in G and g respectively and if F is the foot of perpendicular to the normal at P from the centre C, then the value of \((PF)^2\) is:</p>
<p>(a) \(\frac{a^2b^2}{b^2\sec^2\theta + a^2\tan^2\theta}\)</p>
<p>(b) \(\frac{a^2b^2}{b^2\tan^2\theta + a^2\sec^2\theta}\)</p>
<p>(c) \(\frac{a^2b^2}{b^2\cosec^2\theta + a^2\cot^2\theta}\)</p>
<p>(d) \(\frac{a^2b^2}{b^2\cot^2\theta + a^2\cosec^2\theta}\)</p>
Step-by-Step Solution
Key Concept: The perpendicular distance from the centre to the normal at any point on the hyperbola follows a specific formula involving trigonometric functions.
<p>The foot of perpendicular from the centre C to the normal at P divides the normal in a specific ratio. Using the perpendicular distance formula from a point to a line, we get \((PF)^2 = \frac{a^2b^2}{b^2\tan^2\theta + a^2\sec^2\theta}\).</p>
Correct Answer: B