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 \((PG)^2\) is:</p>
<p>(a) \(\frac{b^2}{a^2}(b^2\cosec^2\theta + a^2\cot^2\theta)\)</p>
<p>(b) \(\frac{a^2}{b^2}(b^2\cosec^2\theta + a^2\cot^2\theta)\)</p>
<p>(c) \(\frac{b^2}{a^2}(b^2\sec^2\theta + a^2\tan^2\theta)\)</p>
<p>(d) \(\frac{b^2}{a^2}(b^2\tan^2\theta + a^2\sec^2\theta)\)</p>
Step-by-Step Solution
Key Concept: The normal to a hyperbola at a point on the curve intersects the axes at specific points whose distances from P can be calculated using coordinate geometry.
<p>The normal at point P on the hyperbola meets the transverse axis at G. Using the distance formula and properties of the normal to a hyperbola, we can calculate PG. The result is \((PG)^2 = \frac{b^2}{a^2}(b^2\cosec^2\theta + a^2\cot^2\theta)\).</p>
Correct Answer: A