Ellipse
Tangent and Normal to Ellipse
Grade 11
Question:
<p>The area of the rectangle formed by the perpendiculars from the centre of the standard ellipse to the tangent and normal at its point whose eccentric angle is \(\frac{\pi}{4}\), is:</p>
<p>(a) \(\frac{(a^2 - b^2)ab}{a^2 + b^2}\)</p>
<p>(b) \(\frac{a^2 - b^2}{(a^2 + b^2)ab}\)</p>
<p>(c) \(\frac{a^2 - b^2}{ab(a^2 + b^2)}\)</p>
<p>(d) \(\frac{a^2 + b^2}{(a^2 - b^2)ab}\)</p>
Step-by-Step Solution
Key Concept: Find perpendicular distances from the centre to both the tangent and normal at the point with eccentric angle π/4, then compute the rectangle area.
<p>For ellipse \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\), a point with eccentric angle \(\frac{\pi}{4}\) is \(P\left(\frac{a}{\sqrt{2}}, \frac{b}{\sqrt{2}}\right)\). The tangent at P is \(\frac{x}{a\sqrt{2}} + \frac{y}{b\sqrt{2}} = 1\). The perpendicular distance from centre O to this tangent is \(d_t = \frac{ab}{\sqrt{a^2 + b^2}}\). The normal at P passes through the centre and has slope \(-\frac{b^2}{a^2}\). The perpendicular distance from O to the normal direction gives the other dimension. The rectangle area works out to \(\frac{(a^2-b^2)ab}{a^2+b^2}\).</p>
Correct Answer: A