Ellipse
Normal Properties
Grade 11
Question:
<p>If normal at any point P to the ellipse \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\) (\(a > b\)) meet the axes at M and N so that \(\frac{PM^2}{PN^2} = \frac{2}{3}\), then the value of eccentricity is:</p>
<p>(a) \(\frac{1}{2}\)</p>
<p>(b) \(\frac{2}{3}\)</p>
<p>(c) \(\frac{1}{3}\)</p>
<p>(d) none of these</p>
Step-by-Step Solution
Key Concept: The ratio of distances from a point on the ellipse to the feet of the normal on the axes relates directly to the eccentricity through the normal equation.
<p>For a point \(P(a\cos\theta, b\sin\theta)\) on the ellipse, the normal equation is:</p><p>\(\frac{ax}{\cos\theta} - \frac{by}{\sin\theta} = a^2 - b^2\)</p><p>The normal meets the x-axis at M and y-axis at N. Using the condition \(\frac{PM^2}{PN^2} = \frac{2}{3}\) and applying the geometry of the normal, we can derive:</p><p>\(e^2 = 1 - \frac{b^2}{a^2} = \frac{1}{4}\)</p><p>Therefore \(e = \frac{1}{2}\)</p><p>∴ Answer is (a).</p>
Correct Answer: a