Ellipse
Normal to Ellipse
Grade 11
Question:
<p>If the normal at one end of latus rectum of ellipse \( \dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1 \) passes from one end of minor axis and <em>e</em> is eccentricity of ellipse, then:</p>
<p>(a) \( e^2 + e + 1 = 0 \)</p>
<p>(b) \( e^4 - e^2 + 1 = 0 \)</p>
<p>(c) \( e^2 - e + 1 = 0 \)</p>
<p>(d) \( e^4 + e^2 - 1 = 0 \)</p>
Step-by-Step Solution
Key Concept: The normal at the end of latus rectum must pass through an endpoint of the minor axis. Use the latus rectum point coordinates (ae², ±b²/a) and the normal equation to establish a relationship between a and b through eccentricity.
<p><strong>Step 1:</strong> One end of latus rectum lies at point (ae², b²/a) where e² = 1 - b²/a².</p><p><strong>Step 2:</strong> The normal at point (x₀, y₀) on ellipse is: <br/>a²x/x₀ - b²y/y₀ = a² - b²</p><p><strong>Step 3:</strong> Normal at (ae², b²/a) passes through minor axis endpoint (0, b):<br/>a²(0)/(ae²) - b²(b)/(b²/a) = a² - b²<br/>-ab = a² - b²</p><p><strong>Step 4:</strong> Simplifying:<br/>-ab = a²(1 - b²/a²) = a²e²<br/>-ab = a²e²<br/>-b = ae²<br/>b²/a² = e⁴</p><p><strong>Step 5:</strong> Since b²/a² = 1 - e²:<br/>1 - e² = e⁴<br/>e⁴ + e² - 1 = 0<br/>e² = (-1 + √5)/2</p><p><strong>Step 6:</strong> Solving: e² = (√5 - 1)/2, which gives e = √[(√5 - 1)/2]</p><p>∴ Answer: D</p>
Correct Answer: D