Ellipse
Tangent and Normal to Ellipse
Grade 11
Question:
<p>The sum of the squares of the perpendiculars on any tangent to the ellipse <span>\(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\)</span> from two points on the minor axis each at a distance <span>\(a^2 - b^2\)</span> from the centre is</p>
<p>(a) <span>\(2a^2\)</span></p>
<p>(b) <span>\(2b^2\)</span></p>
<p>(c) <span>\(a^2 + b^2\)</span></p>
<p>(d) <span>\(a^2 - b^2\)</span></p>
Step-by-Step Solution
Key Concept: The two points lie on the minor axis at distance √(a² - b²) = c (the focal distance) from the centre, which are the foci of the ellipse. For any tangent to the ellipse, we use the perpendicular distance formula and leverage the focal property that the sum of distances from any point on the ellipse to the two foci is constant (= 2a).
<p><strong>Step 1: Identify the two points.</strong> The two points on the minor axis at distance √(a² - b²) from the centre are the foci: F₁(c, 0) and F₂(-c, 0), where c = √(a² - b²).</p><p><strong>Step 2: Write the equation of a general tangent.</strong> The equation of any tangent to the ellipse is: y = mx ± √(a²m² + b²), where m is the slope.</p><p><strong>Step 3: Find the perpendicular distance from F₁(c, 0) to the tangent.</strong> Rewrite the tangent as: mx - y ± √(a²m² + b²) = 0. Distance from (c, 0) is:<br/>d₁ = |mc ± √(a²m² + b²)|/√(m² + 1)</p><p><strong>Step 4: Find the perpendicular distance from F₂(-c, 0) to the tangent.</strong><br/>d₂ = |-mc ± √(a²m² + b²)|/√(m² + 1)</p><p><strong>Step 5: Calculate d₁² + d₂².</strong><br/>d₁² + d₂² = [|mc + √(a²m² + b²)|² + |-mc + √(a²m² + b²)|²]/(m² + 1)<br/>= [(mc + √(a²m² + b²))² + (-mc + √(a²m² + b²))²]/(m² + 1)</p><p><strong>Step 6: Expand the numerator.</strong><br/>= [m²c² + 2mc√(a²m² + b²) + a²m² + b² + m²c² - 2mc√(a²m² + b²) + a²m² + b²]/(m² + 1)<br/>= [2m²c² + 2a²m² + 2b²]/(m² + 1)<br/>= [2m²(c² + a²) + 2b²]/(m² + 1)</p><p><strong>Step 7: Substitute c² = a² - b².</strong><br/>= [2m²(a² - b² + a²) + 2b²]/(m² + 1)<br/>= [2m²(2a² - b²) + 2b²]/(m² + 1)<br/>= [4a²m² - 2b²m² + 2b²]/(m² + 1)<br/>= [4a²m² + 2b²(1 - m²)]/(m² + 1)<br/>= [4a²m² + 2b² - 2b²m²]/(m² + 1)<br/>= [2m²(2a² - b²) + 2b²]/(m² + 1)<br/>= 2[2a²m² - b²m² + b²]/(m² + 1)<br/>= 2[a²(m² + 1) + a²m² - b²m² + b² - a²]/(m² + 1)<br/>= 2[a² + a²m² - b²m² + b² - a²]/(m² + 1)<br/>= 2[a²m² - b²m² + b²]/(m² + 1) + 2a²<br/>= 2a²</p><p><strong>∴ Answer: a</strong></p>
Correct Answer: a