Ellipse
Intersection of Ellipses
Grade 11
Question:
<p>Let \(E_1\) and \(E_2\) be two ellipses \(\dfrac{x^2}{a^2} + y^2 = 1\) and \(x^2 + \dfrac{y^2}{a^2} = 1\) (where \(a\) is a parameter). Then the locus of the points of intersection of the ellipses \(E_1\) and \(E_2\) is a set of curves comprising</p>
<p>(a) Two straight lines</p>
<p>(b) One straight line</p>
<p>(c) One circle</p>
<p>(d) One parabola</p>
Step-by-Step Solution
Key Concept: Subtract the equations of the two ellipses to eliminate the parameter a and find the locus independent of a. The resulting equation represents the actual curve(s) through all intersection points regardless of a's value.
<p><strong>Step 1:</strong> Write the equations of the two ellipses.</p><p>E₁: $\frac{x^2}{a^2} + y^2 = 1$</p><p>E₂: $x^2 + \frac{y^2}{a^2} = 1$</p><p><strong>Step 2:</strong> Subtract E₂ from E₁ to eliminate the parameter a.</p><p>$\left(\frac{x^2}{a^2} + y^2\right) - \left(x^2 + \frac{y^2}{a^2}\right) = 0$</p><p>$\frac{x^2}{a^2} - x^2 + y^2 - \frac{y^2}{a^2} = 0$</p><p>$x^2\left(\frac{1}{a^2} - 1\right) + y^2\left(1 - \frac{1}{a^2}\right) = 0$</p><p><strong>Step 3:</strong> Factor the expression.</p><p>$\left(\frac{1}{a^2} - 1\right)(x^2 - y^2) = 0$</p><p>For $a \neq 1$: $x^2 - y^2 = 0$, which gives $x^2 = y^2$</p><p><strong>Step 4:</strong> Identify the locus.</p><p>This factors as $(x-y)(x+y) = 0$, representing <strong>two straight lines: x = y and x = -y</strong></p><p>∴ Answer: A (A pair of straight lines)</p>
Correct Answer: A