Ellipse
Orthogonal intersection with hyperbola
Grade 11
Question:
<p>An ellipse intersects the hyperbola \(2x^2 - 2y^2 = 1\) orthogonally. The eccentricity of the ellipse is reciprocal of that of the hyperbola. If the axes of the ellipse are along the coordinate axes, then which of the following is correct?</p>
<p>(a) Equation of ellipse is \(x^2 + 2y^2 = 2\)</p>
<p>(b) The foci of ellipse are \((\pm 1, 0)\)</p>
<p>(c) Equation of ellipse is \(x^2 + 2y^2 = 4\)</p>
<p>(d) The foci of ellipse are \((\pm \sqrt{2}, 0)\)</p>
Step-by-Step Solution
Key Concept: Use orthogonality condition for curves (product of slopes = -1) combined with eccentricity relation to determine ellipse parameters.
<p><strong>Step 1:</strong> For hyperbola \(2x^2 - 2y^2 = 1\), rewrite as \(\frac{x^2}{1/2} - \frac{y^2}{1/2} = 1\).</p><p>Here \(a^2 = b^2 = 1/2\), so \(c^2 = 1\), giving eccentricity \(e_h = \sqrt{2}\).</p><p><strong>Step 2:</strong> The ellipse has eccentricity \(e_e = \frac{1}{\sqrt{2}}\).</p><p><strong>Step 3:</strong> Let the ellipse be \(\frac{x^2}{A^2} + \frac{y^2}{B^2} = 1\) with \(A > B\).</p><p>For orthogonal intersection, \(\frac{dy}{dx}_{hyperbola} \cdot \frac{dy}{dx}_{ellipse} = -1\).</p><p>From hyperbola: \(4x - 4y\frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = \frac{x}{y}\).</p><p>From ellipse: \(\frac{2x}{A^2} + \frac{2y}{B^2}\frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = -\frac{xB^2}{yA^2}\).</p><p>For orthogonality: \(\frac{x}{y} \cdot \left(-\frac{xB^2}{yA^2}\right) = -1 \Rightarrow \frac{x^2B^2}{y^2A^2} = 1\).</p><p><strong>Step 4:</strong> Using \(e_e^2 = 1 - \frac{B^2}{A^2} = \frac{1}{2}\), we get \(\frac{B^2}{A^2} = \frac{1}{2}\).</p><p>Substituting into orthogonality condition and solving gives \(A^2 = 2\), \(B^2 = 1\).</p><p><strong>Step 5:</strong> Ellipse equation is \(x^2 + 2y^2 = 2\). Foci are at \((\pm c, 0)\) where \(c^2 = A^2 - B^2 = 1\), so foci are \((\pm 1, 0)\).</p><p>∴ Answers are (a) and (b).</p>
Correct Answer: A, B