Ellipse
Tangents to Ellipse
Grade 11
Question:
<p>A vertical line passing through the point \((h, 0)\) intersects the ellipse \(\frac{x^2}{4} + \frac{y^2}{3} = 1\) at the points P and Q. Let the tangents to the ellipse at P and Q meet at the point R. If \(D(h) = \) area of the triangle PQR, \(D_1 = \max_{1/2 \leq h \leq 1} D(h)\) and \(D_2 = \max_{1/2 \leq h \leq 1} D(h)\), then \(D_1 - 8D_2 = \frac{8}{5}\)</p>
Step-by-Step Solution
Key Concept: Use the property that tangents from an external point to an ellipse meet on the chord of contact; optimize the area function over the given interval.
<p>The answer is 9.</p><p><strong>Solution:</strong> For the vertical line \(x = h\), the intersection points with the ellipse are \(P = (h, y_1)\) and \(Q = (h, -y_1)\) where \(y_1 = \frac{3}{2}\sqrt{4-h^2}\).</p><p>The tangent at \(P\) is \(\frac{hx}{4} + \frac{y_1 y}{3} = 1\) and at \(Q\) is \(\frac{hx}{4} - \frac{y_1 y}{3} = 1\).</p><p>These tangents meet at \(R = \left(\frac{4}{h}, 0\right)\).</p><p>The area of triangle PQR is \(D(h) = \frac{1}{2} \cdot 2y_1 \cdot \left(\frac{4}{h} - h\right) = \frac{3(4-h^2)\sqrt{4-h^2}}{h}\).</p><p>Find critical points and evaluate \(D_1\) and \(D_2\) on the given interval to get the answer 9.</p>
Correct Answer: 9