Ellipse
Tangent to Ellipse
Grade 11
Question:
<p>On the ellipse \(4x^2 + 9y^2 = 1\), the points at which the tangents are parallel to line \(8x = 9y\) are:</p>
<p>(a) \(\left(\frac{2}{5}, \frac{1}{5}\right)\)</p>
<p>(b) \(\left(-\frac{2}{5}, \frac{1}{5}\right)\)</p>
<p>(c) \(\left(-\frac{2}{5}, -\frac{1}{5}\right)\)</p>
<p>(d) \(\left(\frac{2}{5}, -\frac{1}{5}\right)\)</p>
Step-by-Step Solution
Key Concept: Find points on the ellipse where the tangent has the same slope as the given line using the tangent condition.
<p><strong>Step 1:</strong> The slope of the given line \(8x = 9y\) is \(m = \frac{8}{9}\).</p><p><strong>Step 2:</strong> For ellipse \(4x^2 + 9y^2 = 1\), the condition for a tangent with slope \(m\) is that the tangent touches at point \((x, y)\) satisfying the tangent equation.</p><p><strong>Step 3:</strong> Using the general tangent equation and equating slopes, we find the points \(\left(-\frac{2}{5}, \frac{1}{5}\right)\) and \(\left(\frac{2}{5}, -\frac{1}{5}\right)\).</p><p>∴ Answer is (b, d).</p>
Correct Answer: b, d