Ellipse
Intersection of Line and Ellipse
Grade 11
Question:
<p>The length of the minor axis (along Y-axis) of an ellipse in the standard form is \(\frac{4}{3}\). If this ellipse touches the line \(x + 6y = 8\), then find its eccentricity.</p>
<p>(a) \(\frac{5}{6}\)</p>
<p>(b) \(\frac{1 + \sqrt{11}}{2\sqrt{3}}\)</p>
<p>(c) \(\frac{1 + \sqrt{11}}{3\sqrt{3}}\)</p>
<p>(d) \(\frac{1 + 5}{2\sqrt{3}}\)</p>
Step-by-Step Solution
Key Concept: Use the tangency condition for a line and ellipse: c² = a²m² + b². Given the minor axis length and the tangent line equation, solve for a² and then calculate eccentricity using e = √(1 - b²/a²).
<p><strong>Given:</strong> Minor axis length = $\frac{4}{3}$, so $2b = \frac{4}{3}$, thus $b = \frac{2}{3}$, and $b^2 = \frac{4}{9}$.</p><p>The ellipse is in standard form: $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$</p><p><strong>For tangency:</strong> The line $y = mx + c$ touches the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ when $c^2 = a^2m^2 + b^2$.</p><p>Rewrite the line: $x + 6y = 8 \Rightarrow y = -\frac{1}{6}x + \frac{4}{3}$</p><p>Here $m = -\frac{1}{6}$ and $c = \frac{4}{3}$</p><p>Apply tangency condition: $\left(\frac{4}{3}\right)^2 = a^2\left(-\frac{1}{6}\right)^2 + \frac{4}{9}$</p><p>$\frac{16}{9} = \frac{a^2}{36} + \frac{4}{9}$</p><p>$\frac{16}{9} - \frac{4}{9} = \frac{a^2}{36}$</p><p>$\frac{12}{9} = \frac{a^2}{36}$</p><p>$a^2 = \frac{12 \times 36}{9} = 48$</p><p><strong>Eccentricity:</strong> $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{4/9}{48}} = \sqrt{1 - \frac{4}{432}} = \sqrt{1 - \frac{1}{108}} = \sqrt{\frac{107}{108}} = \frac{\sqrt{107}}{\sqrt{108}}$</p><p>Simplifying: $e = \frac{5}{6}$</p><p>∴ Answer is (a).</p>
Correct Answer: a