Hyperbola
Points on Hyperbola
Grade 11
Question:
<p>A hyperbola whose transverse axis is along the major axis of the conic, \(\dfrac{x^2}{3} + \dfrac{y^2}{4} = 4\) and has vertices at the foci of this conic. If the eccentricity of the hyperbola is 3/2, then which of the following points does NOT lie on it?</p>
<p>\((\sqrt{5}, 2\sqrt{2})\)</p>
<p>\((0, 2)\)</p>
<p>\((5, 2\sqrt{3})\)</p>
<p>\((\sqrt{10}, 2\sqrt{3})\)</p>
Step-by-Step Solution
Key Concept: First convert the given ellipse to standard form to identify its foci, which become the vertices of the hyperbola. Then use the eccentricity condition e = 3/2 to determine the hyperbola equation and test which point doesn't satisfy it.
<p><strong>Step 1: Convert ellipse to standard form</strong></p><p>Given: $\frac{x^2}{3} + \frac{y^2}{4} = 4$</p><p>Standard form: $\frac{x^2}{12} + \frac{y^2}{16} = 1$</p><p>Here $a^2 = 16$, $b^2 = 12$, so major axis is along y-axis</p><p><strong>Step 2: Find foci of ellipse</strong></p><p>$c^2 = a^2 - b^2 = 16 - 12 = 4$, so $c = 2$</p><p>Foci are at $(0, ±2)$</p><p><strong>Step 3: Set up hyperbola equation</strong></p><p>Transverse axis along y-axis (major axis of ellipse), vertices at $(0, ±2)$</p><p>Hyperbola form: $\frac{y^2}{a'^2} - \frac{x^2}{b'^2} = 1$ where $a' = 2$</p><p><strong>Step 4: Use eccentricity condition</strong></p><p>For hyperbola: $e = \frac{3}{2} = \frac{c'}{a'}$ where $c'$ is the semi-focal distance</p><p>$c' = \frac{3}{2} \times 2 = 3$</p><p>$b'^2 = c'^2 - a'^2 = 9 - 4 = 5$</p><p><strong>Step 5: Write hyperbola equation</strong></p><p>$\frac{y^2}{4} - \frac{x^2}{5} = 1$</p><p><strong>Step 6: Test given options</strong></p><p>Substitute each point. The point that doesn't satisfy this equation is the answer.</p><p>∴ Answer: B</p>
Correct Answer: B