Ellipse
Tangent to Ellipse
Grade 11
Question:
<p>If <span>\(3x + 4y = 12\sqrt{2}\)</span> is a tangent to the ellipse <span>\(\frac{x^2}{a^2} + \frac{y^2}{9} = 1\)</span> for some <span>\(a \in \mathbb{R}\)</span>, then the distance between the foci of the ellipse is</p>
<p>(a) <span>\(2\sqrt{7}\)</span></p>
<p>(b) <span>\(4\)</span></p>
<p>(c) <span>\(2\sqrt{2}\)</span></p>
<p>(d) <span>\(2\sqrt{5}\)</span></p>
Step-by-Step Solution
Key Concept: Use the condition that a given line is tangent to an ellipse to find the parameter a, then calculate the distance between foci using c² = a² - b².
<p><strong>Step 1:</strong> Since the line <span>\(3x + 4y = 12\sqrt{2}\)</span> is a tangent to the ellipse <span>\(\frac{x^2}{a^2} + \frac{y^2}{9} = 1\)</span>, the equation of tangent to the ellipse having slope <span>\(m\)</span> is:</p><p><span>\(y = mx \pm \sqrt{a^2m^2 + 9}\)</span> ... (i)</p><p><strong>Step 2:</strong> The slope of the line <span>\(3x + 4y = 12\sqrt{2}\)</span> is <span>\(m = -\frac{3}{4}\)</span></p><p><strong>Step 3:</strong> Substituting <span>\(m = -\frac{3}{4}\)</span> into equation (i) and comparing with the given tangent line, we can determine <span>\(a^2\)</span>.</p><p><strong>Step 4:</strong> From the condition that <span>\(3x + 4y = 12\sqrt{2}\)</span> is tangent, we find <span>\(a^2 = 16\)</span>.</p><p><strong>Step 5:</strong> For the ellipse <span>\(\frac{x^2}{16} + \frac{y^2}{9} = 1\)</span>, we have <span>\(a^2 = 16, b^2 = 9\)</span></p><p><strong>Step 6:</strong> Distance between foci = <span>\(2c = 2\sqrt{a^2 - b^2} = 2\sqrt{16 - 9} = 2\sqrt{7}\)</span></p><p>∴ Answer is (a) <span>\(2\sqrt{7}\)</span></p>
Correct Answer: A