Ellipse
Tangents from External Point
Grade 11
Question:
<p>Find the equation of tangents drawn from the point \((2, 3)\) to the ellipse \(\frac{x^2}{16} + \frac{y^2}{9} = 1\).</p>
<p>(a) \(x + y + 5 = 0\)</p>
<p>(b) \(x + y - 5 = 0\)</p>
<p>(c) \(y + 3 = 0\)</p>
<p>(d) \(y - 3 = 0\)</p>
Step-by-Step Solution
Key Concept: Substitute the line equation into the ellipse equation and use the condition that the discriminant of the resulting quadratic equals zero for tangency.
<p>From external point \((2, 3)\), we find tangents to the ellipse \(\frac{x^2}{16} + \frac{y^2}{9} = 1\).</p><p>A line through \((2, 3)\) with slope \(m\) is: \(y - 3 = m(x - 2)\) or \(y = mx - 2m + 3\).</p><p>For this to be tangent to the ellipse: \(\frac{x^2}{16} + \frac{(mx - 2m + 3)^2}{9} = 1\).</p><p>Expanding and simplifying: \((9 + 16m^2)x^2 + 2 \cdot 16m(-2m + 3)x + 16(-2m + 3)^2 - 144 = 0\).</p><p>For tangency, discriminant = 0: \([32m(-2m + 3)]^2 - 4(9 + 16m^2)[16(-2m + 3)^2 - 144] = 0\).</p><p>Solving: \(m = 0\) (giving \(y = 3\)) and \(m = -1\) (giving \(x + y - 5 = 0\)).</p>
Correct Answer: b, d