Parabola
Common Tangent to Parabola and Ellipse
Grade 11
Question:
<p>x - 2y + 4 = 0 is a common tangent to \(y^2 = 4x\) and \(\frac{x^2}{4} + \frac{y^2}{b^2} = 1\). Then the value of b and the other common tangent are given by:</p>
<p>(a) \(b = 3\); \(x + 2y + 4 = 0\)</p>
<p>(b) \(b = \sqrt{3}\); \(x + 2y + 4 = 0\)</p>
<p>(c) \(b = \sqrt{3}\); \(x + 2y - 4 = 0\)</p>
<p>(d) \(b = \sqrt{3}\); \(x - 2y - 4 = 0\)</p>
Step-by-Step Solution
Key Concept: Use tangency conditions for parabola and ellipse separately, then apply symmetry to find the second tangent.
<p>For a line to be tangent to \(y^2 = 4x\), comparing with \(y = mx + c\) form: \(c = \frac{1}{m}\). The line \(x - 2y + 4 = 0\) can be written as \(y = \frac{x+4}{2}\), so \(m = \frac{1}{2}\) and \(c = 2\). Check: \(c = \frac{1}{1/2} = 2\) ✓. For tangency to ellipse \(\frac{x^2}{4} + \frac{y^2}{b^2} = 1\), using tangent condition \(c^2 = 4m^2 + b^2\): \(4 = 4(\frac{1}{4}) + b^2\), so \(b^2 = 3\), thus \(b = \sqrt{3}\). By symmetry about the x-axis, the other common tangent is \(x + 2y + 4 = 0\).</p>
Correct Answer: B