<p>If \(y = mx + 4\) is a tangent to both the parabolas \(y^2 = 4x\) and \(x^2 = 2by\), then \(b\) is equal to</p>
Step-by-Step Solution
Key Concept: Use the condition that a line is tangent to a parabola when it touches at exactly one point (discriminant = 0), and match coefficients for tangent to first parabola.
<p><strong>Step 1:</strong> For parabola \(y^2 = 4x\) with slope \(m\), the tangent is \(y = mx + \frac{1}{m}\)</p><p><strong>Step 2:</strong> Comparing with \(y = mx + 4\):</p><p>\[\frac{1}{m} = 4 \implies m = \frac{1}{4}\]</p><p><strong>Step 3:</strong> So the tangent line is \(y = \frac{1}{4}x + 4\)</p><p><strong>Step 4:</strong> For this to be tangent to \(x^2 = 2by\), substitute \(y = \frac{1}{4}x + 4\):</p><p>\[x^2 = 2b\left(\frac{1}{4}x + 4\right)\]</p><p>\[x^2 = \frac{b}{2}x + 8b\]</p><p>\[x^2 - \frac{b}{2}x - 8b = 0\]</p><p><strong>Step 5:</strong> For tangency, discriminant = 0:</p><p>\[\left(\frac{b}{2}\right)^2 + 32b = 0\]</p><p>\[\frac{b^2}{4} + 32b = 0\]</p><p>\[b^2 + 128b = 0\]</p><p>\[b(b + 128) = 0\]</p><p>Since \(b \neq 0\), we have \(b = -128\)</p><p>∴ Answer is (b).</p>
Correct Answer: B