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JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola

A circle of radius \(2\) unit passes through the vertex and the focus of the parabola \(y^{2}=2 x\) and touches the parabola \(y=\left(x-\frac{1}{4}\right)^{2}+\alpha\), where \(\alpha>0\). Then \((4 \alpha-8)^{2}\) is equal to

  1. A \(60\)
  2. B \(61\)
  3. C \(62\)
  4. D \(63\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(63\)

Step-by-step Solution

Detailed explanation

Vertex and focus of parabola \(y^{2}=2 x\) are \(V (0,0)\) and \(S\left(\frac{1}{2}, 0\right)\) resp. Let equation of circle be \((x-h)^{2}+(y-k)^{2}=4\) \(\because\) Circle passes through \((0,0)\) \(\Rightarrow h ^{2}+ k ^{2}=4 \ldots \ldots(1)\)…
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