The smallest focal chord of curve C is

Engineering

Mathematics

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Question

$\frac{1}{2}$

$\frac{1}{4}$

$\frac{1}{12}$

$\frac{1}{16}$

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$Circumcentre is (\frac{3 p + 4}{2 p} \cdot \frac{4 - 3 p}{2 q})$

let circumcentre be (h, k)

$h = \frac{3 p + 4}{2 p} \Rightarrow p = \frac{4}{2 h - 3}$

$k = \frac{4 - 3 p}{2 q} \Rightarrow q = \frac{4 h - 12}{k (2 h - 3)}$

Now, p² = 4q

${(h - \frac{9}{4})}^{2} = \frac{1}{2} (k + \frac{9}{8})$

$locus {(x - \frac{9}{4})}^{2} = \frac{1}{2} (y + \frac{9}{8})$ .