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Q.

A polynomial equation $x^{4} + p x^{2} + q x + r = 0; (p, q, r \in R - {0})$ has 3 repeated real roots. If all possible points having coordinates $(p, r)$ in 2D plane lie on a part of conic, then the length of its semi latus rectum is

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a

2

b

12

c

6

d

3

answer is C.

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Detailed Solution

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Let repeated roots be $α$
$α^{4} + p α^{2} + q α + r = 0____(1)$
$4 α^{3} + 2 p α + q = 0____(2)$
$12 α^{2} + 2 p = 0____(3)$
So $α^{2} = - \frac{p}{6}$
Putting in (2), $\frac{4 p α}{3} = - q$
Putting in (1), $\frac{p^{2}}{36} - \frac{p^{2}}{6} - \frac{4 p}{3} \times (- \frac{p}{6}) + r = 0$
$\Rightarrow p^{2} = - 12 r$
Hence semilatus rectum = 6.

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