Let $S$ be the set of values of paramenter $a$ for which the points of intersection of the parabolas $y^{2} = 3 a x$ and $y = \frac{1}{2} (x^{2} + a x + 5)$ are concyclic, then $S$ can be

(- 2, \infty)

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(- \infty, 2)

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(- \infty, - 2)

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(2, \infty)

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Solution

The correct options are
C $(- \infty, - 2)$
D $(2, \infty)$
Any curve througth intersection of given curves will be of form
$y^{2} - 3 a x + λ (x^{2} + a x + 5 - 2 y) = 0 \dots (1)$
This should be a real circle $\Rightarrow λ = 1$ and radius $> 0$
$\Rightarrow x^{2} + y^{2} - 2 a x - 2 y + 5 = 0 \Rightarrow (x - a)^{2} + (y - 1)^{2} = a^{2} - 4$
$\Rightarrow a^{2} - 4 > 0 \Rightarrow a > 2 or a < - 2$

Suggest Corrections

Similar questions

S

a

y^{2} = 3 a x

y = \frac{1}{2} (x^{2} + a x + 5)

S

a x + b y = a b a n d b x + a y = a b

\frac{x}{a} + \frac{y}{b} = 1

\frac{x}{c} + \frac{y}{d} = 1

a^{2} + c^{2} = b^{2} + d^{2}

(a, b, c, d > 0)

a x^{2} + 4 x y + 2 y^{2} + x + y + 5 = 0

a x^{2} + 6 x y + 5 y^{2} + 2 x + 3 y + 8 = 0

a

y^{2} = 4 a x

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