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Double Ordinate

The parabola ...

Question

The parabola $y^{2} = 4 a x$ passes through the centre of the circle $4 x^{2} + 4 y^{2} - 8 x + 12 y - 7 = 0$ . The directrix of the parabola will be

4 x + 9 = 0

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4 x + 15 = 0

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16 x + 9 = 0

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16 x - 7 = 0

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Solution

The correct option is C $16 x + 9 = 0$
Circle : $4 x^{2} + 4 y^{2} - 8 x + 12 y - 7 = 0$
$\Rightarrow x^{2} + y^{2} - 2 x + 3 y - \frac{7}{4} = 0$
Centre of this circle $C \equiv (1, - \frac{3}{2})$

Given parabola passes through this centre so,
${(- \frac{3}{2})}^{2} = 4 a (1)$
$\Rightarrow a = \frac{9}{16}$
Directrix : $x = - a$
$x = - \frac{9}{16} \Rightarrow 16 x + 9 = 0$

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Double Ordinate

Standard XIII Mathematics

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The parabola y2=4ax passes through the centre of the circle 4x2+4y2−8x+12y−7=0. The directrix of the parabola will be

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The parabola $y^{2} = 4 a x$ passes through the centre of the circle $4 x^{2} + 4 y^{2} - 8 x + 12 y - 7 = 0$ . The directrix of the parabola will be