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Two parabolas with a common vertex and wih axes along x-axis and y-axis, respectively, , intersect each other in the first quadrant. if the length of the latus rectum of each parabola is $3,$ then the equation of the common tangent to the two parabolas is

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$4 (x + y) + 3 = 0$

$8 (2 x + y) + 3 = 0$

$3 (x + y) + 4 = 0$

$x + 2 y + 3 = 0$

answer is A.

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Detailed Solution

Length of latus rectum $= 3$

$\Rightarrow$ Equations of parabolas are $y^{2} = 3 x, x^{2} = 3 y$

The equation of common tangent to $y^{2} = 4 a x, x^{2} = 4 b y$ is

$a^{1 / 3} x + b^{1 / 3} y + a^{2 / 3} b^{2 / 3} = 0$

$∴ Required common tangent is {(\frac{3}{4})}^{1 / 3} x + {(\frac{3}{4})}^{1 / 3} y + {(\frac{3}{4})}^{4 / 3} = 0$

$\Rightarrow x + y + \frac{3}{4} = 0 \Rightarrow 4 (x + y) + 3 = 0$ .

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