Question

Two parabolas with a common vertex and with 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 :

• A. $4(x+y)+3=0$
• B. $3(x+y)+4=0$
• C. $8(2x+y)+3=0$
• D. $x+2y+3=0$

Answer 1

Answer: (A)

Equation two parabola are $Y^2=3x$ and $x^2=3y$
Let equation of tangent to $y^2=3x$ is y = mx $+\frac{3}{4m}$
is also tangent to $x^2=3y$
$\Rightarrow x^2=3mx+\frac{9}{4m}$
$\Rightarrow 4m{x}^2-12m^{2}x-9=0$ have equal roots
$\Rightarrow D=0$
$\Rightarrow 144m^4=4\left(4m\right)\left(-9\right)$
$\Rightarrow m^4+m=0\Rightarrow m=-1$
Hence common tangent is $y=-x-\frac{3}{4}$
$4\left(x+y\right)+3=0$