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

The correct option is A $4(x+y)+3=0$


The parabolas will be,
$x^2=3y{y}^2=3x$
Tangent in slope form to the parabola $y^2=3x$,
$y=mx+\frac{3}{4m}$
This will also be tangent to the parabola $x^2=3y$
$x^2=3(mx+\frac{3}{4m})\Rightarrow x^2-3mx-\frac{9}{4m}=0$
So the quadratic equation must have equal roots,
$D=0\Rightarrow 9m^2+\frac{4\times 9}{4m}=0\Rightarrow m^3+1=0;m\ne 0\Rightarrow m=-1$

Hence the tangent will be,
$y=-x-\frac{3}{4}\Rightarrow 4(x+y)+3=0$