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

Answer 1

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

Origin $(0,0)$ is the only point common to x-axis and y-axis.

$\Rightarrow$ Origin $(0,0)$ is the common vertex

Let the equation of $2$ parabola be $y^2=4ax$ and $x^2=4by$

Latus rectum$=3$

$\Rightarrow 4a=4b=3$

$\Rightarrow a=b=\frac{3}{4}$

$\therefore$ The $2$ parabolas are $y^2=3x$ and $x^2=3y$

Let $y=mx+c$ be the common tangent

$y^2=3x$

$\Rightarrow (mx+c{)}^2=3x$

$\Rightarrow m^2{x}^2+(2mc-3)x+c^2=0$

The tangent touches at only one point

$\Rightarrow b^2-4ac=0$

$\Rightarrow (2mc-3{)}^2-4m^2{c}^2=0$

$\Rightarrow 4m^2{c}^2+9-12mc-4m^2{c}^2=0$

$\Rightarrow c=\frac{9}{12m}=\frac{3}{4m}$ ………$\left(1\right)$

$m^2=-c=\frac{-3}{4m}$

$x^2=3y$

$\Rightarrow x^2=3(mx+c)$

$\Rightarrow x^2-3mx-3c=0$

Tangent touches at only one point

$\Rightarrow b^2-4ac=0$

$\Rightarrow 9m^2-4\left(1\right)(-3c)=0$

$\Rightarrow 9m^2=-12c$ …………$\left(2\right)$

From $\left(1\right)$ and $\left(2\right)$

$m^2=\frac{-4c}{3}=\frac{-4}{3}\left(\frac{3}{4m}\right)$

$\Rightarrow m^3=-1$

$\Rightarrow m=-1$

$\Rightarrow c=\frac{-3}{4}$

$\therefore y=mx+c=-x-\frac{3}{4}$

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