Question

The equations of common tangents to the parabolas $y=x^2$ and $y=-(x-2{)}^2$ are

• A. $y=0$
• B. $y=4(x-1)$
• C. $y+4(x-1)=0$
• D. $y+30x+50=0$

Answer 1

Answer: (A), (B)

$y=4(x-1)$
$y=0$

Equation of tangent to the parabola $x^2=y$ with slope $m$ is given by,
$x=my+\frac{1}{4m}..\left(1\right)$, since here $a=\frac{1}{4}$
And equation of tangent to the parabola $(x-2{)}^2=-y$ with slope $m^{\prime }$ is given by,
$(x-2)=m^{\prime }(-y)+\frac{1}{4m^{\prime }}$
$\Rightarrow x=-m^{\prime }y+(\frac{1}{4m^{\prime }}+2)..\left(2\right)$
For common tangent both lines are same,
$\Rightarrow m=-m^{\prime }$ and $\frac{1}{4m}=\frac{1}{4m^{\prime }}+2$
$\Rightarrow m=\frac{1}{4}$
Hence common tangent to the given parabola's is $y=4(x-1)$
Also $y=0$ is vertex of both the parabola, hence it is also common tangent.