Question

The point of intersection of the lines which are common tangent to both the curves $y=x^2$ and $x^2+y+1=0$ is

• A. $\left(0,-\frac{1}{4}\right)$
• B. $\left(0,-\frac{1}{3}\right)$
• C. $\left(0,-\frac{1}{5}\right)$
• D. $\left(0,-\frac{1}{2}\right)$

Answer 1

Answer: (D)

For $y=x^2$ ,
equation of the tangent is $y=mx-\frac{m^2}{4}$
For this line to be tangent to $x^2+y+1=0$
$\Rightarrow x^2+mx-\frac{m^2}{4}+1=0$ will have $D=0$
$\Rightarrow m^2+4\left(\frac{m^2}{4}-1\right)=0\Rightarrow m=\pm \sqrt{2}$
$\Rightarrow$ Tangents are $y=\sqrt{2}x-\frac{1}{2}$ and $y=-\sqrt{2}x-\frac{1}{2}$
$\Rightarrow$ Point of intersection of the tangents is $\left(0,-\frac{1}{2}\right)$