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Q.
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
NTA AbhyasNTA Abhyas 2020Conic Sections
Solution:
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 $\left(0 , - \frac{1}{2}\right)$