Question

Slope of common tangents of parabola $(x-1{)}^2=4(y-2)$ and ellipse $(x-1{)}^2+\frac{(y-2{)}^2}{2}=1$ are $m_1$ and $m_2$ then $m_1^2+m_2^2$ is equal to

• A. $2$
• B. $3$
• C. $4$
• D. $6$

Answer 1

Answer: (C)

$4$

We have

${\left(x-1\right)}^2=4\left(y-2\right)$

${\left(x-1\right)}^2+\frac{{\left(y-2\right)}^2}{2}=1$

Now,the slope of the given parabola and ellipse

$X^2=4Y$

$y=mx=m^2$

And,

$X^2+\frac{Y^2}{2}=1$

$y=mx-\sqrt{m^2+2}$

Now, the common slope of the parabola and the ellipse

$m^2=\sqrt{m^2+2}$

$m^4=m^2+2$

$m^4-m^2-2=0$

$m^4-2m^2+m^2-2=0$

$\left(m^2+1\right)\left(m^2-2\right)=0$

$m=\pm \sqrt{2}$

Therefore,

$m_1^2+m_2^2=2+2=4$

Hence, the option $\left(C\right)$ is the correct answer.