Slope of common tangents of parabola (x -1)^2 | vedclass

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Question

$\mathrm{X}^{2}=4 \mathrm{Y}, \mathrm{X}^{2}+\frac{\mathrm{Y}^{2}}{2}=1$

$\mathrm{y}=\mathrm{mx}-\mathrm{m}^{2}$

$y=m x-\sqrt{m^{2}+2}$

$\mat

Vedclass · Accepted Answer

$4$

Vedclass · Answer

$\mathrm{X}^{2}=4 \mathrm{Y}, \mathrm{X}^{2}+\frac{\mathrm{Y}^{2}}{2}=1$

$\mathrm{y}=\mathrm{mx}-\mathrm{m}^{2}$

$y=m x-\sqrt{m^{2}+2}$

$\mathrm{m}^{2}=\sqrt{\mathrm{m}^{2}+2}$

$\Rightarrow m^{4}-m^{2}-2=0$

$\mathrm{m}^{2}=2,-1 \Rightarrow \mathrm{m}$

$\mathrm{m}_{1}^{2}+\mathrm{m}_{2}^{2}=4$

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

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