The slopes of the common tangents to the para | vedclass

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(C) Let $X = x - 1$ and $Y = y - 2$. The equations become $X^2 = 4Y$ and $X^2 + \frac{Y^2}{2} = 1$.The equation of a tangent to the parabola $X^2 = 4Y

Vedclass · Accepted Answer

$4$

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(C) Let $X = x - 1$ and $Y = y - 2$. The equations become $X^2 = 4Y$ and $X^2 + \frac{Y^2}{2} = 1$.The equation of a tangent to the parabola $X^2 = 4Y$ with slope $m$ is $Y = mX - am^2$,where $a = 1$. Thus,$Y = mX - m^2$.The equation of a tangent to the ellipse $X^2 + \frac{Y^2}{2} = 1$ with slope $m$ is $Y = mX \pm \sqrt{a^2m^2 + b^2}$,where $a^2 = 1$ and $b^2 = 2$. Thus,$Y = mX \pm \sqrt{m^2 + 2}$.For the tangents to be common,the constant terms must be equal: $-m^2 = \pm \sqrt{m^2 + 2}$.Squaring both sides,we get $m^4 = m^2 + 2$,which implies $m^4 - m^2 - 2 = 0$.Let $t = m^2$. Then $t^2 - t - 2 = 0$,which factors as $(t - 2)(t + 1) = 0$.Since $m^2$ must be non-negative,$m^2 = 2$. The slopes $m_1$ and $m_2$ are $\sqrt{2}$ and $-\sqrt{2}$.Therefore,$m_1^2 + m_2^2 = 2 + 2 = 4$.

The slopes of the common tangents to the parabola $(x - 1)^2 = 4(y - 2)$ and the ellipse $\frac{(x - 1)^2}{1} + \frac{(y - 2)^2}{2} = 1$ are $m_1$ and $m_2$. Then,$m_1^2 + m_2^2$ is equal to:

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