If P(-3, 2) is an end point of the focal chor | vedclass

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(A) The equation of the parabola is $y^2 + 4x + 4y = 0$,which can be written as $(y + 2)^2 = -4(x - 1)$.Comparing this with $Y^2 = 4AX$,we have $Y = y

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$\frac{-1}{2}$

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(A) The equation of the parabola is $y^2 + 4x + 4y = 0$,which can be written as $(y + 2)^2 = -4(x - 1)$.Comparing this with $Y^2 = 4AX$,we have $Y = y + 2$,$X = x - 1$,and $4A = -4$,so $A = -1$.The parametric coordinates are $X = At^2$ and $Y = 2At$,which gives $x - 1 = -t^2$ and $y + 2 = -2t$.For point $P(-3, 2)$,we have $-3 - 1 = -t^2$ $\Rightarrow t^2 = 4$ $\Rightarrow t = \pm 2$. Also $2 + 2 = -2t \Rightarrow t = -2$.Since $P$ and $Q$ are ends of a focal chord,their parameters $t_1$ and $t_2$ satisfy $t_1 t_2 = -1$. Thus,$t_2 = \frac{-1}{-2} = \frac{1}{2}$.The coordinates of $Q$ are $x = 1 - (\frac{1}{2})^2 = \frac{3}{4}$ and $y = -2(\frac{1}{2}) - 2 = -3$.The slope of the tangent at $Q$ is given by differentiating $y^2 + 4x + 4y = 0$: $2y \frac{dy}{dx} + 4 + 4 \frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = \frac{-2}{y + 2}$.At $Q(\frac{3}{4}, -3)$,the slope of the tangent is $m_T = \frac{-2}{-3 + 2} = \frac{-2}{-1} = 2$.The slope of the normal $m_N$ is $-\frac{1}{m_T} = \frac{-1}{2}$.

If $P(-3, 2)$ is an end point of the focal chord $PQ$ of the parabola $y^2 + 4x + 4y = 0$,then the slope of the normal drawn at $Q$ is

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