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The coordinates of the end point of the latus rectum of the parabola $$\left ( y-1 \right )^{2}= 2\left ( x+2 \right )$$, which does not lie on the line $$2x + y + 3 = 0$$ are 


A
(2,1)
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B
(32,1)
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C
(32,2)
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D
(32,0)
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Solution

The correct option is B $$\displaystyle \left ( -\dfrac32,2 \right )$$
We know coordinate of the ends of the latus rectum of parabola $$Y^2=4aX$$ are given by,
$$X=a, Y=2a$$ and $$X=a, Y=-2a$$
Here $$X  =x+2, Y=y-1$$ and $$a = \dfrac{1}{2}$$
Thus ends of the latus rectum of the given parabola are $$\left(-\dfrac{3}{2},0\right)$$ and $$\left(-\dfrac{3}{2},2\right)$$
Clearly point $$\left(-\dfrac{3}{2},0\right)$$ lies on the line $$2x+y+3=0$$

Mathematics

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