A) \[\left( -\frac{1}{2},-\frac{3}{2} \right)\]
B) \[\left( \frac{1}{2},-\frac{3}{2} \right)\]
C) \[(0,-1)\]
D) \[\left( 0,-\frac{3}{2} \right)\]
Correct Answer: D
Solution :
Given parabola is\[{{x}^{2}}=-2y\] Coordinates of end points of latusrectum are \[A\left( 1-\frac{1}{2} \right),B=\left( -1,-\frac{1}{2} \right)\] Now, \[2x=-2\frac{dy}{dx}\Rightarrow \frac{dy}{dx}=-x\] and slope of normal is \[-\frac{dx}{dy}=\frac{1}{x}\] The equations of normals at points A and B are \[y+\frac{1}{2}=\frac{1}{1}(x-1)\] \[\Rightarrow \] \[2y-2x=-3\] ...(i) and \[y+\frac{1}{2}=-\frac{1}{1}(x+1)\] \[\Rightarrow \] \[2y+2x=-3\] ...(ii) On solving Eqs. (i) and (ii), we get \[x=0,y=-\frac{3}{2}\]
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