A) \[{{y}^{2}}=-104x\]
B) \[{{y}^{2}}=14x\]
C) \[{{y}^{2}}=26x\]
D) \[{{y}^{2}}=-14x\]
Correct Answer: A
Solution :
Ellipse is\[\frac{{{x}^{2}}}{16}+\frac{{{y}^{2}}}{3}=1\] Now, equation of normal at (2, 3/2) is \[\frac{16x}{2}-\frac{3y}{3/2}=16-3\]\[\Rightarrow \]\[8x-2y=13\] \[\Rightarrow \]\[y=4x-\frac{13}{2}\] Let \[y=4x-\frac{13}{2}\]touches a parabola\[{{y}^{2}}=4ax.\] We know, a straight \[y=mx+c\] touchesa parabola \[{{y}^{2}}=4ax\] if\[a-mc=0\] \[\therefore \]\[a-\left( 4 \right)\left( -\frac{13}{2} \right)=0\Rightarrow a=-26\] Hence, required equation of parabola is \[{{y}^{2}}=4(-26)x=-104x\]
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