A) \[x+2y=4\]
B) \[2y-x=2\]
C) \[4x-2y=1\]
D) \[4x+2y=7\]
Correct Answer: C
Solution :
Eccentricity of ellipse \[=\frac{1}{2}\] Now, \[-\frac{a}{e}=-4\Rightarrow a=4\times \frac{1}{2}=2\] \[\therefore \] \[{{b}^{2}}={{a}^{2}}(1-{{e}^{2}})={{a}^{2}}\left( 1-\frac{1}{4} \right)=3\] \[\therefore \]Equation of ellipse \[\frac{{{x}^{2}}}{4}+\frac{{{y}^{2}}}{3}=1\] \[\Rightarrow \] \[\frac{x}{2}+\frac{2y}{3}\times y'=0\,\Rightarrow y'=-\frac{3x}{4y}\] \[y'{{|}_{(1,3/2)}}=-\frac{3}{4}\times \frac{2}{3}=-\frac{1}{2}\] \[\therefore \]Equation of normal at \[\left( 1,\frac{3}{2} \right)\] \[y-\frac{3}{2}=2(x-1)\Rightarrow 2y-3=4x-4\] \[\therefore \] \[4x-2y=1\]You need to login to perform this action.
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