A) (6, -9)
B) (6, 9)
C) (-6, -9)
D) (-6, 9)
Correct Answer: C
Solution :
Equation of normal will be \[\frac{x-2}{2+2}=\frac{y-3}{3+3}\] \[\Rightarrow \]\[3x-2y=0\Rightarrow x=\frac{2y}{3}\] Thus \[{{\left( \frac{2y}{3} \right)}^{2}}+{{y}^{2}}+4\left( \frac{2y}{3} \right)+6y-39=0\] \[\Rightarrow y=3,\ -9\Rightarrow x=2,\ -6\] Hence another point will be \[(-6,\,-9)\].You need to login to perform this action.
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