A) \[\left( \frac{9}{5},\frac{8}{5}, \right)\]
B) \[\left( \frac{8}{5},-\frac{9}{5}, \right)\]
C) \[\left( -\frac{9}{5},\frac{8}{5}, \right)\]
D) \[\left( -\frac{8}{5},\frac{9}{5} \right)\]
Correct Answer: C
Solution :
Given ellipse is \[4{{x}^{2}}+9{{y}^{2}}=36\] \[\Rightarrow \]\[\frac{{{x}^{2}}}{9}+\frac{{{y}^{2}}}{4}=1\] Normal at the point is parallel to the line \[4x-2y-5=0\]Slope of normal = 2 Slope of tangent \[=\frac{-1}{2}\] Point of contact to ellipse\[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1\]and line is\[\left( \frac{{{a}^{2}}m}{\sqrt{{{a}^{2}}{{m}^{2}}+{{b}^{2}}}},\frac{b}{\sqrt{{{a}^{2}}{{m}^{2}}+{{b}^{2}}}} \right)\] Now,\[{{a}^{2}}=9,{{b}^{2}}=4\] \[\therefore \]Point \[=\left( \frac{-9}{5},\frac{8}{5} \right)\]
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