A) \[\left( \frac{8}{5},-\frac{-9}{5} \right)\]
B) \[\left( \frac{-9}{5},\frac{8}{5} \right)\]
C) \[\left( \frac{9}{5},\frac{8}{5} \right)\]
D) \[\left( \frac{8}{5},\frac{9}{5} \right)\]
Correct Answer: C
Solution :
Let \[P({{x}_{1}},\,\,{{y}_{1}})\] be the point on curve \[4{{x}^{2}}+9{{y}^{2}}=36\] where the normal is parallel to line\[4x-2y=5\]. \[\therefore \]\[{{\left. \frac{dy}{dx} \right|}_{P({{x}_{1}},\,\,{{y}_{1}})}}=\left( \frac{-4{{x}_{1}}}{9{{y}_{1}}} \right)\] Now, \[\frac{9{{y}_{1}}}{4{{x}_{1}}}=2\Rightarrow 9{{y}_{1}}=8{{x}_{1}}\] ?(i) Also, \[4x_{1}^{2}+9y_{1}^{2}=36\] ?(ii) \[\therefore \] On solving (1) and (2), we get \[P\left( {{x}_{1}}=\frac{9}{5},\,\,{{y}_{1}}=\frac{8}{5} \right)\]or\[\left( {{x}_{1}}=\frac{-9}{5},\,\,{{y}_{1}}=\frac{-8}{5} \right)\]
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