A) \[5(y-3)=2\left( x-\frac{\sqrt{117}}{2} \right)\]
B) \[2x-5y+10-2\sqrt{18}=0\]
C) \[2x-5y-10-2\sqrt{18}=0\]
D) None of the above
Correct Answer: D
Solution :
\[4{{x}^{2}}-9{{y}^{2}}=36\] On differentiating w.r.t.\[x,\]we get \[\Rightarrow \] \[8x-18y\frac{dy}{dx}=0\] \[\Rightarrow \] \[\frac{dy}{dx}=\frac{4x}{9y}\] Slope of the tangent\[=\frac{4x}{9y}\] \[\therefore \]For this tangent to be perpendicular to the straight line\[5x+2y-10=0,\] we must have \[\frac{4x}{9y}\times \frac{5}{-2}=-1\] \[\Rightarrow \] \[y=\frac{10x}{9}\] Putting this value of y in\[4{{x}^{2}}-9{{y}^{2}}=36,\]we get \[-64{{x}^{2}}=324,\]which does not have real roots. Hence, at no points on the given curve can the tangent be perpendicular to given line.
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