• # question_answer The equation of tangent to $4{{x}^{2}}-9{{y}^{2}}=36$ which are perpendicular to straight line$5x+2y-10=0$are A) $5(y-3)2=\left( x-\frac{\sqrt{117}}{2} \right)$B) $2y-5x+10-2\sqrt{18}=0$C) $2y-5x-10-2\sqrt{18}=0$D)  None of these

Slope of the equations$4{{x}^{2}}-9{{y}^{2}}=36$ $8x-18y\frac{dy}{dx}=0\Rightarrow \frac{dy}{dx}=\frac{4x}{9y}$or${{m}_{1}}=\frac{4x}{9y}$ Slope of the straight line,$5x+2y-10=0$is${{m}_{2}}=-\frac{5}{2}$ Therefore, for the perpendicularity,${{m}_{1}}{{m}_{2}}=-1$ Now,     $\frac{4x}{9y}\times \frac{-5}{2}=-1\Rightarrow y=\frac{10x}{9}$ Putting $y=\frac{10x}{9}$ in $4{{x}^{2}}-9{{y}^{2}}=36$ gives imaginary roots resulting in no tangents.