• # question_answer On the ellipse $4{{x}^{2}}+9{{y}^{2}}=1$, the points at which the tangents are parallel to the line $8x=9y$ are          [IIT 1999] A)            $\left( \frac{2}{5},\ \frac{1}{5} \right)$                                     B)            $\left( -\frac{2}{5},\ \frac{1}{5} \right)$ C)            $\left( -\frac{2}{5},\ -\frac{1}{5} \right)$                                  D)            $\left( \frac{2}{5},\ -\frac{1}{5} \right)$

Ellipse is $\frac{{{x}^{2}}}{\frac{1}{4}}+\frac{{{y}^{2}}}{\frac{1}{9}}=1$Þ ${{a}^{2}}=\frac{1}{4}$, ${{b}^{2}}=\frac{1}{9}$            The equation of its tangent is $4xx'+9yy'=1$            \$m=-\frac{4x'}{9y'}=\frac{8}{9}$Þ$x'=-2y'$            and $4x{{'}^{2}}+9y{{'}^{2}}=1$Þ$4x{{'}^{2}}+9\frac{x{{'}^{2}}}{4}=1$Þ$x'=\pm \frac{2}{5}$            When $x'=\frac{2}{5},\,\,y'=-\frac{1}{5}$and when $x'=-\frac{2}{5},\,\,\,y'=\frac{1}{5}$                    Hence points are $\left( \frac{2}{5},\,\,-\frac{1}{5} \right)\,$and $\left( -\frac{2}{5},\,\,\frac{1}{5} \right)$.