UPSC NDA Math Model Paper 8

Given: Equation of curve is $4 x^{2}+9 y^{2}=1$ and the tangents to the given curve are perpendicular to the line 2y + x = 0. Let $(x_1, y_1)$ be the point of contact. The given equation of line 2y + x = 0 can be written as: y = (-1/2) x. Now by comparing the equation y = (-1/2) x with y = mx + c we get: m = - 1/2 and c = 0 So, the slope of the given line 2y + x = 0 is $m_1 = -\frac{1}{2}$ As we know that slope of the tangent at any point say $(x_1, y_1)$ to a curve is given by:$m=\left[\frac{dy}{dx}\right](x_{1}, y_{1})$ Now by differentiating the equation of curve $4 x^{2}+9 y^{2}=1$ with respect to $x$ we get $\Rightarrow 8x+18y \frac{dy}{dx}=0$ $\Rightarrow \frac{dy}{dx}=-\frac{4x}{9y}$ $\Rightarrow \left[\frac{dy}{dx}\right]_{(x_{1}, y_{1})}=-\frac{4x_{1}}{9y_{1}}$ So, the slope of the tangent is $m_{2}=-\frac{4x_{1}}{9y_{1}}$ It is given that the tangents to the given curve are perpendicular to the line 2y + x = 0. As we know that, if $L_{1}$ and $L_{2}$ are two lines with slope $m_{1}$ and $m_{2}$ respectively and if they are perpendicular to each other then $m_{1} \times m_{2} = -1$ $\Rightarrow m_{1} \times m_{2} = -\frac{1}{2} \cdot -\frac{4x_{1}}{9y_{1}} = -1$ $\Rightarrow y_{1} = -\frac{2}{9} x_{1}$ $\because$ The point $(x_{1}, y_{1})$ lies on the curve $4 x^{2}+9 y^{2}=1$ $\Rightarrow 4 x_{1}^{2}+9 y_{1}^{2}=1$. By substituting $y_{1}=-\frac{2}{9} x_{1}$ in the above equation we get $\Rightarrow 4 x_{1}^{2}+9 \times \frac{4}{81} \times x_{1}^{2}=1$ $\Rightarrow x_{1}^{2}=\frac{9}{40}$ $\Rightarrow x_{1}=\pm \frac{3}{2\sqrt{10}}$ When $x_{1}=\frac{3}{2\sqrt{10}}$ then $y_{1}=-\frac{2}{9} x_{1} = -\frac{2}{9} \cdot \frac{3}{2\sqrt{10}} = \frac{1}{3\sqrt{10}}$ Similarly when $x_{1}=-\frac{3}{2\sqrt{10}}$ then $y_{1}=-\frac{2}{9}, x_{1}=-\frac{2}{9}$ $\frac{-3}{2\sqrt{10}} = \frac{-1}{3\sqrt{10}}$ Hence, the points of contact are $\left(\frac{3}{2\sqrt{10}}, \frac{1}{3\sqrt{10}}\right)$ and $\left(\frac{-3}{2\sqrt{10}}, \frac{-1}{3\sqrt{10}}\right)$