Q107 / 160 · AP EAMCET 2023 (18 May Shift 1) · Mathematics
The slope of the tangent at any point $(x, y)$ on the curve, is equal to the product of the coordinates of that point. If the equation of the normal to the curve at the point $(\sqrt{2}, e)$ is $a x+b y=1$, then $\frac{b}{a}=$
- A$\frac{1}{\sqrt{2} \mathrm{e}}$
- B$\frac{\mathrm{e}}{\sqrt{2}}$
- Correct answerC$\sqrt{2} \mathrm{e}$
- D$\frac{\sqrt{2}}{\mathrm{e}}$
Solution
Slope of tangent at point $(\sqrt{2}, \mathrm{e})$ is $\mathrm{m}=\mathrm{e} \sqrt{2}$
Then, slope of normal, $m^{\prime}=-\frac{1}{m}=-\frac{1}{e \sqrt{2}}$
Equation of normal at point $(\sqrt{2} \mathrm{e})$ :
$(y-e)=m^{\prime}(x-\sqrt{2})$
$\begin{aligned} & \Rightarrow y-e=-\frac{1}{e \sqrt{2}}(x-\sqrt{2}) \\ & \Rightarrow e \sqrt{2} y-e^2 \sqrt{2}=-x+\sqrt{2} \\ & \Rightarrow x+e \sqrt{2} y=\sqrt{2}+e^2 \sqrt{2}\end{aligned}$
$\Rightarrow \frac{x}{\sqrt{2}\left(e^2+1\right)}+\frac{e \sqrt{2}}{\sqrt{2}\left(e^2+1\right)} y=1$ ...(i)
Given equation of normal is, $a x+b y=1$ ...(ii)
Then $a=\frac{1}{\sqrt{2}\left(e^2+1\right)}$ and $b=\frac{e \sqrt{2}}{\sqrt{2}\left(e^2+1\right)}$
$\frac{\mathrm{b}}{\mathrm{a}}=\frac{\mathrm{e} \sqrt{2}}{1}=\sqrt{2} \mathrm{e}$
Hence, the correct answer is option: 'C'.