Question

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 $ax+by=1$, then $\dfrac{b}{a}=$

• A. $\dfrac{1}{\sqrt{2}e}$
• B. $\dfrac{e}{\sqrt{2}}$
• C. $\sqrt{2}e$
• D. $\dfrac{\sqrt{2}}{e}$

Hint:

Tangents and Normals to Curves:

• Slope of normal: $m_N = -\frac1m_T$
• Use point-slope: $y - y_1 = m_N(x - x_1)$
• Rearrange to $ax + by = 1$ to compare and find $\fracba$

Answer 1

The Correct Option is C

Given: $\dfrac{dy}{dx} = xy$ At $(\sqrt{2}, e)$, slope of tangent: $\sqrt{2}e$ $\Rightarrow$ slope of normal: $-\dfrac{1}{\sqrt{2}e}$ Equation of normal: \[ y - e = -\frac{1}{\sqrt{2}e}(x - \sqrt{2}) \Rightarrow x + \sqrt{2}ey = \sqrt{2}(1 + e^2) \] Divide both sides by $\sqrt{2}(1+e^2)$: \[ \frac{x}{\sqrt{2}(1+e^2)} + \frac{e}{1+e^2} y = 1 \] Compare with $ax + by = 1$: \[ a = \frac{1}{\sqrt{2}(1+e^2)}, \quad b = \frac{e}{1+e^2} \Rightarrow \frac{b}{a} = \sqrt{2}e \]