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 $ax + by = 1$: \[ a = \frac{1}{\sqrt{2}(1+e^2)}, \quad b = \frac{e}{1+e^2} \Rightarrow \frac{b}{a} = \sqrt{2}e \]

Answer 2

Step 1: Understand the problem
The slope of the tangent at any point \((x,y)\) on the curve is given by:
\[ \frac{dy}{dx} = xy \]
We need to find \(\frac{b}{a}\) if the equation of the normal at point \((\sqrt{2}, e)\) is \(ax + by = 1\).

Step 2: Find the slope of the tangent at the given point
At \((x, y) = (\sqrt{2}, e)\),
\[ m_{\text{tangent}} = x \times y = \sqrt{2} \times e = \sqrt{2} e \]

Step 3: Find the slope of the normal
The normal is perpendicular to the tangent, so its slope is:
\[ m_{\text{normal}} = -\frac{1}{m_{\text{tangent}}} = -\frac{1}{\sqrt{2} e} \]

Step 4: Write the equation of the normal
Using point-slope form:
\[ y - e = m_{\text{normal}} (x - \sqrt{2}) = -\frac{1}{\sqrt{2} e}(x - \sqrt{2}) \]
Multiply both sides by \(\sqrt{2} e\):
\[ \sqrt{2} e (y - e) = -(x - \sqrt{2}) \]
\[ \sqrt{2} e y - \sqrt{2} e^2 = -x + \sqrt{2} \]
Rearranging:
\[ x + \sqrt{2} e y = \sqrt{2} + \sqrt{2} e^2 \]

Step 5: Express the normal equation in form \(ax + by = 1\)
Divide entire equation by \(\sqrt{2} + \sqrt{2} e^2\):
\[ \frac{x}{\sqrt{2} + \sqrt{2} e^2} + \frac{\sqrt{2} e}{\sqrt{2} + \sqrt{2} e^2} y = 1 \]
Thus,
\[ a = \frac{1}{\sqrt{2} + \sqrt{2} e^2}, \quad b = \frac{\sqrt{2} e}{\sqrt{2} + \sqrt{2} e^2} \]

Step 6: Calculate \(\frac{b}{a}\)
\[ \frac{b}{a} = \frac{\frac{\sqrt{2} e}{\sqrt{2} + \sqrt{2} e^2}}{\frac{1}{\sqrt{2} + \sqrt{2} e^2}} = \sqrt{2} e \]

Final answer:
\[ \boxed{\sqrt{2} e} \]