Question

The area (in sq units) of the triangle formed by the normal at point $(1,0)$ on the curve $x = e^{\sin y}$ with coordinate axes is:

• A. $1$
• B. $\dfrac{1}{4}$
• C. $\dfrac{1}{2}$
• D. $\dfrac{3}{8}$

Hint:

Normal and Triangle Area:

• Tangent slope $\Rightarrow$ normal slope $= -1/m$
• Normal + axes intersection $\Rightarrow$ triangle
• Area = $\frac12 \cdot x_\textint \cdot y_\textint$

Answer 1

The Correct Option is C

Given: $x = e^{\sin y} \Rightarrow \frac{dx}{dy} = e^{\sin y} \cos y$ \[ \Rightarrow \frac{dy}{dx} = \frac{1}{e^{\sin y} \cos y} \Rightarrow \left. \frac{dy}{dx} \right|_{(1,0)} = 1 \Rightarrow \text{slope of normal } = -1 \] Equation of normal: \[ y - 0 = -1(x - 1) \Rightarrow x + y = 1 \] Intercepts: $(1,0)$ and $(0,1) \Rightarrow \text{Area} = \frac{1}{2}(1)(1) = \boxed{\frac{1}{2}}$

Answer 2

Step 1: Understand the problem
We have the curve \( x = e^{\sin y} \) and the point \( (1,0) \) on it.
We need to find the area of the triangle formed by the normal at this point with the coordinate axes.

Step 2: Verify the point lies on the curve
At \( y=0 \), \(\sin 0 = 0\), so \( x = e^0 = 1 \).
Hence, the point \( (1,0) \) lies on the curve.

Step 3: Find \(\frac{dx}{dy}\)
Differentiate \( x = e^{\sin y} \) with respect to \( y \):
\[ \frac{dx}{dy} = e^{\sin y} \cos y \]
At \( y=0 \):
\[ \frac{dx}{dy} \Big|_{y=0} = e^0 \times \cos 0 = 1 \times 1 = 1 \]

Step 4: Find slope of tangent \(\frac{dy}{dx}\) at \(y=0\)
Since \( \frac{dx}{dy} = 1 \), by reciprocal rule:
\[ \frac{dy}{dx} = \frac{1}{\frac{dx}{dy}} = \frac{1}{1} = 1 \]
So slope of tangent \(m_t = 1\).

Step 5: Find slope of normal \(m_n\)
Slope of normal is negative reciprocal of slope of tangent:
\[ m_n = -\frac{1}{m_t} = -1 \]

Step 6: Write equation of normal at point \( (1,0) \)
Using point-slope form:
\[ y - 0 = -1(x - 1) \implies y = -x + 1 \]

Step 7: Find intercepts of the normal line with coordinate axes
- \(x\)-intercept: set \(y=0\)
\[ 0 = -x + 1 \implies x = 1 \]
- \(y\)-intercept: set \(x=0\)
\[ y = -0 + 1 = 1 \]

Step 8: Calculate area of triangle formed by intercepts
Area \(= \frac{1}{2} \times \text{(x-intercept)} \times \text{(y-intercept)} = \frac{1}{2} \times 1 \times 1 = \frac{1}{2}\)

Final answer:
\[ \boxed{\frac{1}{2}} \]