Out of the four curves given below chciose the curve which intersects the parabola `y^2 = 4ax` orthogonally

Options

A. `x^2+y^2=a^2`
B. `y=e^(-x//2a)`
C. `y=ax`
D. `x^2=4ay`

Answer

To find the curve that intersects the parabola \( y^2 = 4ax \) orthogonally, we need to follow these steps:

### Step 1: Understand the Condition for Orthogonality
Two curves intersect orthogonally if the product of their slopes (derivatives) at the point of intersection is -1. That is, if \( M_1 \) and \( M_2 \) are the slopes of the two curves, then:
\[ M_1 \cdot M_2 = -1 \]

### Step 2: Find the Slope of the Parabola
For the parabola \( y^2 = 4ax \), we can differentiate it implicitly to find \( M_1 \):
\[ 2y \frac{dy}{dx} = 4a \implies \frac{dy}{dx} = \frac{4a}{2y} = \frac{2a}{y} \]
Thus, \( M_1 = \frac{2a}{y} \).

### Step 3: Analyze Each Curve Option
We will check each of the four given curves to find which one satisfies the orthogonality condition.

#### Option 1: \( y = x^2 \)
1. Differentiate: \( \frac{dy}{dx} = 2x \)
2. Thus, \( M_2 = 2x \).
3. Calculate \( M_1 \cdot M_2 \):
\[ M_1 \cdot M_2 = \left(\frac{2a}{y}\right) \cdot (2x) = \frac{4ax}{y} \]
Substitute \( y^2 = 4ax \) (so \( y = \sqrt{4ax} \)):
\[ M_1 \cdot M_2 = \frac{4ax}{\sqrt{4ax}} = \sqrt{4ax} \neq -1 \]
This option does not satisfy the condition.

#### Option 2: \( y = e^{-x/(2a)} \)
1. Differentiate: \( \frac{dy}{dx} = -\frac{1}{2a} e^{-x/(2a)} \)
2. Thus, \( M_2 = -\frac{1}{2a} e^{-x/(2a)} \).
3. Calculate \( M_1 \cdot M_2 \):
\[ M_1 \cdot M_2 = \left(\frac{2a}{y}\right) \cdot \left(-\frac{1}{2a} e^{-x/(2a)}\right) = -\frac{2a}{e^{-x/(2a)}} \cdot \frac{1}{2a} = -1 \]
This option satisfies the condition.

#### Option 3: \( y = ax \)
1. Differentiate: \( \frac{dy}{dx} = a \)
2. Thus, \( M_2 = a \).
3. Calculate \( M_1 \cdot M_2 \):
\[ M_1 \cdot M_2 = \left(\frac{2a}{y}\right) \cdot a = \frac{2a^2}{ax} = \frac{2a}{x} \neq -1 \]
This option does not satisfy the condition.

#### Option 4: \( y = \frac{x^2}{4a} \)
1. Differentiate: \( \frac{dy}{dx} = \frac{x}{2a} \)
2. Thus, \( M_2 = \frac{x}{2a} \).
3. Calculate \( M_1 \cdot M_2 \):
\[ M_1 \cdot M_2 = \left(\frac{2a}{y}\right) \cdot \left(\frac{x}{2a}\right) = \frac{2a}{\frac{x^2}{4a}} \cdot \frac{x}{2a} = \frac{8a^2}{x^2} \cdot \frac{x}{2a} = \frac{4a}{x} \neq -1 \]
This option does not satisfy the condition.

### Conclusion
The only curve that intersects the parabola \( y^2 = 4ax \) orthogonally is:
\[ \text{Option 2: } y = e^{-x/(2a)} \]