The length of the subnormal to the parabola `y^(2)=4ax` at any point is equal to

Options

A. `asqrt2`
B. `2sqrt2a`
C. `a/sqrt2`
D. `2a`

Answer

To find the length of the subnormal to the parabola \( y^2 = 4ax \) at any point, we can follow these steps:

### Step 1: Identify the point on the parabola
Let the point on the parabola be given in parametric form as \( P(t) = (at^2, 2at) \).

### Step 2: Find the y-coordinate
From the point \( P(t) \), the y-coordinate is \( y = 2at \).

### Step 3: Differentiate the equation of the parabola
We differentiate the equation \( y^2 = 4ax \) with respect to \( x \):
\[ \frac{d}{dx}(y^2) = \frac{d}{dx}(4ax) \]
Using the chain rule, we have:
\[ 2y \frac{dy}{dx} = 4a \]
Thus, we can solve for \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = \frac{4a}{2y} = \frac{2a}{y} \]

### Step 4: Substitute the y-coordinate into the derivative
Now, substituting \( y = 2at \) into the derivative:
\[ \frac{dy}{dx} = \frac{2a}{2at} = \frac{1}{t} \]

### Step 5: Use the formula for the length of the subnormal
The formula for the length of the subnormal \( L \) is given by:
\[ L = |y| \cdot \frac{dy}{dx} \]
Substituting the values we found:
\[ L = |2at| \cdot \frac{1}{t} \]
This simplifies to:
\[ L = 2a \]

### Conclusion
Thus, the length of the subnormal to the parabola \( y^2 = 4ax \) at any point is equal to \( 2a \).

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