The straight line `y=x+2a` touches the parabola `y^2 = 4a (x+ a)` at the point

Options

A. `(-a,a)`
B. `(0,2a)`
C. `(-2a,0)`
D. `(a,3a)`

Answer

To find the point of tangency between the straight line \( y = x + 2a \) and the parabola \( y^2 = 4a(x + a) \), we can follow these steps:

### Step 1: Substitute the line equation into the parabola equation

We start with the equations:
- Line: \( y = x + 2a \)
- Parabola: \( y^2 = 4a(x + a) \)

First, we express \( x \) in terms of \( y \) from the line equation:
\[ x = y - 2a \]

Now, substitute this expression for \( x \) into the parabola equation:
\[ y^2 = 4a((y - 2a) + a) \]
This simplifies to:
\[ y^2 = 4a(y - a) \]

### Step 2: Expand and rearrange the equation

Expanding the right-hand side gives:
\[ y^2 = 4ay - 4a^2 \]

Now, rearranging the equation:
\[ y^2 - 4ay + 4a^2 = 0 \]

### Step 3: Factor the quadratic equation

The quadratic equation can be factored as:
\[ (y - 2a)^2 = 0 \]

This means that the equation has a double root, indicating that the line is tangent to the parabola.

### Step 4: Solve for \( y \)

From the factored equation, we find:
\[ y - 2a = 0 \implies y = 2a \]

### Step 5: Substitute \( y \) back to find \( x \)

Now, substitute \( y = 2a \) back into the line equation to find \( x \):
\[ x = 2a - 2a = 0 \]

### Step 6: Write the point of tangency

Thus, the point of tangency is:
\[ (0, 2a) \]

### Final Answer

The straight line \( y = x + 2a \) touches the parabola \( y^2 = 4a(x + a) \) at the point \( (0, 2a) \).

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