The number of chords drawn from point (a, a) on the circle `x^(2)+y"^(2)=2a^(2)`, which are bisected by the parabola `y^(2)=4ax`, is

Options

A. 1
B. 4
C. 2
D. 0

Explanation

To solve the problem of finding the number of chords drawn from the point (a, a) on the circle \(x^2 + y^2 = 2a^2\) that are bisected by the parabola \(y^2 = 4ax\), we will follow these steps:

### Step 1: Understand the given equations
We have:
1. Circle: \(x^2 + y^2 = 2a^2\)
2. Parabola: \(y^2 = 4ax\)

### Step 2: Parameterize the point on the parabola
A point \(P\) on the parabola can be parameterized as:
\[ P(t) = (at^2, 2at) \]

### Step 3: Write the equation of the chord
The equation of a chord of the circle that is bisected at point \(P(t)\) is given by:
\[ x x_1 + y y_1 = x_1^2 + y_1^2 \]
Substituting \(x_1 = at^2\) and \(y_1 = 2at\), we get:
\[ x(at^2) + y(2at) = (at^2)^2 + (2at)^2 \]
This simplifies to:
\[ at^2 x + 2aty = a^2 t^4 + 4a^2 t^2 \]

### Step 4: Substitute the point (a, a) into the chord equation
Since the chord passes through the point (a, a), we substitute \(x = a\) and \(y = a\):
\[ at^2(a) + 2at(a) = a^2 t^4 + 4a^2 t^2 \]
This simplifies to:
\[ at^3 + 2at^2 = a^2 t^4 + 4a^2 t^2 \]
Dividing through by \(a\) (assuming \(a \neq 0\)):
\[ t^3 + 2t^2 = at^4 + 4at^2 \]

### Step 5: Rearranging the equation
Rearranging gives:
\[ at^4 + (4a - 1)t^2 - t^3 = 0 \]
This can be rewritten as:
\[ t^4 + \left(\frac{4a - 1}{a}\right)t^2 - \frac{1}{a}t^3 = 0 \]

### Step 6: Analyze the polynomial
Let \(f(t) = t^4 - t^3 + (4a - 1)t^2\). We need to find the number of real roots of this polynomial.

### Step 7: Check for roots
We can check for the number of real roots using the derivative:
\[ f'(t) = 4t^3 - 3t^2 + 2(4a - 1)t \]
Since \(f(t)\) is a polynomial of degree 4, it can have at most 4 real roots.

### Step 8: Evaluate the function at specific points
Evaluate \(f(0)\):
\[ f(0) = 0 \]
Evaluate \(f(1)\):
\[ f(1) = 1 - 1 + (4a - 1) = 4a - 1 \]
- If \(a > \frac{1}{4}\), \(f(1) > 0\).
- If \(a < \frac{1}{4}\), \(f(1) < 0\).

### Step 9: Conclusion
Since \(f(t)\) is continuous and changes sign between \(0\) and \(1\), there is at least one root in that interval. Given that \(f(t)\) is a polynomial of degree 4, we can conclude that there are exactly two real roots, which correspond to two chords.

Thus, the number of chords drawn from the point (a, a) on the circle that are bisected by the parabola is **2**.