The point `(p, p+1)` lies on the locus of the point which moves such that its distance from the point (1, 0) is twice the distance from (0,1). The value of `1//p^(2)+1//p^(4)` is equal to

Answer

To solve the problem step by step, we start with the information given in the question.

### Step 1: Set up the distance equations
We are given that the distance from the point \( (x, y) \) to the point \( A(1, 0) \) is twice the distance from the point \( (x, y) \) to the point \( B(0, 1) \).

Using the distance formula, we can express this mathematically:
\[ d(A) = \sqrt{(x - 1)^2 + (y - 0)^2} \]
\[ d(B) = \sqrt{(x - 0)^2 + (y - 1)^2} \]
According to the problem, we have:
\[ d(A) = 2 \cdot d(B) \]

### Step 2: Square both sides to eliminate the square roots
Squaring both sides gives us:
\[ ((x - 1)^2 + y^2) = 4((x^2) + (y - 1)^2) \]

### Step 3: Expand both sides
Now we expand both sides:
\[ (x - 1)^2 + y^2 = 4(x^2 + (y^2 - 2y + 1)) \]
Expanding the left side:
\[ x^2 - 2x + 1 + y^2 = 4(x^2 + y^2 - 2y + 1) \]
Expanding the right side:
\[ x^2 - 2x + 1 + y^2 = 4x^2 + 4y^2 - 8y + 4 \]

### Step 4: Rearranging the equation
Rearranging gives us:
\[ x^2 - 2x + 1 + y^2 = 4x^2 + 4y^2 - 8y + 4 \]
\[ 0 = 4x^2 + 4y^2 - 8y - x^2 - y^2 + 2x + 3 \]
Combining like terms:
\[ 0 = 3x^2 + 3y^2 - 8y + 2x + 3 \]

### Step 5: Substitute the point (p, p + 1)
Now we substitute \( x = p \) and \( y = p + 1 \):
\[ 0 = 3p^2 + 3(p + 1)^2 - 8(p + 1) + 2p + 3 \]
Expanding \( (p + 1)^2 \):
\[ 0 = 3p^2 + 3(p^2 + 2p + 1) - 8p - 8 + 2p + 3 \]
\[ 0 = 3p^2 + 3p^2 + 6p + 3 - 8p - 8 + 2p + 3 \]
Combining terms:
\[ 0 = 6p^2 + (6p - 8p + 2p) + (3 - 8 + 3) \]
\[ 0 = 6p^2 + 0p - 2 \]
This simplifies to:
\[ 6p^2 = 2 \implies p^2 = \frac{1}{3} \]

### Step 6: Calculate \( \frac{1}{p^2} + \frac{1}{p^4} \)
Now we need to find \( \frac{1}{p^2} + \frac{1}{p^4} \):
\[ \frac{1}{p^2} = 3 \quad \text{(since } p^2 = \frac{1}{3}\text{)} \]
\[ \frac{1}{p^4} = \frac{1}{(p^2)^2} = \frac{1}{\left(\frac{1}{3}\right)^2} = 9 \]
Thus,
\[ \frac{1}{p^2} + \frac{1}{p^4} = 3 + 9 = 12 \]

### Final Answer
The value of \( \frac{1}{p^2} + \frac{1}{p^4} \) is \( 12 \).

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