A line through the point (3, 0) meets the variable line `y=tx` at right angle at the point P. Find, in terms of t, the co-ordinates of P.
Find the value of k for which Plies on the curve `x^(2)+y^(2)=kx`.

Answer

To solve the problem step by step, we will follow the outlined approach to find the coordinates of point P and the value of k.

### Step 1: Understand the problem
We have a line through the point (3, 0) that meets the variable line \( y = tx \) at a right angle at point P. We need to find the coordinates of P in terms of t and then find the value of k for which P lies on the curve \( x^2 + y^2 = kx \).

### Step 2: Determine the slopes
The slope of the line \( y = tx \) is \( t \). For two lines to be perpendicular, the product of their slopes must equal -1. If we denote the slope of the line through (3, 0) as \( m_2 \), we have:
\[ t \cdot m_2 = -1 \implies m_2 = -\frac{1}{t} \]

### Step 3: Write the equation of the line through (3, 0)
Using the point-slope form of the equation of a line, we can write the equation of the line through the point (3, 0) with slope \( m_2 \):
\[ y - 0 = -\frac{1}{t}(x - 3) \]
This simplifies to:
\[ y = -\frac{1}{t}x + \frac{3}{t} \]

### Step 4: Set up the equations for intersection
We have two equations:
1. \( y = tx \) (the variable line)
2. \( y = -\frac{1}{t}x + \frac{3}{t} \) (the line through (3, 0))

To find the intersection point P, we set these two equations equal to each other:
\[ tx = -\frac{1}{t}x + \frac{3}{t} \]

### Step 5: Solve for x
Rearranging the equation gives:
\[ tx + \frac{1}{t}x = \frac{3}{t} \]
Factoring out \( x \):
\[ x\left(t + \frac{1}{t}\right) = \frac{3}{t} \]
Thus,
\[ x = \frac{3/t}{t + \frac{1}{t}} = \frac{3}{t^2 + 1} \]

### Step 6: Find y-coordinate
Now substituting \( x \) back into the equation \( y = tx \):
\[ y = t\left(\frac{3}{t^2 + 1}\right) = \frac{3t}{t^2 + 1} \]

### Step 7: Coordinates of point P
The coordinates of point P in terms of t are:
\[ P\left(\frac{3}{t^2 + 1}, \frac{3t}{t^2 + 1}\right) \]

### Step 8: Find the value of k
We need to find the value of k such that point P lies on the curve \( x^2 + y^2 = kx \). Substituting the coordinates of P into the curve equation:
\[ \left(\frac{3}{t^2 + 1}\right)^2 + \left(\frac{3t}{t^2 + 1}\right)^2 = k\left(\frac{3}{t^2 + 1}\right) \]

### Step 9: Simplify the left side
Calculating the left side:
\[ \frac{9}{(t^2 + 1)^2} + \frac{9t^2}{(t^2 + 1)^2} = \frac{9(1 + t^2)}{(t^2 + 1)^2} \]
So we have:
\[ \frac{9(1 + t^2)}{(t^2 + 1)^2} = k\left(\frac{3}{t^2 + 1}\right) \]

### Step 10: Cross-multiply and solve for k
Cross-multiplying gives:
\[ 9(1 + t^2) = 3k(t^2 + 1) \]
Dividing both sides by \( 3(t^2 + 1) \):
\[ k = 3 \]

### Final Answers
1. The coordinates of point P are \( \left(\frac{3}{t^2 + 1}, \frac{3t}{t^2 + 1}\right) \).
2. The value of k is \( 3 \).