AB is a line of fixed length, 6 units, joining the points A (t, 0) and B which lies on the positive y-axis. P is a point on AB distant 2 units from A. Express the co-ordinates of B and P in terms of t. Find the locus of P as t varies.

Answer

To solve the problem step by step, we will follow the given information and derive the required coordinates and locus of point P.

### Step 1: Identify the coordinates of points A and B
- Point A is given as \( A(t, 0) \).
- Point B lies on the positive y-axis, so we can denote its coordinates as \( B(0, k) \), where \( k > 0 \).

### Step 2: Use the distance formula to find the relationship between A and B
The length of line segment AB is given as 6 units. Using the distance formula:
\[ AB = \sqrt{(t - 0)^2 + (0 - k)^2} = 6 \]
This simplifies to:
\[ \sqrt{t^2 + k^2} = 6 \]
Squaring both sides, we get:
\[ t^2 + k^2 = 36 \quad \text{(1)} \]

### Step 3: Determine the coordinates of point P
Point P is located on line segment AB and is 2 units away from A. Since AB is 6 units long, the distance from P to B will be \( 6 - 2 = 4 \) units.

Using the section formula, since P divides AB in the ratio \( 1:2 \) (1 part from A to P and 2 parts from P to B), the coordinates of P can be calculated as follows:
\[ P\left(\frac{1 \cdot 0 + 2 \cdot t}{1 + 2}, \frac{1 \cdot k + 2 \cdot 0}{1 + 2}\right) = P\left(\frac{2t}{3}, \frac{k}{3}\right) \]

### Step 4: Express k in terms of t using equation (1)
From equation (1):
\[ k^2 = 36 - t^2 \]
Taking the square root:
\[ k = \sqrt{36 - t^2} \]

### Step 5: Substitute k into the coordinates of P
Now substituting \( k \) into the coordinates of P:
\[ P\left(\frac{2t}{3}, \frac{\sqrt{36 - t^2}}{3}\right) \]

### Step 6: Find the locus of P as t varies
To find the locus of point P, we need to eliminate t from the coordinates of P. We have:
\[ x = \frac{2t}{3} \quad \Rightarrow \quad t = \frac{3x}{2} \]
Substituting this into the equation for y:
\[ y = \frac{\sqrt{36 - t^2}}{3} = \frac{\sqrt{36 - \left(\frac{3x}{2}\right)^2}}{3} \]
Calculating \( t^2 \):
\[ t^2 = \left(\frac{3x}{2}\right)^2 = \frac{9x^2}{4} \]
Thus,
\[ y = \frac{\sqrt{36 - \frac{9x^2}{4}}}{3} \]
Simplifying further:
\[ y = \frac{1}{3}\sqrt{\frac{144 - 9x^2}{4}} = \frac{1}{6}\sqrt{144 - 9x^2} \]
Squaring both sides:
\[ 36y^2 = 144 - 9x^2 \]
Rearranging gives:
\[ 9x^2 + 36y^2 = 144 \]
Dividing through by 144:
\[ \frac{x^2}{16} + \frac{y^2}{4} = 1 \]
This is the equation of an ellipse.

### Final Answer:
The locus of point P as t varies is given by:
\[ \frac{x^2}{16} + \frac{y^2}{4} = 1 \]