Let `A(a,0) and B(b,0)` be fixed distinct points on the x-axis, none of which coincides with the `O(0,0)`, and let C be a point on the y-axis. Let L be a line through the `O(0,0)` and perpendicular to the line AC. The locus of the point of intersection of the lines L and BC if C varies along is (provided `c^2 +ab != 0`)

Options

A. (a) `(x^(2))/(a)+(y^(2))/(b) =x`
B. (b) `(x^(2))/(a)+(y^(2))/(b) =y`
C. (c) `(x^(2))/(b)+(y^(2))/(a)=x`
D. (d) `(x^(2))/(b)+(y^(2))/(a) =y`

Explanation

To solve the problem step by step, we will derive the locus of the point of intersection of the lines L and BC as point C varies along the y-axis.

### Step 1: Define the points and their coordinates
Let:
- Point A be \( A(a, 0) \)
- Point B be \( B(b, 0) \)
- Point C be \( C(0, k) \) where \( k \) varies along the y-axis.
- The origin \( O(0, 0) \)

### Step 2: Find the equation of line AC
The slope of line AC can be calculated as follows:
\[ \text{slope of AC} = \frac{k - 0}{0 - a} = -\frac{k}{a} \]
Using the point-slope form, the equation of line AC can be written as:
\[ y - 0 = -\frac{k}{a}(x - a) \]
Rearranging gives:
\[ y = -\frac{k}{a}x + k \]

### Step 3: Find the slope of line L
Line L is perpendicular to line AC. The slope of line L is the negative reciprocal of the slope of line AC:
\[ \text{slope of L} = \frac{a}{k} \]
Since line L passes through the origin, its equation is:
\[ y = \frac{a}{k}x \]

### Step 4: Find the equation of line BC
The slope of line BC can be calculated as follows:
\[ \text{slope of BC} = \frac{k - 0}{0 - b} = -\frac{k}{b} \]
Using the point-slope form, the equation of line BC can be written as:
\[ y - 0 = -\frac{k}{b}(x - b) \]
Rearranging gives:
\[ y = -\frac{k}{b}x + k \]

### Step 5: Find the point of intersection of lines L and BC
To find the intersection point (let's call it point N), we need to solve the equations of lines L and BC:
1. \( y = \frac{a}{k}x \) (Equation of line L)
2. \( y = -\frac{k}{b}x + k \) (Equation of line BC)

Setting these two equations equal to each other:
\[ \frac{a}{k}x = -\frac{k}{b}x + k \]

### Step 6: Solve for x
Rearranging gives:
\[ \frac{a}{k}x + \frac{k}{b}x = k \]
Factoring out \( x \):
\[ x\left(\frac{a}{k} + \frac{k}{b}\right) = k \]
Thus,
\[ x = \frac{k}{\frac{a}{k} + \frac{k}{b}} = \frac{kb}{a + \frac{k^2}{b}} = \frac{kb^2}{ab + k^2} \]

### Step 7: Substitute x back to find y
Substituting \( x \) back into the equation of line L:
\[ y = \frac{a}{k}\left(\frac{kb^2}{ab + k^2}\right) = \frac{ab^2}{ab + k^2} \]

### Step 8: Find the locus
The coordinates of point N are:
\[ N\left(\frac{kb^2}{ab + k^2}, \frac{ab^2}{ab + k^2}\right) \]
Let \( x = \frac{kb^2}{ab + k^2} \) and \( y = \frac{ab^2}{ab + k^2} \).

Cross-multiplying gives:
\[ x(ab + k^2) = kb^2 \]
\[ abx + k^2x = kb^2 \]
Rearranging gives:
\[ k^2x = kb^2 - abx \]
Dividing by \( k \) (assuming \( k \neq 0 \)):
\[ kx = b^2 - ab\frac{x}{k} \]
This leads to:
\[ \frac{x^2}{b} + \frac{y^2}{a} = x \]

### Conclusion
The locus of the point of intersection of lines L and BC as point C varies along the y-axis is given by:
\[ \frac{x^2}{b} + \frac{y^2}{a} = x \]