The locus of point of intersection of lines `(xt)/(a)-(y)/(b)+t=0` and `(x)/(at)+(y)/(b)=(1)/(t)`

Options

A. striaght line
B. circle
C. parabola
D. an ellipse

Answer

To find the locus of the point of intersection of the lines given by the equations \(\frac{xt}{a} - \frac{y}{b} + t = 0\) and \(\frac{x}{at} + \frac{y}{b} = \frac{1}{t}\), we will eliminate the parameter \(t\) step by step.

### Step 1: Rearranging the first equation
Starting with the first equation:
\[ \frac{xt}{a} - \frac{y}{b} + t = 0 \]
We can rearrange it to isolate \(t\):
\[ t\left(\frac{x}{a} + 1\right) = \frac{y}{b} \]
Thus, we can express \(t\) as:
\[ t = \frac{y/b}{(x/a) + 1} = \frac{y}{b} \cdot \frac{a}{x + a} = \frac{ay}{b(x + a)} \]

### Step 2: Rearranging the second equation
Now, we take the second equation:
\[ \frac{x}{at} + \frac{y}{b} = \frac{1}{t} \]
Multiplying through by \(t\) gives:
\[ \frac{x}{a} + \frac{y}{b}t = 1 \]
Rearranging for \(t\) yields:
\[ \frac{y}{b}t = 1 - \frac{x}{a} \]
Thus, we can express \(t\) as:
\[ t = \frac{b(1 - \frac{x}{a})}{y} = \frac{b(a - x)}{ay} \]

### Step 3: Equating the two expressions for \(t\)
Now we have two expressions for \(t\):
1. \(t = \frac{ay}{b(x + a)}\)
2. \(t = \frac{b(a - x)}{ay}\)

Setting these equal to each other:
\[ \frac{ay}{b(x + a)} = \frac{b(a - x)}{ay} \]

### Step 4: Cross-multiplying to eliminate \(t\)
Cross-multiplying gives:
\[ a^2y^2 = b^2(a - x)(x + a) \]

### Step 5: Expanding and rearranging
Expanding the right-hand side:
\[ a^2y^2 = b^2(ax + a^2 - x^2 - ax) = b^2(a^2 - x^2) \]
Rearranging gives:
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]

### Conclusion
This is the equation of an ellipse. Therefore, the locus of the point of intersection of the given lines is:
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]