PA and PB are tangents drawn from a point P to the ellipse `(x^2)/(a^2)+(y^2)/(b^2)=1`. The area of the triangle formed by the chord of contact AB and axes of co-ordinates is constant. Then locus of P is:

Options

A. ellipse
B. circle
C. hyperbola
D. straight

Answer

To solve the problem, we need to find the locus of the point \( P(h, k) \) from which tangents \( PA \) and \( PB \) are drawn to the ellipse given by the equation:

\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]

### Step 1: Equation of the Chord of Contact
The equation of the chord of contact from the point \( P(h, k) \) to the ellipse is given by:

\[ \frac{hx}{a^2} + \frac{ky}{b^2} = 1 \]

### Step 2: Finding Points of Intersection with Axes
To find the area of the triangle formed by the chord of contact and the coordinate axes, we first find the points where the chord intersects the axes.

1. **Intersection with the y-axis** (set \( x = 0 \)):
\[ \frac{h(0)}{a^2} + \frac{ky}{b^2} = 1 \implies y = \frac{b^2}{k} \]
So, the point of intersection is \( M(0, \frac{b^2}{k}) \).

2. **Intersection with the x-axis** (set \( y = 0 \)):
\[ \frac{hx}{a^2} + \frac{k(0)}{b^2} = 1 \implies x = \frac{a^2}{h} \]
So, the point of intersection is \( N(\frac{a^2}{h}, 0) \).

### Step 3: Area of the Triangle
The area \( A \) of triangle \( OMN \) (where \( O \) is the origin) can be calculated using the formula for the area of a triangle formed by the axes:

\[ A = \frac{1}{2} \times \text{base} \times \text{height} \]

Here, the base is \( \frac{a^2}{h} \) and the height is \( \frac{b^2}{k} \):

\[ A = \frac{1}{2} \times \frac{a^2}{h} \times \frac{b^2}{k} = \frac{a^2b^2}{2hk} \]

### Step 4: Setting Area to be Constant
We are given that the area \( A \) is constant, say \( C \):

\[ \frac{a^2b^2}{2hk} = C \]

Rearranging gives:

\[ hk = \frac{a^2b^2}{2C} \]

### Step 5: Locus of Point \( P \)
Let \( k = \frac{a^2b^2}{2C} \cdot \frac{1}{h} \). If we let \( k = \frac{a^2b^2}{2C} \cdot \frac{1}{x} \), we can rewrite this as:

\[ xy = \frac{a^2b^2}{2C} \]

This is the equation of a hyperbola.

### Conclusion
Thus, the locus of the point \( P \) is given by:

\[ xy = k \quad \text{(where \( k = \frac{a^2b^2}{2C} \) is a constant)} \]