Find the locus of the intersection of tangents of ellipse if the lines joining the points of contact to the centre be perpendicular.

Answer

To find the locus of the intersection of tangents of an ellipse when the lines joining the points of contact to the center are perpendicular, we can follow these steps:

### Step 1: Define the Ellipse
The standard equation of the ellipse is given by:
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]
where \( (0, 0) \) is the center of the ellipse.

### Step 2: Identify Points of Contact
Let \( P(h, k) \) be the point from which tangents are drawn to the ellipse. The points of contact on the ellipse can be denoted as \( A(x_1, y_1) \) and \( B(x_2, y_2) \).

### Step 3: Equation of the Chord of Contact
The equation of the chord of contact from point \( P(h, k) \) to the ellipse is given by:
\[ \frac{hx}{a^2} + \frac{ky}{b^2} = 1 \]

### Step 4: Condition for Perpendicularity
The lines joining the points of contact \( A \) and \( B \) to the center \( (0, 0) \) are perpendicular. This means that the slopes of the lines \( OA \) and \( OB \) must satisfy the condition:
\[ \text{slope of } OA \cdot \text{slope of } OB = -1 \]
If \( A(x_1, y_1) \) and \( B(x_2, y_2) \) are the points of contact, this translates to:
\[ \frac{y_1}{x_1} \cdot \frac{y_2}{x_2} = -1 \]

### Step 5: Homogenization
To find the locus, we need to homogenize the ellipse equation with the chord of contact. We replace \( 1 \) in the ellipse equation with \( \frac{hx}{a^2} + \frac{ky}{b^2} \):
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} - \left(\frac{hx}{a^2} + \frac{ky}{b^2}\right) = 0 \]

### Step 6: Rearranging the Equation
After substituting and rearranging, we get:
\[ x^2 - \frac{h^2x^2}{a^4} + y^2 - \frac{k^2y^2}{b^4} + \frac{2hxy}{a^2b^2} = 0 \]

### Step 7: Condition for Perpendicular Lines
Since the lines are perpendicular, we can use the condition:
\[ \frac{1}{a^2} - \frac{h^2}{a^4} + \frac{1}{b^2} - \frac{k^2}{b^4} = 0 \]
This leads to:
\[ \frac{b^4 + a^4}{a^4b^4} = \frac{h^2}{a^4} + \frac{k^2}{b^4} \]

### Step 8: Final Locus Equation
After simplification, we arrive at the locus of the point \( P(h, k) \):
\[ a^2b^2(h^2 + k^2) = a^4 + b^4 \]

### Conclusion
The locus of the intersection of the tangents of the ellipse, given that the lines joining the points of contact to the center are perpendicular, is:
\[ a^2b^2(h^2 + k^2) = a^4 + b^4 \]