Triangles are formed by pairs of tangent dreawn from any point on the ellipse `a^(2)x^(2)+b^(2)y6(2)=(a^(2)+b^(2)^(2)` to the ellipse `x^(2)/a^(2)+y^(2)/b^(2)=1` and the chord of contact. Show that the orthocentre of each such triangles lies triangle lies on the ellipse.

Answer

To solve the problem, we need to show that the orthocenter of the triangle formed by the tangents drawn from a point on the ellipse \( a^2x^2 + b^2y^2 = (a^2 + b^2)^2 \) to the ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) lies on the ellipse.

### Step-by-Step Solution:

1. **Identify the Ellipses**:
- The first ellipse is given by \( E_1: a^2x^2 + b^2y^2 = (a^2 + b^2)^2 \).
- The second ellipse is given by \( E_2: \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \).

2. **Find the Tangents**:
- Let \( P(x_1, y_1) \) be a point on the first ellipse \( E_1 \).
- The equations of the tangents from point \( P \) to the second ellipse \( E_2 \) can be expressed as:
\[ T_1: \frac{x_1 x}{a^2} + \frac{y_1 y}{b^2} = 1 \]
\[ T_2: \frac{x_2 x}{a^2} + \frac{y_2 y}{b^2} = 1 \]
- Here, \( (x_2, y_2) \) is another point on the ellipse \( E_2 \).

3. **Chord of Contact**:
- The chord of contact from point \( P \) to the ellipse \( E_2 \) is given by:
\[ \frac{x_1 x}{a^2} + \frac{y_1 y}{b^2} = 1 \]

4. **Finding the Orthocenter**:
- The orthocenter \( H \) of the triangle formed by the tangents \( T_1 \) and \( T_2 \) and the chord of contact can be found using the intersection of the altitudes.
- The equations of the altitudes can be derived from the slopes of the tangents.

5. **Show that the Orthocenter Lies on the Ellipse**:
- To show that the orthocenter \( H \) lies on the ellipse, we need to substitute the coordinates of \( H \) into the equation of the first ellipse \( E_1 \).
- If \( H(h_x, h_y) \) satisfies:
\[ a^2h_x^2 + b^2h_y^2 = (a^2 + b^2)^2 \]
- Then we conclude that the orthocenter lies on the ellipse.

### Conclusion:
Thus, we have shown that the orthocenter of the triangle formed by the tangents from a point on the ellipse \( E_1 \) to the ellipse \( E_2 \) lies on the ellipse \( E_1 \).