If the chords of contact of tangents from two points to the ellipse are a right angles, then show that `(x_(1)x_(2))/(y_(1)y_(2))=-(a^(4))/(b^(4))`

Options

A. `-16`
B. 4
C. 6
D. `4/3`

Answer

To solve the problem, we need to show that if the chords of contact of tangents from two points to the ellipse are at right angles, then:

\[ \frac{x_1 x_2}{y_1 y_2} = -\frac{a^4}{b^4} \]

### Step 1: Write the equation of the ellipse
The standard form of the ellipse is given by:

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

### Step 2: Write the equations of the chords of contact
For a point \((x_1, y_1)\) outside the ellipse, the chord of contact is given by:

\[ \frac{x_1 x}{a^2} + \frac{y_1 y}{b^2} = 1 \]

Similarly, for the point \((x_2, y_2)\), the chord of contact is:

\[ \frac{x_2 x}{a^2} + \frac{y_2 y}{b^2} = 1 \]

### Step 3: Find the slopes of the tangents
The slope of the tangent line from the point \((x_1, y_1)\) can be derived from the chord of contact equation. Rearranging the equation gives:

\[ y = -\frac{b^2}{y_1} \cdot \frac{x_1}{a^2} x + \frac{b^2}{y_1} \]

Thus, the slope \(m_1\) of the tangent from point \((x_1, y_1)\) is:

\[ m_1 = -\frac{b^2 x_1}{a^2 y_1} \]

Similarly, for point \((x_2, y_2)\):

\[ m_2 = -\frac{b^2 x_2}{a^2 y_2} \]

### Step 4: Use the condition for perpendicular lines
Since the tangents are at right angles, we have:

\[ m_1 \cdot m_2 = -1 \]

Substituting the slopes:

\[ \left(-\frac{b^2 x_1}{a^2 y_1}\right) \cdot \left(-\frac{b^2 x_2}{a^2 y_2}\right) = -1 \]

This simplifies to:

\[ \frac{b^4 x_1 x_2}{a^4 y_1 y_2} = -1 \]

### Step 5: Rearranging the equation
Rearranging gives:

\[ \frac{x_1 x_2}{y_1 y_2} = -\frac{a^4}{b^4} \]

### Conclusion
Thus, we have shown that:

\[ \frac{x_1 x_2}{y_1 y_2} = -\frac{a^4}{b^4} \]