If the chord of contact of the tangents drawn from the point `(alpha,beta)` to the ellipse `x^2//a^2 + y^2//b^2 = 1` touches the circle `x^2 + y^2 = c^2`, then the point `(alpha,beta)` lies on the ellipse `x^2//a^4 + y^2//b^4 = 1//c^2`. T or F?

Answer

To determine whether the statement is true or false, we will follow a systematic approach to analyze the conditions given in the problem.

### Step 1: Write the equation of the ellipse and the circle
The equation of the ellipse is given by:
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]
The equation of the circle is given by:
\[ x^2 + y^2 = c^2 \]

### Step 2: Find the chord of contact from the point \((\alpha, \beta)\)
The chord of contact of the tangents drawn from the point \((\alpha, \beta)\) to the ellipse can be expressed using the formula:
\[ \frac{\alpha x}{a^2} + \frac{\beta y}{b^2} = 1 \]

### Step 3: Determine the condition for the chord of contact to touch the circle
For the chord of contact to touch the circle, the distance from the center of the circle (which is at the origin \((0, 0)\)) to the line represented by the chord of contact must equal the radius \(c\) of the circle.

The distance \(d\) from a point \((x_0, y_0)\) to the line \(Ax + By + C = 0\) is given by:
\[ d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \]

In our case, the line can be rewritten as:
\[ \frac{\alpha x}{a^2} + \frac{\beta y}{b^2} - 1 = 0 \]
Here, \(A = \frac{\alpha}{a^2}\), \(B = \frac{\beta}{b^2}\), and \(C = -1\).

Substituting \((x_0, y_0) = (0, 0)\):
\[ d = \frac{|-1|}{\sqrt{\left(\frac{\alpha}{a^2}\right)^2 + \left(\frac{\beta}{b^2}\right)^2}} = \frac{1}{\sqrt{\frac{\alpha^2}{a^4} + \frac{\beta^2}{b^4}}} \]

### Step 4: Set the distance equal to the radius
Since the distance \(d\) must equal the radius \(c\):
\[ \frac{1}{\sqrt{\frac{\alpha^2}{a^4} + \frac{\beta^2}{b^4}}} = c \]

### Step 5: Square both sides to eliminate the square root
Squaring both sides gives:
\[ \frac{1}{\frac{\alpha^2}{a^4} + \frac{\beta^2}{b^4}} = c^2 \]

### Step 6: Rearranging the equation
Rearranging gives:
\[ \frac{\alpha^2}{a^4} + \frac{\beta^2}{b^4} = \frac{1}{c^2} \]

### Step 7: Rewrite the equation
This can be rewritten as:
\[ \frac{\alpha^2}{a^4} + \frac{\beta^2}{b^4} = \frac{1}{c^2} \]
This is the equation of the ellipse:
\[ \frac{x^2}{a^4} + \frac{y^2}{b^4} = \frac{1}{c^2} \]

### Conclusion
Since we have shown that the point \((\alpha, \beta)\) lies on the ellipse \(\frac{x^2}{a^4} + \frac{y^2}{b^4} = \frac{1}{c^2}\), the statement is **True**.