Ifchord ofcontact ofthe 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 locus of the point ( α , β ) is

Answer

To find the locus of the point \((\alpha, \beta)\) such that the chord of contact of the tangents drawn from this point to the ellipse \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\) touches the circle \(x^2 + y^2 = c^2\), we can follow these steps:

### Step 1: Write the equation of the chord of contact
The chord of contact of tangents drawn from the point \((\alpha, \beta)\) to the ellipse is given by the equation:
\[ \frac{\alpha x}{a^2} + \frac{\beta y}{b^2} - 1 = 0 \]

### Step 2: Identify the center and radius of the circle
The given circle is \(x^2 + y^2 = c^2\). The center of this circle is at the origin \((0, 0)\) and its radius is \(c\).

### Step 3: Calculate the perpendicular distance from the center of the circle to the chord of contact
Using the formula for the perpendicular distance \(d\) from a point \((x_1, y_1)\) to the line \(Ax + By + C = 0\):
\[ d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \]
In our case, the line is \(\frac{\alpha x}{a^2} + \frac{\beta y}{b^2} - 1 = 0\), which can be rewritten as:
\[ \frac{\alpha}{a^2} x + \frac{\beta}{b^2} y - 1 = 0 \]
Here, \(A = \frac{\alpha}{a^2}\), \(B = \frac{\beta}{b^2}\), and \(C = -1\). The coordinates of the center of the circle are \((0, 0)\).

Thus, the perpendicular distance \(d\) is:
\[ d = \frac{\left| -1 \right|}{\sqrt{\left(\frac{\alpha}{a^2}\right)^2 + \left(\frac{\beta}{b^2}\right)^2}} = \frac{1}{\sqrt{\frac{\alpha^2}{a^4} + \frac{\beta^2}{b^4}}} \]

### Step 4: Set the distance equal to the radius of the circle
Since the chord of contact touches the circle, the perpendicular distance 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 this equation leads to:
\[ \frac{\alpha^2}{a^4} + \frac{\beta^2}{b^4} = \frac{1}{c^2} \]

### Conclusion: The locus of the point \((\alpha, \beta)\)
The equation \(\frac{\alpha^2}{a^4} + \frac{\beta^2}{b^4} = \frac{1}{c^2}\) represents an ellipse. Therefore, the locus of the point \((\alpha, \beta)\) is an ellipse.