Find the locus of the point intersection of the tangents at the end of chord of elipse `(x^(2))/(a^(2))+(y^(2))/(b^(2))=1`, which substends an angle `alpha` at origin.

Answer

To find the locus of the point of intersection of the tangents at the endpoints of a chord of the ellipse \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\) that subtends an angle \(\alpha\) at the origin, we can follow these steps:

### Step 1: Understand the Geometry
We have an ellipse centered at the origin. The endpoints of the chord are points \(A(x_1, y_1)\) and \(B(x_2, y_2)\) on the ellipse. The angle subtended by the chord \(AB\) at the origin \(O(0, 0)\) is \(\alpha\).

### Step 2: Equation of the Chord
The equation of the chord can be expressed using the parameterization of the ellipse. If \(A\) and \(B\) are points on the ellipse, we can use the parametric equations:
\[ x_1 = a \cos \theta_1, \quad y_1 = b \sin \theta_1 \]
\[ x_2 = a \cos \theta_2, \quad y_2 = b \sin \theta_2 \]
where \(\theta_1\) and \(\theta_2\) are the angles corresponding to points \(A\) and \(B\).

### Step 3: Condition for Angle Subtended
The angle \(\alpha\) subtended at the origin by the chord can be expressed using the tangent of half the angle:
\[ \tan\left(\frac{\alpha}{2}\right) = \frac{y_1 - y_2}{x_1 - x_2} \]
Substituting the coordinates of points \(A\) and \(B\):
\[ \tan\left(\frac{\alpha}{2}\right) = \frac{b \sin \theta_1 - b \sin \theta_2}{a \cos \theta_1 - a \cos \theta_2} \]

### Step 4: Equation of Tangents
The equations of the tangents at points \(A\) and \(B\) can be derived from the general tangent equation to the ellipse:
\[ \frac{xx_1}{a^2} + \frac{yy_1}{b^2} = 1 \quad \text{(for point A)} \]
\[ \frac{xx_2}{a^2} + \frac{yy_2}{b^2} = 1 \quad \text{(for point B)} \]

### Step 5: Intersection of Tangents
Let the intersection point of the tangents be \(P(h, k)\). The equations of the tangents can be rewritten as:
\[ \frac{hx_1}{a^2} + \frac{ky_1}{b^2} = 1 \]
\[ \frac{hx_2}{a^2} + \frac{ky_2}{b^2} = 1 \]

### Step 6: Eliminate Parameters
From the above equations, we can eliminate \(x_1\) and \(y_1\) in terms of \(h\) and \(k\). This will yield a relationship between \(h\) and \(k\).

### Step 7: Locus Equation
After manipulating the equations and substituting the values, we can derive the locus of point \(P(h, k)\) in terms of \(h\) and \(k\). The final equation will be:
\[ \frac{h^2}{a^2 \cos^2 \alpha} + \frac{k^2}{b^2 \sin^2 \alpha} = 1 \]

### Final Result
The locus of the point of intersection of the tangents at the endpoints of the chord of the ellipse that subtends an angle \(\alpha\) at the origin is given by:
\[ \frac{x^2}{a^2 \cos^2 \alpha} + \frac{y^2}{b^2 \sin^2 \alpha} = 1 \]