A line intersects the ellipe `(x^(2))/(a^(2))+(y^(2))/(b^(2))=1` at P and Q and the parabola `y^(2)=4d(x+a)` at R and S. The line segment PQ subtends a right angle at the centre of the ellipse. Find the locus of the point intersection of the tangents to the parabola at R and S.

Answer

To solve the problem step by step, we will find the locus of the point of intersection of the tangents to the parabola at points R and S, given that the line segment PQ subtends a right angle at the center of the ellipse.

### Step 1: Understand the Given Shapes
We have an ellipse defined by the equation:
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]
and a parabola defined by:
\[ y^2 = 4d(x + a) \]

### Step 2: Set Up the Line Equation
Let the line intersecting the ellipse be given in the slope-intercept form:
\[ y = mx + c \]
This line intersects the ellipse at points P and Q.

### Step 3: Condition for Right Angle at the Center
The line segment PQ subtends a right angle at the center of the ellipse (0,0). For this to happen, the product of the slopes of the tangents at points P and Q must equal -1.

### Step 4: Find Intersection Points
Substituting the line equation into the ellipse equation:
\[ \frac{x^2}{a^2} + \frac{(mx + c)^2}{b^2} = 1 \]
This leads to a quadratic equation in x. The roots of this equation will give us the x-coordinates of points P and Q.

### Step 5: Find the Slopes of the Tangents
From the quadratic equation, we can find the slopes of the tangents at points P and Q. Let the slopes be \(m_1\) and \(m_2\). The condition for the right angle gives us:
\[ m_1 \cdot m_2 = -1 \]

### Step 6: Tangents to the Parabola
The tangents to the parabola at points R and S can be found using the formula for the tangent to a parabola:
\[ yy_0 = 2d(x + x_0) \]
where \( (x_0, y_0) \) are the coordinates of points R and S.

### Step 7: Find the Intersection of Tangents
Let the points of tangency be R and S. The equations of the tangents at these points can be expressed, and we need to find their intersection point.

### Step 8: Locus of the Intersection Point
Let the intersection point of the tangents be \( (h, k) \). We need to express \(h\) and \(k\) in terms of the parameters of the ellipse and parabola.

### Step 9: Substitute and Simplify
Using the relations derived from the previous steps, we can substitute \(h\) and \(k\) into the equations and simplify to find the locus.

### Step 10: Final Locus Equation
After simplification, we will arrive at the locus equation, which will typically be in the form of a conic section (in this case, a hyperbola).

### Conclusion
The locus of the point of intersection of the tangents to the parabola at points R and S is given by the derived equation.