Problem: Find the area of the triangle PQR formed by the intersection of lines x y = 1 and 3y = x-3 with the ellipse x^2/9 + y^2 = 1.

Solution:

To solve the problem, we need to find the points P, Q, and R of intersection between the given lines and the ellipse.

Step 1: Find the points of intersection of the ellipse and the line x y = 1.

Substituting y = 1/x in the equation of the ellipse, we get:

x^2/9 + (1/x)^2 = 1

Multiplying both sides by x^2, we get:

x^4/9 + 1 = x^2

Bringing all terms to one side, we get:

x^4 - 9x^2 + 9 = 0

This is a quadratic equation in x^2. Solving for x^2 using the quadratic formula, we get:

x^2 = (9 ± 3√5)/2

Since x^2 is positive, we take the positive root and get:

x^2 = (9 + 3√5)/2

Taking the square root, we get:

x = √[(9 + 3√5)/2]

Substituting this value of x in the equation x y = 1, we get:

y = 1/x = √[2/(9 + 3√5)]

So, the point of intersection of the ellipse and the line x y = 1 is:

P = (√[(9 + 3√5)/2], √[2/(9 + 3√5)])

Similarly, we can find the point of intersection of the ellipse and the line 3y = x-3.

Step 2: Find the area of the triangle PQR.

Let Q be the point of intersection of the ellipse and the line 3y = x-3.

Substituting y = (x-3)/3 in the equation of the ellipse, we get:

x^2/9 + ((x-3)/3)^2 = 1

Simplifying, we get:

4x^2 - 36x + 36 = 0

Solving for x using the quadratic formula, we get:

x = 3/2 ± √(5)/2

Since the lines x y = 1 and 3y = x-3 intersect at the point (3, 0), we can easily see that the point of intersection of the ellipse and the line 3y = x-3 is:

R = (3/2 - √(5)/2, 3/2)

Now, we can find the lengths of the sides of the triangle PQR using the distance formula.

PR = √[(√[(9 + 3√5)/2] - 3/2 + √(5)/2)^2 + (√[2/(9 + 3√5)] - 3/2)^2]

QR = √[(3/2 + √(5)/2 - (3/2 - √(5)/2))^2 + (3/2 - 0)^2]