The centre of a circle C is at the centre of the ellipse E:a2x2+b2y2=1, a>b. Let C pass through the foci F1 and F2 of E such that the circle C and the ellipse E intersect at four points. Let P be one of these four points. If the area of the triangle △PF1F2 is 30 and the length of the major axis of E is 17, then the distance between the foci of E is:
Get the complete, step-by-step math solution for: "The centre of a circle C is at the centre of the ellipse E: (x²)/(a²)+(y²)/(b²)=1, a>b. Let C pass through the foci F_1 and F_2 of E such that the cir...". Powered by SolveForX AI math tutor.
Step-by-Step Solution
Step 1: Identify key properties of the ellipse and circle
For an ellipse given by the equation a2x2+b2y2=1 with a>b, the foci are located at F1=(−ae,0) and F2=(ae,0), where e is the eccentricity. The center of the ellipse is at the origin (0,0). The problem states that the circle C is centered at the origin and passes through the foci F1 and F2. Therefore, the radius of the circle C is the distance from the origin to either focus, which is ae.
Step 2: Use the definition of an ellipse
A fundamental property of an ellipse is that for any point P on the ellipse, the sum of the distances from P to the two foci F1 and F2 is constant and equal to the length of the major axis, 2a. We are given that the length of the major axis is 17, so 2a=17.
Step 3: Calculate the area of triangle PF1F2
The base of the triangle △PF1F2 is the distance between the foci, which is F1F2=2ae. The height of the triangle is the absolute value of the y -coordinate of point P, denoted as ∣yP∣, since F1 and F2 lie on the x -axis. We are given that the area of △PF1F2 is 30.
Step 4: Relate point P to the circle and ellipse equations
Since point P(xP,yP) lies on the circle C with center (0,0) and radius ae, its coordinates satisfy xP2+yP2=(ae)2. Also, since P lies on the ellipse, its coordinates satisfy the ellipse equation a2xP2+b2yP2=1. We know that b2=a2(1−e2).
Step 5: Solve for yP2
Substitute xP2 from the circle equation into the ellipse equation. Simplify the expression using b2=a2(1−e2) and a2−b2=a2e2. This allows us to solve for yP2 in terms of a, b, and e.
Step 6: Substitute yP into the area equation and solve for ae
Now substitute the expression for ∣yP∣ into the area equation ae∣yP∣=30. This simplifies directly to b2=30.
Step 7: Calculate the distance between the foci
We have 2a=17, so a=217. We found b2=30. The distance between the foci is 2ae. We use the relationship a2e2=a2−b2 to find ae. Substitute the values of a and b2 to calculate ae, and then multiply by 2 to get the final distance.