Question

An ellipse and a hyperbola have the same foci, A and B, and intersect at four points. The ellipse has major axis 50, and minor axis 40. The hyperbola has conjugate axis of length 20. Let P be a point on both the hyperbola and ellipse. What is PA cdot PB?

Answer 1

An ellipse is a closed curve in which the sum of the distances from any point on the curve to the two foci is constant. The ellipse in this question has a major axis of 50 and a minor axis of 40. To find the product of PA and PB, the coordinates of point P need to be determined using the equations of the ellipse and hyperbola. Once the coordinates are known, PA and PB can be calculated using the distance formula, and their product can be found.

Step-by-step explanation:

An ellipse is a closed curve in which the sum of the distances from any point on the curve to the two foci is constant. In this question, the ellipse has a major axis of 50 and a minor axis of 40. The hyperbola, on the other hand, has a conjugate axis (which is the distance between the vertices) of length 20. Since the ellipse and hyperbola intersect at four points, we can conclude that these points lie on both curves and therefore share the same foci.

To find the product of PA and PB, we need to determine the coordinates of the point P that lies on both the hyperbola and ellipse.

Let's assume the center of the ellipse and hyperbola is at the origin (0,0). Since the major axis of the ellipse is 50, the semi-major axis is 25. The coordinates of P can be determined by substituting different values for x and solving for y in the equation of the ellipse: x²/25² + y²/20² = 1.

Once we have the coordinates of P, we can find the distance PA and PB using the distance formula, and then calculate the product of PA and PB.

Answer 2

PA.PB=600

Step-by-step explanation:

Given Data

Major axis=50

Minor axis=40

conjugate axis of length=20

P be a point on both the hyperbola and ellipse

PA.PB=?

Solution

Taken definition of ellipse

PA+PB=50

Minor axis of ellipse is 40 then a=25 and b=20

Which makes focal distance c=15

For hyperbola

`a^(2)+b^(2)=c^(2)`

2b is conjugate axis

2a is the transverse axis

So for this hyperbola 2b=20

b=20/2

b=10

and c=15

then

`a=\sqrt{c^(2)-b^(2)  }\\ a=\sqrt{(15)^(2) -(10)^(2) }  \\a=√(25)\\ a=5`

that makes

2a=10 Distance between vertices

If P is point

So for this hyperbola

|PA-PB|=10

`(PA-PB)^(2) =100`

We know for ellipse

PA+PB=50

`(PA+PB)^(2)=2500`

Since

`(PA+PB)^(2)-(PA-PB)^(2)\\  4PA.PB=2500-100\\PA.PB=(2400)/(4)\\ PA.PB=600`