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# Example 6 of finding the equation of an ellipse

Question:

The equation of the circle is x2 + y2 = 81. P is a point in the circle. PD is perpendicular to the x-axis at point D. E is a point on PD and PE = (1/3) PD. If point P moves in the circle, what is the trajectory of the point E? Solution:

When point P moves in the circle, PD moves as point P moves. So, point E moves, we need to find the equation for trajectory of the point E.

Given: PE = (1/3) PD, so, ED = (2/3)PD

Let point E has the coordinate (x, y), point P has the coordinate (x0, y0). Because point P lies on the circle, so point P satisfy the circle equation, x02 + y02 = 81

Because PD is perpendicular to the x-axis, so, the x-coordinate of point E is equal to the x-coordinate of the point P. That is, x = x0.

Because ED = (2/3)PD. Y coordinate of the point E is y and y-coordinate of the point P is y0, so, y = (2/3)y0.

Now, we express x0, y0 in terms of x, y.

x0 = x and y0 = (3/2)y, now, substitute xo, y0 into the circle equation.

x2 + [(3/2)y]2 = 81
x2 + (9/4)y2 = 81
each term divided by 81
x2/81 + [(9/4)y2]/81 = 1
x2/81 + y2/36 = 81
x2/92 + y2/62 = 1

Let y = 0, then x = +- 9, the ellipse interests x-axis at +9 and -9. Let x = 0, then y = +-6, the ellipse intersects the y-axis at +6 and -6. Please watch the video for more details.