Given the ellipse: . Finding the points of the ellipse whose distances to the focus left is
Answer:
So we need to find the coordinates of the left focus. Then with (x,y) as any arbitrary point on the ellipse, we equate the distance between the left focus and point (x,y) to 5/2.
7x^2 +16y^2 = 112 ----------(i)
Divide both sides by 112 to make the RHS equal to 1,
(x^2)/16 +(y^2)/7 = 1
Or,
(x^2)/(4^2) +(y^2)/[(sqrt(7))^2] = 1 ---the same ellipse as (i).
a = 4 and b= sqrt(7).
Center of ellipce is at (0,0)
The leftmost end of the ellipse is at (-4,0).
The left focus is at (-c,0).
c^2 = a^2 -b^2
c^2 = 4^2 -(sqrt(7))^2 = 9
c = 3
Hence, the left focus is at (-3,0).
Then,
(5/2)^2 = (x-(-3))^2 +(y-0)62
25/4 = x^2 +6x +9 +y^2
From (i),
y^2 = (112 -7x^2)/16
so,
25/4 = x^2 +6x +9 +(112 -7x^2)/16
Clear the fractions, multiply both sides by 16,
100 = 16x^2 +96x +144 +112 -7x^2
0 = 9x^2 +96x +156
0 = 3x^2 +32x +52
0 = (3x +26)(x+2)
x = -26/3, or -2
When x = -26/3 = - 8.667, this point is not on the ellipse because the leftmost point on the ellipse is at x = -4 only
When x = -2,
y^2 = (112 -7x^2)/16
y^2 = (112 -7(4))/16 = 84/16 = 21/4
y = +,-sqrt(21) /2
Therefore, two points on the ellipse, (-2, sqrt(21) /2) and (-2, -sqrt(21) /2), are 5/2 units away from the left focus. ----answer.