Answer:
B. (-0.125, 2.98)
Step-by-step explanation:
You want the point on ellipse 9x² +y² = 9 that is farthest from the point (1, 0).
Circle
A plot of the ellipse and the given point shows the point lies on a co-vertex, and the ellipse vertices are (0, ±3). The largest circle tangent to the ellipse will necessarily have tangent points to the left of the y-axis.
This tells us a couple of things:
- the largest y-coordinate of any point on the ellipse will be 3, which eliminates choice C
- the point farthest from the right co-vertex will have a negative x-coordinate, which eliminates choices A and D.
The only viable choice for the point of interest is choice B, (-0.125, 2.98).
Analytic solution
A circle centered at (1, 0) with radius r will have the equation ...
(x -1)² +y² = r²
Subtracting this equation from that of the ellipse gives ...
(9x² +y²) -((x -1)² +y²) = (9) -(r²)
8x² +2x +(r² -10) = 0 . . . . . . . . . . . simplify to standard form
This will have real solutions when the discriminant is non-negative:
b² -4ac ≥ 0
2² -4(8)(r² -10) ≥ 0
324 ≥ 32r²
r² ≤ 10.125
Now we can write the equation for x as ...
8x² +2x +(10.125 -10) = 0 . . . . . . for maximum r
x² +1/4x +1/64 = 0 . . . . . . . . . . divide by 8
(x +1/8)² = 0 ⇒ x = -1/8
Then y can be found from the ellipse equation:
9x² +y² = 9
y² = 9(1 -x²) = 9(63/64)
y = ±(3/8)√63 ≈ ±2.97647 ≈ ±2.98
One of the points of interest is (-0.125, 2.98), choice B.
