To find the points on the ellipse defined by the equation 4x² + y² = 4 that are farthest from the point (1, 0), we can follow a systematic approach that involves both geometric intuition and optimization techniques.
Step 1: Understand the Ellipse
The ellipse represented by the equation can be rewritten in standard form:
x²/1 + y²/4 = 1
This tells us that the ellipse is centered at the origin (0, 0), with a semi-major axis of 2 (along the y-axis) and a semi-minor axis of 1 (along the x-axis).
Step 2: Distance from the Point
The next step is to express the distance from any point on the ellipse to the point (1, 0). The distance D between a point (x, y) on the ellipse and the point (1, 0) is given by the distance formula:
D = √[(x – 1)² + (y – 0)²] = √[(x – 1)² + y²]
Since we want to maximize the distance, it’s easier to maximize the squared distance instead:
D² = (x – 1)² + y²
Step 3: Substitute for y²
From the ellipse’s equation, we can express y² in terms of x:
y² = 4 – 4x²
Substituting this into the distance formula gives:
D² = (x – 1)² + (4 – 4x²)
Which simplifies to:
D² = (x – 1)² + 4 – 4x²
Step 4: Simplifying D²
Further simplify the expression:
D² = x² – 2x + 1 + 4 – 4x²
D² = -3x² – 2x + 5
Step 5: Finding the Maximum
Now, we need to find the maximum value of this quadratic function. Since it opens downwards (the coefficient of x² is negative), the maximum can be found at the vertex, which occurs at:
x = -b/(2a) = -(-2)/(2*-3) = 1/3
Step 6: Finding Corresponding y Values
To find the corresponding y values, substitute back into the ellipse’s equation:
4(1/3)² + y² = 4
Solving for y²: y² = 4 – 4/9 = 36/9 – 4/9 = 32/9
Thus, y = ±√(32/9) = ±(4√2)/3
Step 7: Points on the Ellipse
The points on the ellipse that are farthest from (1, 0) are:
- (1/3, (4√2)/3)
- (1/3, -(4√2)/3)
In summary, the points on the ellipse that are farthest from the point (1, 0) are (1/3, 4√2/3) and (1/3, -4√2/3).