Let C be a point that varies on the ellipse
x^2/4 + y^2 = 1.
Let H be the orthocenter of triangle ABC, where A = (-4,0) and B = (4,0). Then H traces a closed curve as C varies over the ellipse. Find the area inside the curve.
First, we need to find the equation of the ellipse in standard form:
x^2/4 + y^2 = 1
Multiplying both sides by 4, we get:
x^2 + 4y^2 = 4
Dividing both sides by 4, we get:
x^2/4 + y^2/1 = 1
So, the semi-major axis a = 2 and the semi-minor axis b = 1.
Next, we need to find the equation of the line AB:
The slope of AB is 0, so it is a horizontal line passing through (0,0). Therefore, the equation of AB is y = 0.
Now, we need to find the equations of the altitudes from points A and B to the ellipse.
For point A, we need to find the perpendicular line passing through A and intersecting the ellipse. The slope of this line is undefined, so it is a vertical line passing through (-4,0). Therefore, the equation of this line is x = -4.
Substituting this into the equation of the ellipse, we get:
(-4)^2/4 + y^2 = 1
Simplifying, we get:
y^2 = 3
So, the coordinates of the foot of the altitude from A are (-4, ±sqrt(3)).
For point B, we do the same thing. The perpendicular line passing through B is y = 0, so substituting this into the equation of the ellipse, we get:
x^2/4 + 0^2 = 1
Simplifying, we get:
x^2 = 4
So, the coordinates of the foot of the altitude from B are (±2, 0).
Now, we need to find the equation of the altitude from point C. Let the coordinates of point C be (x,y).
The altitude from C must pass through the midpoint of AB, which is (0,0). Therefore, the equation of this line is y = mx + 0, where m is the slope of AB.
The slope of AB is 0, so the equation of the altitude from C is simply x = k, where k is a constant.
Substituting this into the equation of the ellipse, we get:
k^2/4 + y^2 = 1
Solving for y, we get:
y^2 = 1 - k^2/4
So, the coordinates of the foot of the altitude from C are (k, ±sqrt(1-k^2/4)).
Now, we need to find the intersection points of the altitudes from A, B, and C.
The altitude from A intersects the altitude from B at (±2, ±sqrt(3)).
Substituting these coordinates into the equation of the altitude from C, we get:
k = ±2
So, the altitude from C intersects the altitude from A at (-2, ±sqrt(3)/2) and (2, ±sqrt(3)/2), and it intersects the altitude from B at (-2, 0) and (2, 0).
Thus, the vertices of the orthic triangle ABC are (-2, ±sqrt(3)/2), (2, ±sqrt(3)/2), and (-4, 0) or (4, 0).
To find the orthocenter H, we need to find the intersection points of the altitudes of the orthic triangle.
The altitude from (-2, ±sqrt(3)/2) to AB is y = ±sqrt(3)/2, and the altitude from (2, ±sqrt(3)/2) to AB is also y = ±sqrt(3)/2.
The altitude from (-4, 0) (or (4, 0)) to the line passing through (-2, ±sqrt(3)/2) and (2, ±sqrt(3)/2) has slope equal to the negative reciprocal of the slope of that line, which is
±sqrt(3)/2 / 2 = ±sqrt(3)/4.
So, the equation of this altitude is y = ±sqrt(3)/4 x ± sqrt(3)/2.
Setting these two equations equal to each other, we get:
±sqrt(3)/2 = ±sqrt(3)/4 x ± sqrt(3)/2
Simplifying, we get:
x = ±2
So, the intersection points of the altitudes are (-2, 0) and (2, 0). Therefore, the orthocenter H is (0,0).
Since H is a fixed point, the closed curve traced by H as C varies over the ellipse is simply a circle with radius equal to the distance from H to any vertex of the ellipse. The distance from (0,0) to the nearest vertex (which is (-2,0) or (2,0)) is 2, so the area inside the curve is
π(2)^2 = 4π.
Therefore, the area inside the curve is 4π.
To find the area inside the curve traced by the orthocenter H as point C varies on the ellipse, we need to understand some properties of the orthocenter and the given ellipse.
Step 1: Find the equation of the line passing through A = (-4,0) and B = (4,0):
Since A and B lie on the x-axis, the line passing through A and B is also the x-axis, which can be represented by the equation y = 0.
Step 2: Find the equation of the line passing through A = (-4,0) and C = (x, y):
The slope of the line passing through A and C can be calculated using the formula:
m = (y2 - y1) / (x2 - x1)
Plugging the values:
m = (y - 0) / (x - (-4))
m = y / (x + 4)
Since the line is perpendicular to the line passing through A and B, let's find the negative reciprocal of the slope:
m_perpendicular = -1 / m
m_perpendicular = -(x + 4) / y
Using point-slope form of the line:
y - 0 = -(x + 4) / y (x - (-4))
y^2 = -(x + 4) (x + 4)
y^2 = -(x^2 + 8x + 16)
y^2 = -x^2 - 8x - 16
Step 3: Substitute the equation of the line passing through A and C into the equation of the ellipse:
x^2/4 + y^2 = 1
Substituting y^2 = -x^2 - 8x - 16:
x^2/4 + (-x^2 - 8x - 16) = 1
x^2 + 4(-x^2 - 8x - 16) = 4
x^2 - 4x^2 - 32x - 64 = 4
-3x^2 - 32x - 68 = 0
Step 4: Solve the quadratic equation -3x^2 - 32x - 68 = 0:
We can either use factoring, completing the square, or the quadratic formula to solve the equation. Let's use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
Plugging the values:
x = (-(-32) ± √((-32)^2 - 4(-3)(-68))) / (2(-3))
x = (32 ± √(1024 - 816)) / (-6)
x = (32 ± √208) / (-6)
Simplifying:
x = (16 ± √52) / (-3)
Step 5: Determine the limits of x to find the interval on the ellipse:
To find the limits of x, we need to determine when the values inside the square root become zero:
(16 ± √52)^2 = 0
(16 ± √52)^2 - 52 = 0
(16 ± √52 + √52)(16 ± √52 - √52) = 0
(16 ± 2√13)(16 ± √13) = 0
Since we only consider real values, we can ignore the ± sign. Therefore:
(16 + 2√13)(16 + √13) = 0
Simplifying the expressions in the brackets to find the limits of x:
16 + 2√13 = 0, x = (16 ± √52) / (-3)
16 + 2√13 = -32
2√13 = -48
√13 = -24, which is not possible since a square root should always be positive.
16 + √13 = 0, x = (16 - √52) / (-3)
We found a limit for x:
x = (16 - √52) / (-3)
Step 6: Calculate the area inside the curve:
To find the area inside the curve, we integrate the equation of the ellipse from x = -4 to x = (16 - √52) / (-3):
Area = ∫[from -4 to (16 - √52) / (-3)] (√(1 - x^2/4)) dx
Evaluating the integral will give us the area inside the curve traced by the orthocenter H as point C varies on the ellipse.