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.
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.
1 answer
1 answer