diff --git a/exercises/hyperbola_intuition.html b/exercises/hyperbola_intuition.html index 9183c2e97..d0c9e7beb 100644 --- a/exercises/hyperbola_intuition.html +++ b/exercises/hyperbola_intuition.html @@ -2,7 +2,7 @@
-The equation of a north-south opening hyperbola is :
- \qquad \dfrac{(x - x_1)^2}{b^2} - \dfrac{(y - y_1)^2}{a^2} = 1
.
+ \qquad \dfrac{(y - y_1)^2}{a^2} - \dfrac{(x - x_1)^2}{b^2} = 1
.
Find the foci of the hyperbola below by moving the orange points to their correct positions. @@ -54,7 +54,7 @@ graph.H = H; graph.K = K; graph.C = C; - + initAutoscaledGraph([[-20, 20], [-20, 20]], {}); addMouseLayer(); @@ -161,7 +161,7 @@
The center of a hyperbola is at the midpoint of its two foci.
\left(\dfrac{H + H}{2}, \dfrac{K+C + K-C}{2}\right) = (H, K)
The absolute value of the distance from one focus minus the distance to the other focus is equal to 2a
The absolute value of the distance from one focus minus the distance to the other focus is equal to 2a
.
\qquad\begin{align*}
2a &= 2 * A \\
@@ -191,18 +191,18 @@
randFromArray([[3, 4, 5], [6, 8, 10], [9, 12, 15], [5, 12, 13]])
A * A
B * B
- 18 - C
- randRange(-5, 5)
- randRange(-MAX_Y, MAX_Y)
+ 18 - C
+ randRange(-MAX_X, MAX_X)
+ randRange(-5, 5)
The equation of an east-west opening hyperbola is:
- \qquad \dfrac{(x - x_1)^2}{b^2} - \dfrac{(y - y_1)^2}{a^2} = 1
.
+ \qquad \dfrac{(x - x_1)^2}{a^2} - \dfrac{(y - y_1)^2}{b^2} = 1
.
Find the foci of the hyperbola below by moving the orange points to their correct positions.
- Then use that information to find the values of y_1
, x_1
, a
and b
.
+ Then use that information to find the values of x_1
, y_1
, a
and b
.
x_1 = |
- K | -||
y_1 = |
H | ||
b = |
- B | +y_1 = |
+ K |
a = |
A | ||
b = |
+ B | +
One focus is (K - C, H)
and the other is (K + C, H)
.
One focus is (H - C, K)
and the other is (H + C, K)
.
x_1
and y_1
are the coordinates of the center of the hyperbola.
The center of a hyperbola is at the midpoint of its two foci.
-\left(\dfrac{K - C + K + C}{2}, \dfrac{H + H}{2}\right) = (K, H)
\left(\dfrac{H - C + H + C}{2}, \dfrac{K + K}{2}\right) = (H, K)
The absolute value of the distance from one focus minus the distance to the other focus is equal to 2a
The absolute value of the distance from one focus minus the distance to the other focus is equal to 2a
.
\qquad\begin{align*}
2a &= 2 * A \\
@@ -359,7 +359,7 @@
So the equation of the hyperbola is
- \dfrac{(x - H)^2}{B^2} + \dfrac{(y - K)^2}{A^2} = 1
.
+ \dfrac{(x - H)^2}{A^2} - \dfrac{(y - K)^2}{B^2} = 1
.