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 @@ - Hyperbola and ellipse intuition + Hyperbola intuition @@ -20,7 +20,7 @@

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.

@@ -232,7 +232,7 @@ graph.H = H; graph.K = K; graph.C = C; - + initAutoscaledGraph([[-20, 20], [-20, 20]], {}); addMouseLayer(); @@ -258,12 +258,12 @@ $("#problemarea span.focus2-y-label").html("<code>" + coordY + "</code>").tex(); }; - graph.func1 = addInteractiveFn(function(x) { - return K + sqrt(A2 / B2 * (x - H) * (x - H) + A2); + graph.func1 = addInteractiveFn(function(y) { + return H + sqrt(A2 / B2 * (y - K) * (y - K) + A2); }, { swapAxes: true }); - graph.func2 = addInteractiveFn(function(x) { - return K - sqrt(A2 / B2 * (x - H) * (x - H) + A2); + graph.func2 = addInteractiveFn(function(y) { + return H - sqrt(A2 / B2 * (y - K) * (y - K) + A2); }, { swapAxes: true }); doHyperbolaInteraction(graph.func1, graph.focus1, graph.focus2); @@ -290,9 +290,9 @@ return ""; } - return guess[0][1] === H && guess[1][1] === H && - (guess[0][0] === K - C && guess[1][0] === K + C) || - (guess[0][0] === K + C && guess[1][0] === K - C) + return guess[0][1] === K && guess[1][1] === K && + (guess[0][0] === H - C && guess[1][0] === H + C) || + (guess[0][0] === H + C && guess[1][0] === H - C)
graph.focus1.setCoord(guess[0]); @@ -305,20 +305,20 @@ - - - - - - + + + + + +
x_1 = K
y_1 = H
b = By_1 = K
a = A
b = B

@@ -329,17 +329,17 @@ There is only one way to arrange the two foci such that this is true.

-

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.