The hyperbola and the conics together · Grade 12 · 5 units
In The hyperbola and the conics together (Grade 12 · 5 units) you practise 6 skills: The vertices and the foci · The equation of a hyperbola · The difference of the distances from the foci · The asymptotes · A hyperbola and a line · Which conic. A hyperbola is the locus of all the points whose distances from two fixed points, the foci, differ by a fixed number less than the distance between the foci. That difference equals the distance between the two vertices. With the centre at the origin and the foci on the horizontal axis, the standard equation is: x squared over the square of a vertex’s distance from the centre, minus y squared over a second positive number, equals one. The square of a focus’s distance from the centre is the sum of the two denominators.
A hyperbola has two branches and two asymptotes: lines through the centre that the branches approach and never meet. The gradients of the asymptotes are plus and minus the square root of the denominator under y squared divided by the denominator under x squared. When the y squared term is the positive one, the branches open upwards and downwards, and the foci lie on the vertical axis. A line parallel to an asymptote meets the hyperbola at exactly one point, and the asymptote itself never meets it. Almost every second degree equation with no product of x and y describes a conic: when both squares appear with the same sign it is a circle or an ellipse, when their signs differ it is a hyperbola or a pair of lines, and when only one square appears it is a parabola.
Tip: On a hyperbola a focus is further from the centre than a vertex, so add the denominators to get the square of its distance. On an ellipse you subtract.
What you learn
The vertices and the foci
For example: The hyperbola x²/9 − y²/100 = 1 is written in the form x² / a² − y² / b² = 1 , with a and b positive. What is a ?
Answer and solution
The answer: 3
Here a² = 9 , so a = 3 .
The equation of a hyperbola
For example: The foci of a hyperbola are (5, 0) and (−5, 0) , and the distances of each of its points from the foci differ by 8 . What is the equation of the hyperbola?
- y²/16 − x²/9 = 1
- x²/16 + y²/9 = 1
- x²/16 − y²/25 = 1
- x²/16 − y²/9 = 1
Answer and solution
The answer: x²/16 − y²/9 = 1
Here 2a = 8 , so a = 4 and b² = 25 − 16 = 9 . The foci are on the x axis, so the equation is x²/16 − y²/9 = 1 .
The difference of the distances from the foci
For example: The foci of the hyperbola x²/16 − y²/9 = 1 are F(5, 0) and G(−5, 0) . The point P is on its left branch, and PF = 10 . What is PG ?
Answer and solution
The answer: 2
Here a = 4 . The left branch is nearer to G , so PF − PG = 8 . Then PG = 10 − 8 = 2 .
The asymptotes
For example: One asymptote of the hyperbola x²/9 − y²/100 = 1 has a positive gradient. What is that gradient?
Answer and solution
The answer: 10/3
Here a = 3 and b = 10 , so the asymptotes are 10x − 3y = 0 and 10x + 3y = 0 . The positive gradient is m = 10/3 .
A hyperbola and a line
For example: The hyperbola x²/16 − y²/64 = 1 and the line y = −2x are given. How many common points do they have?
- One common point
- Two common points
- No common points
Answer and solution
The answer: No common points
The line is an asymptote of the hyperbola. Substituting gives 0 = 1024 , which has no solution, so there are no common points.
Which conic
For example: Which curve does the equation (x + 5)² = y² describe?
- A circle
- An ellipse
- A hyperbola
- A pair of lines
Answer and solution
The answer: A pair of lines
Rearranging gives (x + 5)² − y² = 0 . The signs differ and the other side is zero, so it is a pair of lines.
Not on the site yet
Parts of this topic that are not built yet: The reciprocal function as a hyperbola (enrichment) · Conic sections (enrichment).
The exercises are made by rules and the answers are worked out in code, so there are new exercises every time. The topic follows the Israeli Ministry of Education’s high school maths curriculum at 5 units (grades 10 to 12).