If the light from a lamp casts a hyperbolic shadow pattern on the wall, calculate the equation of the hyperbola if the distance between the vertices is 2 feet and the foci are half a foot from the vertices.
The Correct Answer and Explanation is:
Correct Answer:
The equation of the hyperbola is:x212−y2(0.5)2=1\frac{x^2}{1^2} – \frac{y^2}{(0.5)^2} = 112×2−(0.5)2y2=1
or simplified:x21−y20.25=1\frac{x^2}{1} – \frac{y^2}{0.25} = 11×2−0.25y2=1
Explanation:
A hyperbola is a set of all points such that the difference of the distances to two fixed points (called foci) is constant. The standard form of a horizontal hyperbola centered at the origin is:x2a2−y2b2=1\frac{x^2}{a^2} – \frac{y^2}{b^2} = 1a2x2−b2y2=1
In this equation:
- 2a2a2a is the distance between the vertices (so aaa is half that),
- ccc is the distance from the center to each focus,
- bbb is calculated using the formula:
c2=a2+b2c^2 = a^2 + b^2c2=a2+b2
Given:
- Distance between vertices = 2 feet, so a=1a = 1a=1
- Foci are half a foot from each vertex, so c=a+0.5=1.5c = a + 0.5 = 1.5c=a+0.5=1.5
Now use the relationship:c2=a2+b2c^2 = a^2 + b^2c2=a2+b2(1.5)2=12+b2(1.5)^2 = 1^2 + b^2(1.5)2=12+b22.25=1+b22.25 = 1 + b^22.25=1+b2b2=2.25−1=1.25b^2 = 2.25 – 1 = 1.25b2=2.25−1=1.25
However, if we re-interpret the problem carefully:
If the foci are half a foot from the vertices, and the vertices are 1 foot from the center, then the foci are 1.5 feet from the center, so c=1.5c = 1.5c=1.5, as above.
Then:b2=c2−a2=(1.5)2−12=2.25−1=1.25b^2 = c^2 – a^2 = (1.5)^2 – 1^2 = 2.25 – 1 = 1.25b2=c2−a2=(1.5)2−12=2.25−1=1.25
So the final equation becomes:x21−y21.25=1\frac{x^2}{1} – \frac{y^2}{1.25} = 11×2−1.25y2=1
Alternatively, if the foci are only 0.5 feet from the vertices and not added to the vertex distance, then c=1+0.5=1.5c = 1 + 0.5 = 1.5c=1+0.5=1.5 as above.
Thus, the correct equation is:x21−y21.25=1\frac{x^2}{1} – \frac{y^2}{1.25} = 11×2−1.25y2=1
This represents a hyperbola opening left and right, where the light source causes the shadow boundary to take this shape.
