Start by drawing the graph of the hyperbola. Write an equation for the line on which the light travels before it hits the mirror.
(40 - 24 sqrt(5)/5, 45 - 12 sqrt(5)/5)
Practice makes perfect
We see that the given equation is an equation of a vertical hyperbola centered at the origin (0,0).
y^2/9-x^2/16=1
To draw the graph of the hyperbola, we need to identify the values of a, b, and c. From the given equation, a and b can be identified directly.
y^2/a^2-x^2/b^2 = 1
⇒
y^2/3^2-x^2/4^2 = 1
We see that a= 3 and b= 4. The value of c is calculated by the following formula.
c^2=a^2+b^2
Let's substitute a= 3 and b= 4 into the formula.
Now, we are ready to find the coordinates of the vertices, the foci, and the equations of the asymptotes.
Vertices (V_(1,2))
Foci (F_(1,2))
Asymptotes (l_(1,2))
Formula
(0, ± a)
(0,± c )
y=± a/bx
Substitution
(0, ± 3)
(0,± 5)
y=± 3/4x
We will plot the points and draw the asymptotes.
Next, we will use the asymptotes as a guide to draw the hyperbola that passes through the vertices.
We want to find a point on the upper branch of the hyperbola so that the light emitted from the light source at the point (- 10,0) will be reflected to (0,- 5). Since (0,- 5) is the focus, the light rays should have a direction to the focus of the mirror.
In order for the light to be reflected to (0,- 5), the light rays should be on the line that passes through the points (- 10,0) and (0,5).
To write the equation of this line, we will first find its slope.
Using the slope-intercept form, we can write the equation of the line since we know that the y-intercept is (0, 5) and the slope is 0.5.
y= mx+ b
⇕
y= 0.5x+ 5
We want to find the point where the line and the upper branch of the hyperbola intersect. To find it, we need to solve the system of equations.
y^29- x^216=1 & (I) y=0.5x+5 & (II)
We will use the Substitution Method to solve the system. Let's substitute 0.5x+5 for y in the first equation.
The solutions for this equation are x= 40 ± 24sqrt(5)5. From the graph, we know that the point of intersection is in the 2^\text{nd} quadrant, so we will substitute the negative solution, 40 - 24 sqrt(5)5, into the second equation to find the y-coordinate of the point.
Therefore, the point ( 40 - 24 sqrt(5)5, 45 - 12 sqrt(5)5) is a solution to the system of equation. This point is where the light should hit the mirror.
