Chapter 10.4, Problem 80AYU

Hyperbolic Mirrors Hyperbolas have interesting reflective properties that make them useful for lenses and mirrors. For example, if a ray of light strikes a convex hyperbolic mirror on a line that would (theoretically) pass through its rear focus, it is reflected through the front focus. This property, and that of the parabola, were used to develop the Cassegrain telescope in 1672. The focus of the parabolic mirror and the rear focus of the hyperbolic mirror are the same point. The rays are collected by the parabolic mirror, then are reflected toward the (common) focus, and thus are reflected by the hyperbolic mirror through the opening to its front focus, where the eyepiece is located. If the equation of the hyperbola is $\frac{y^2}{9}-\frac{x^2}{16}=1$ and the focal length (distance from the vertex to the focus) of the parabola is 6, find the equation of the parabola.

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To determine

To find: The equation of the parabola if details of hyperbola are given.

Answer to Problem 80AYU

Solution:

$x^2=24\left(y+1\right)$

Explanation of Solution

Given:

About hyperbolic mirrors: Hyperbolas have interesting reflective properties that make them useful for lenses and mirrors. For example, if a ray of light strikes a convex hyperbolic mirror on a line that would (theoretically) pass through its rear focus, it is reflected through the front focus. This property, and that of the parabola, were used to develop the Cassegrain telescope in 1672. The focus of the parabolic mirror and the rear focus of the hyperbolic mirror are the same point. The rays are collected by the parabolic mirror, then are reflected toward the (common) focus, and thus are reflected by the hyperbolic mirror through the opening to its front focus, where the eyepiece is located.

The equation of the hyperbola is $\frac{y^2}{9}-\frac{x^2}{16}=1$ and the focal length (distance from the vertex to the focus) of the parabola is 6.

Formula used:

Equation of the parabola is $x^2=4a\left(y-k\right)$.

Calculation:

Let us assume that the center of hyperbola is origin.

From the given equation we see that the transverse axis is parallel to $y\text{-axis}$.

Let the foci of the hyperbola be $\left(0,\pm c\right)$.

$c^2=a^2+b^2=9+16=25$ or $c=5$

Therefore, foci of the hyperbola are at $\left(0,\pm 5\right)$.

Let us assume that the parabola opens up,

The focus is at $\left(0,5\right)$, then the equation of the parabola will be ,

$x^2=4a\left(y-k\right)$

The focal length is given as 6.

We know that the distance focus of the parabola is located at $\left(0,k+a\right)=\left(0,5\right)$.

Hence,

$k+a=5$

$k+6=a$

$k=-1$

The equation of the parabola becomes,

$x^2=4a\left(y-k\right)$

$x^2=4*6*\left(y+1\right)$

$x^2=24\left(y+1\right)$