Chapter 8.2, Problem 76E

Solution

How the given graph is similar to and different to a full-fledged ellipse.

It has been explained how the given graph is similar to and different to a full-fledged ellipse.

Given:

The graph of the equation, $\frac{{\left(x-h\right)}^2}{a^2}+\frac{{\left(y-k\right)}^2}{b^2}=0$ is considered to be a degenerate ellipse.

Concept used:

An ellipse is formed when a plane intersects a double-sided cone in a single nappe obliquely.

Calculation:

The given equation is $\frac{{\left(x-h\right)}^2}{a^2}+\frac{{\left(y-k\right)}^2}{b^2}=0$ .

Simplifying,

  ${\left(\frac{x-h}{a}\right)}^2+{\left(\frac{y-k}{b}\right)}^2=0$

Now, the sum of two square quantities is zero if and only if each of the square quantity is zero.

Then, it follows that,

  ${\left(\frac{x-h}{a}\right)}^2=0$ and ${\left(\frac{y-k}{b}\right)}^2=0$

Simplifying,

  $\frac{x-h}{a}=0$ and $\frac{y-k}{b}=0$

On further simplification,

  $x-h=0$ and $y-k=0$

Finally,

  $x=h$ and $y=k$

The above represents a single point $\left(h,k\right)$ .

So, the graph of the given equation is a single point $\left(h,k\right)$ .

Now, as discussed, an ellipse is formed when a plane intersects a double-sided cone in a single nappe obliquely.

This is true for both a full-fledged hyperbola and the above equation.

The difference lies in the fact that the graph of the above equation is formed when the plane passes through the vertex of the cone.

Conclusion:

It has been explained how the given graph is similar to and different to a full-fledged ellipse.