Chapter 11, Problem 6P

(a)

To determine

To show: The equality of the tangents $P{F}_1=P{Q}_1$ and $P{F}_2=P{Q}_2$.

Explanation of Solution

At the points $F_1$ and $F_2$, the spheres and planes are in contact. That is, $F_1$ on the sphere 1 and $F_2$ on the sphere 2.

The arbitrary point P on the curve is on the surface of the cylinder and lies outside the spheres. The point $Q_1$ lies on the sphere 1 and $Q_2$ lies on the sphere 2.

Since, the points $F_1$ and $Q_1$ on the sphere 1, the point $P$ lies outside of the sphere, the lines joining $F_1$ and $Q_1$ with $P$ are tangents to the sphere 1. That is, $P{F}_1$ and $P{Q}_1$ are tangents to the sphere 1 from an external point $P$.

Recall the fact that, the set of all tangents to a sphere from a point outside of it are equal in length.

Hence by the fact that, the tangents to the sphere from a point outside of it are equal. Therefore, $P{F}_1=P{Q}_1$.

Similarly, the points $F_2$ and $Q_2$ on the sphere 2, the point $P$ lies outside the spheres, the lines joining $F_2$ and $Q_2$ with $P$ are tangents to the sphere 1. That is, $P{F}_2$ and $P{Q}_2$ are tangents to the sphere1 from an external point $P$.

Hence by the fact that, the tangents to the sphere from a point outside of it are equal. Therefore, $P{F}_2=P{Q}_2$.

Therefore, it is shown that, $P{F}_1=P{Q}_1$ and $P{F}_2=P{Q}_2$.

(b)

To determine

To explain: The reason why $P{Q}_1+P{Q}_2$ is same for all points $P$ on the curve.

Explanation of Solution

Since, $P$ is an arbitrary point on the curve and the curve is between the spheres. Also, the points $Q_1$ and $Q_2$ on the surface of the cylinder as well as on the equator of the spheres 1 and 2 respectively. Therefore, the point $P$ is between the line joining $Q_1$ and $Q_2$.

Recall the definition of equator of a sphere that the equator of a sphere is a great circle which divides the sphere in to two equal halves.

By the definition of equator of a sphere the vertical distance between $Q_1$ and $Q_2$ remains the same. Hence the arbitrary point $P$ always touches the cylinder and lies between $Q_1$ and $Q_2$, $P{Q}_1+P{Q}_2$ remains same for all the points $P$ on the curve. That is, $P{Q}_1+P{Q}_2=$ constant.

(c)

To determine

To show: The addition of tangent lines $P{F}_1+P{F}_2$ is same for all points $P$ on the curve.

Explanation of Solution

Definition used:

Equator of a sphere:

“The equator of a sphere is a great circle which divides the sphere in to two equal halves”.

From part (b), $P{Q}_1+P{Q}_2=$ constant.

From part (a), $P{Q}_1=P{F}_1$ and $P{Q}_2=P{F}_2$.

Substitute $P{Q}_1=P{F}_1$ and $P{Q}_2=P{F}_2$ in $P{Q}_1+P{Q}_2=$ constant.

$P{F}_1+P{F}_2=$ constant

Thus, it is shown that $P{F}_1+P{F}_2\text{ is same}$ for all points $P$ on the curve.

(d)

To determine

To conclude: The curve is an ellipse with foci $F_1$ and $F_2$.

Explanation of Solution

Definition used:

“If $P$ is an arbitrary point on the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ with foci $F_1\left(-c,0\right)$ and $F_2\left(c,0\right)$ then $F_{1}P+F_{2}P=2a$ (constant)”.

From section (c), it is true that $P{F}_1+P{F}_2=$ constant.

That is, $P{F}_1+P{F}_2=$ constant with $F_1$ and $F_2$ as fixed points and $P$ is an arbitrary point on the curve.

By the definition stated above, the curve is an ellipse.

Thus, it is concluded that the curve is an ellipse with foci $F_1$ and $F_2$.