a) Determine the equation of the hyperbola with a focus at (5,0) and a vertex (3,0) . [2] b) Find the direction cosines of the line joining the points (1,2,3) and (4,5,7) . [2] c) Show that the…

a) Equation of the hyperbola

The given data

• focus : $F(5,0)$
• vertex : $V(3,0)$

Both points lie on the $x$-axis, therefore the transverse axis of the hyperbola is horizontal and the centre is on the $x$-axis.
For a hyperbola whose centre is at the origin the standard form is

$$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1,c^2=a^2+b^2$$

where

• $a$ = distance from centre to a vertex,
• $c$ = distance from centre to a focus.

Because the centre is taken at the origin (the simplest choice that satisfies the data),

$$a=|x\text{-coordinate of vertex}|=3,c=|x\text{-coordinate of focus}|=5.$$

Now compute $b$:

$$b^2=c^2-a^2=5^2-3^2=25-9=16⟹b=4.$$

Substituting $a^2=9$ and $b^2=16$ in the standard form gives

$$\boxed{\frac{x^2}{9}-\frac{y^2}{16}=1}.$$

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b) Direction cosines of the line joining $(1,2,3)$ and $(4,5,7)$

1. Form the direction vector

$$\vec{d}=(4-1,5-2,7-3)=(3,3,4).$$

2. Find its magnitude

$$|\vec{d}|=\sqrt{3^2+3^2+4^2}=\sqrt{9+9+16}=\sqrt{34}.$$

3. Direction cosines are the cosines of the angles that $\vec{d}$ makes with the positive $x$, $y$ and $z$ axes:

$$\ell =\frac{3}{\sqrt{34}},m=\frac{3}{\sqrt{34}},n=\frac{4}{\sqrt{34}}.$$

Hence the required direction cosines are

$$\boxed{\left(\frac{3}{\sqrt{34}},\frac{3}{\sqrt{34}},\frac{4}{\sqrt{34}}\right)}.$$

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c) Collinearity of the three vectors

$$\vec{a}=\hat{i}+2\hat{j}+4\hat{k}=(1,2,4),\vec{b}=2\hat{i}+5\hat{j}-\hat{k}=(2,5,-1),\vec{c}=3\hat{i}+8\hat{j}-6\hat{k}=(3,8,-6).$$

Three vectors in space are collinear (i.e. linearly dependent) iff the determinant formed by their components is zero.

$$\Delta =\left|\begin{array}{ccc}1& 2& 4\\ 2& 5& -1\\ 3& 8& -6\end{array}\right|$$

Calculate the determinant:

$$\begin{array}{rl}\Delta & =1(5(-6)-(-1)8)-2(2(-6)-(-1)3)+4(2\cdot 8-5\cdot 3)\\ & =1(-30+8)-2(-12+3)+4(16-15)\\ & =-22-2(-9)+4(1)\\ & =-22+18+4\\ & =0.\end{array}$$

Since $\Delta =0$, the three vectors are linearly dependent; consequently they lie on the same straight line.

$$\boxed{\text{The vectors }\hat{i}+2\hat{j}+4\hat{k},2\hat{i}+5\hat{j}-\hat{k},3\hat{i}+8\hat{j}-6\hat{k}\text{ are collinear}.}$$