a) Find the foci and centre of the hyperbola with equation 9x^2 16y^2 144 = 0 [2] b) Find the direction cosines of the plane passing through the points (2, 0, 0) and (0, 2,

a) Find the foci and centre of the hyperbola with equation $9x^2-16y^2-144=0$

Step 1: Rewrite the equation in standard form

The given equation of the hyperbola is: $$9x^2-16y^2-144=0$$

First, move the constant term to the other side: $$9x^2-16y^2=144$$

Divide the entire equation by 144 to get the standard form of a hyperbola: $$\frac{9x^2}{144}-\frac{16y^2}{144}=1$$ $$\frac{x^2}{16}-\frac{y^2}{9}=1$$

This is the standard form of a hyperbola centered at the origin with a horizontal transverse axis: $$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$$

Step 2: Identify $a$ and $b$

From the standard form: $$a^2=16⟹a=4$$ $$b^2=9⟹b=3$$

Step 3: Find the distance to the foci

For hyperbolas, the distance to the foci $c$ is given by: $$c^2=a^2+b^2$$ $$c^2=16+9=25⟹c=5$$

Step 4: Determine the foci and centre

The foci are located along the transverse axis (the x-axis in this case) at $(\pm c,0)$: $$\text{Foci}=(5,0)\text{ and }(-5,0)$$

The centre of the hyperbola is at the origin: $$\text{Centre}=(0,0)$$

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b) Find the direction cosines of the plane passing through the points (2, 0, 0) and (0, 2, 0)

Step 1: Identify the third point

The question seems to be incomplete, but assuming the plane passes through three points: (2, 0, 0), (0, 2, 0), and (0, 0, 2), we can proceed. (If the third point is different, the answer will vary.)

Step 2: Find two vectors in the plane

Let the points be: $$A=(2,0,0),B=(0,2,0),C=(0,0,2)$$

Vectors $\overrightarrow{AB}$ and $\overrightarrow{AC}$ are: $$\overrightarrow{AB}=B-A=(0-2,2-0,0-0)=(-2,2,0)$$ $$\overrightarrow{AC}=C-A=(0-2,0-0,2-0)=(-2,0,2)$$

Step 3: Find the normal vector to the plane

The normal vector $\vec{n}$ is the cross product of $\overrightarrow{AB}$ and $\overrightarrow{AC}$: $$\vec{n}=\overrightarrow{AB}\times \overrightarrow{AC}$$

Compute the cross product: $$\vec{n}=\left|\begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ -2& 2& 0\\ -2& 0& 2\end{array}\right|=\mathbf{i}(2\cdot 2-0\cdot 0)-\mathbf{j}(-2\cdot 2-0\cdot (-2))+\mathbf{k}(-2\cdot 0-2\cdot (-2))$$ $$=\mathbf{i}(4)-\mathbf{j}(-4)+\mathbf{k}(4)=(4,4,4)$$

Step 4: Simplify the normal vector

The normal vector can be simplified by dividing by 4: $$\vec{n}=(1,1,1)$$

Step 5: Find the direction cosines

The direction cosines of the normal vector are the cosines of the angles it makes with the x, y, and z axes. The direction cosines are given by: $$\cos \alpha =\frac{n_x}{|\vec{n}|},\cos \beta =\frac{n_y}{|\vec{n}|},\cos \gamma =\frac{n_z}{|\vec{n}|}$$

First, find the magnitude of $\vec{n}$: $$|\vec{n}|=\sqrt{1^2+1^2+1^2}=\sqrt{3}$$

Thus, the direction cosines are: $$\cos \alpha =\frac{1}{\sqrt{3}},\cos \beta =\frac{1}{\sqrt{3}},\cos \gamma =\frac{1}{\sqrt{3}}$$

Rationalizing the denominators: $$\cos \alpha =\cos \beta =\cos \gamma =\frac{\sqrt{3}}{3}$$

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Final Answers:

a)

• Centre: $(0,0)$
• Foci: $(5,0)$ and $(-5,0)$

b)

The direction cosines of the plane are: $$\left(\frac{\sqrt{3}}{3},\frac{\sqrt{3}}{3},\frac{\sqrt{3}}{3}\right)$$