Learning Objectives

# Problem 1 of 3

Given: 4x^{2} - y^{2} - 2y + 3 = 0

(a) Write in standard form.

(b) Sketch the graph.

(c) State the domain, range and equations of the asymptotes.

**Solution**

a)

4x^{2} - y^{2} - 2y + 3 = 0

4x^{2} - y^{2} - 2y = -3

(4x^{2}) - (y^{2} - 2y + 1) = -3-1

(4x^{2})/-4 - (y + 1)^{2}/-4 = -4/-4

-x^{2} + (y + 1)^{2}/-4 = 1

(y + 1)^{2}/-4 -x^{2} = 1

'by bringing the 3 to the other side of the equal side, it becomes a -3

'complete the square for the "y" Considering you can't do anything for the "x", instead of adding a 1 on the right side you subract a 1. This is because the y is negative.

'divide everything by -4

The 4x reduces to -x and the -4 reduces to 1

Since "y is positive, it should be written first and then x. But this is optional

b)

click here to see the graph

First we need to build a box. from the centre,the horizontal distance is 1 unit away, and the vertical distance is 2 units away. Then draw the asymptotes which are broken diagonal lines that go from corner to corner. After that, draw your hyperbolas that go vertical.

c)

Domain : (- infinite, infinite) 'thinking <----------> this way, the graphy goes on forever horizontally

Range : (3 , infinite) thinking vertically, the lowest value it can reach is 3 and the highest value goes on forever aka to infinite

eq'n of asymptopes : y = 4/1, -4/1 'thinking the slope formula, (rise over run) rise is 4, and run is 1

# Problem 2 of 3

A semicircular tunnel has a radius of 9 m. The roadway is a diameter of the circle.

(a) Determine the equation for the circle, part of which forms the tunnel. Choose the x-axis along the base of the tunnel and the y-axis up the centre of the tunnel.

(b) Can a truck 8.3 m high carrying a load 6 m wide pass through the tunnel if the truck is allowed to drive in the middle of the road? Supporting evidence must be shown.

**Solution**

(a) The equation of a circle is (x-h)^2 + (y-k)^2 = r^2

We know that the center of the tunnel is located at the origin since the roadway of the tunnel which is the diameter of the circle is placed along the x-axis. The center of the circle is up along the y-axis and the radius is 9.

Here's a diagram to better explain

By knowing where the center of the circle is and the value of the radius we can input those values into the equation.

center is at (0,0) and the radius is 9.

(x-0)^2 + (y-0)^2 = 9^2

x^2 + y^2 = 81

So the equation of the circle is x^2 + y^2 = 81.

(b)We know that the truck is allowed to drive in the middle road. Given that the truck's width is 6m, for it to drive in the middle of the road half of truck will be on the right side and the other half on the left. Half of 6 is 3 so 3 m of the truck will be on the right side and that 3 is a point on the x-axis.

Here's another diagram to better explain.

To solve this question we need a point along the x-axis that will help us solve for the height. From looking at the diagram we know that 3 will be that point. We now have an x point and all we have to do is look for the y point along the semi-circle that will give us the height of the tunnel. In other words we have the coordinates (3,y) and we are looking for y. To do that we substitute the value of 3 for x in the circle equation and solve for y.

3^2 + y^2 = 81

9 + y^2 = 81

y^2 = 81-9

y^2 = 72

y = square root of 72

y = 8.4853

Since the height of the tunnel is greater than the height of the truck since the height of the truck is 8.3 m and the height of the tunnel is 8.4853 this enables the truck to pass through the tunnel if the truck drove in the middle of the road.

# Problem 3 of 3

A parabolic antenna is 320 cm wide at a distance of 50 cm from its vertex.

(a) Find an equation of the parabola. **Hint: ***Drawing a diagram will help. Put the vertex where you want it.*

(b) Determine the distance of the focus from the vertex.

(c) Sketch the graph and label the focus and the equation of the directrix.

**Solution**

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