Hyperbolas

Exeter Library Hyperbola by Flickr user DrEquation

There is a major intersection in L.A. city that has the most accidents a year.

There is a major intersection in L.A. city that has the most accidents a year.

Just recently there was a 5 car pile up and 3 people were killed. Since the intersection was built in 1982, 297 people have died.

There is a major intersection in L.A. city that has the most accidents a year.

Just recently there was a 5 car pile up and 3 people were killed. Since the intersection was built in 1982, 297 people have died.

The people of L.A. say this is unacceptable and that this intersection must be redesigned.

There is a major intersection in L.A. city that has the most accidents a year.

Just recently there was a 5 car pile up and 3 people were killed. Since the intersection was built in 1982, 297 people have died.

The people of L.A. say this is unacceptable and that this intersection must be redesigned.

However, no one knows what to do. After weeks of discussion on this problem, the city councilors of L.A. decide to ask you for some help. They hope that your math knowledge might be able to aid them with their problem.

This major intersection currently looks like this....

You suggest a hyperbola. Why don’t they make the intersection like a hyperbola that looks

like this....

The city councilors say they want the road running N-S to be 1.5 km long and the road running E-W to be 4 km long before it branches off.

The city councilors say they want the road running N-S to be 1.5 km long and the road running E-W to be 4 km long before it branches off.

This means the transverse axis is 1.5 km long because the hyperbola is vertical. Which means the conjugate axis is 4 km long.

The city councilors say they want the road running N-S to be 1.5 km long and the road running E-W to be 4 km long before it branches off.

The city councilors say they want the road running N-S to be 1.5 km long and the road running E-W to be 4 km long before it branches off.

The semi-conjugate axis is the conjugate axis divided by 2. 4/2=2. Therefore the semi-conjugate axis is 2.

The city councilors say they want the road running N-S to be 1.5 km long and the road running E-W to be 4 km long before it branches off.

The city councilors say they want the road running N-S to be 1.5 km long and the road running E-W to be 4 km long before it branches off.

This means the transverse axis is 1.5 km long because the hyperbola is vertical. Which means the conjugate axis is 4 km long.

The semi-transverse axis is the transverse axis divided by 2. 1.5/2= 0.75. Therefore the semi-transverse axis is 0.75.

The city councilors say they want the road running N-S to be 1.5 km long and the road running E-W to be 4 km long before it branches off.

The city councilors say they want the road running N-S to be 1.5 km long and the road running E-W to be 4 km long before it branches off.

This means the transverse axis is 1.5 km long because the hyperbola is vertical. Which means the conjugate axis is 4 km long.

The semi-conjugate axis is the conjugate axis divided by 2. 4/2=2. Therefore the semi-conjugate axis is 2.

The city councilors say they want the road running N-S to be 1.5 km long and the road running E-W to be 4 km long before it branches off.

The city councilors say they want the road running N-S to be 1.5 km long and the road running E-W to be 4 km long before it branches off.

This means the transverse axis is 1.5 km long because the hyperbola is vertical. Which means the conjugate axis is 4 km long.

The semi-conjugate axis is the conjugate axis divided by 2. 4/2=2. Therefore the semi-conjugate axis is 2.

The semi-transverse axis is the transverse axis divided by 2. 1.5/2= 0.75. Therefore the semi-transverse axis is 0.75.

With this information you can draw the hyperbola.

Assume the centre of the “box” of the hyperbola is at (0,0). Now draw the conjugate and transverse axis.

Assume the centre of the “box” of the hyperbola is at (0,0). Now draw the conjugate and transverse axis.

0.75 km

0.75 km

Assume the centre of the “box” of the hyperbola is at (0,0). Now draw the conjugate and transverse axis.

0.75 km

0.75 km

Transverse Axis

Assume the centre of the “box” of the hyperbola is at (0,0). Now draw the conjugate and transverse axis.

0.75 km

0.75 km

Transverse Axis

2 km2 km

0.75 km

Assume the centre of the “box” of the hyperbola is at (0,0). Now draw the conjugate and transverse axis.

0.75 km

Transverse Axis

2 km 2 km

Conjugate Axis

Now draw the box and the asymptotes.

2 km 2 km

0.75 km

0.75 km

Transverse AxisConjugate Axis

Now draw the box and the asymptotes.

2 km 2 km

0.75 km

0.75 km

Transverse AxisConjugate Axis

Now draw the box and the asymptotes.

2 km 2 km

0.75 km

0.75 km

Transverse AxisConjugate Axis

Now draw the box and the asymptotes.

2 km 2 km

0.75 km

0.75 km

Transverse AxisConjugate Axis

Now draw the box and the asymptotes.

2 km 2 km

0.75 km

0.75 km

Transverse AxisConjugate Axis

Asymptotes

You know the hyperbola opens up and down so draw it in. The vertices of the hyperbola are at the endpoints of the transverse axis.

You know the hyperbola opens up and down so draw it in. The vertices of the hyperbola are at the endpoints of the transverse axis.

Vertices

Congratulations! You’ve designed a major intersection for one of the biggest cities in the world. The city councilors of L.A. are thoroughly pleased. They think you’re right about the usefulness of math.