Which line will have no solution with the parabola?

In this section we investigate under what conditions a line and a parabola might meet in the plane. It is easy to see that there are three main cases to consider.

A line may:

• intersect a parabola at two points
• touch a parabola at one point (in this case we say the line is tangent to the parabola)
• not intersect the parabola at all.

The three cases are shown in the diagrams below.

There is also a special case when the line is parallel to the \(y\)-axis. In this case the line will cut the parabola once. Having noted this case, we will not need to consider it further, since the intersection point is easily found.

Suppose the line has the form \(y=mx+d\) and the parabola is \(y=ax^2+bx+c\). Equating, we have

This gives a quadratic equation. Hence we can conclude:

• A line can meet a parabola in at most two points.
• If the discriminant of the resulting quadratic is negative, the line does not meet the parabola at all.
• If the discriminant of the resulting quadratic is positive, the line meets the parabola in two distinct points.
• If the discriminant of the resulting quadratic is zero, then the line is tangent to the parabola.

Example

Discuss the intersection of the line \(y=3x+2\) and the parabola \(y=4x^2-9x+11\).

Solution

Equating the two equations, we have

Now the discriminant of this quadratic is \(b^2-4ac = 144-4\times 4\times 9 = 0\) and so the equation has only one solution. Hence the line is tangent to the parabola.

I don't understand the two solutions part as there is many ways to explain this, can someone help me get started?

Algebra

3 Answers

David G.

Jul 7, 2017

Thinking graphically, these correspond to the graph of the straight line (a) missing the graph of the parabola entirely, (b) kissing the parabola at one point or (c) cutting across the parabola and coinciding with it at 2 points.

The graphs in Fleur's answer provide a lovely illustration!

Explanation:

Graphically, a quadratic equation is a parabola and a linear equation is a straight line.

We can imagine situations where the straight line never crosses the parabola, passing above or below it (depending on the sign of the#x^2#term) without touching it. This corresponds to there being no solution: there is no point or set of points that lies on both the straight line and the parabola.

We can imagine a situation where the straight line is a tangent to the parabola, 'kissing' it at only one point. That single point lies in both curves, and represents a single solution to the system of equations.

Finally, if the line cuts across the parabola, it will cut the parabola at 2 points (and no more than 2), and those two points will represent two solutions to the system of equations.

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When you throw a ball (or shoot an arrow, fire a missile or throw a stone) it goes up into the air, slowing as it travels, then comes down again faster and faster ...

... and a Quadratic Equation tells you its position at all times!

Example: Throwing a Ball

A ball is thrown straight up, from 3 m above the ground, with a velocity of 14 m/s. When does it hit the ground?

Ignoring air resistance, we can work out its height by adding up these three things:
(Note: t is time in seconds)

Add them up and the height h at any time t is:

h = 3 + 14t − 5t2

And the ball will hit the ground when the height is zero:

3 + 14t − 5t2 = 0

Which is a Quadratic Equation !

In "Standard Form" it looks like:

−5t2 + 14t + 3 = 0

It looks even better when we multiply all terms by −1:

5t2 − 14t − 3 = 0

Let us solve it ...

There are many ways to solve it, here we will factor it using the "Find two numbers that multiply to give a×c, and add to give b" method in Factoring Quadratics:

a×c = −15, and b = −14.

The factors of −15 are: −15, −5, −3, −1, 1, 3, 5, 15

By trying a few combinations we find that −15 and 1 work (−15×1 = −15, and −15+1 = −14)

Rewrite middle with −15 and 1:5t2 − 15t + t − 3 = 0

Factor first two and last two:5t(t − 3) + 1(t − 3) = 0

Common Factor is (t − 3):(5t + 1)(t − 3) = 0

And the two solutions are:5t + 1 = 0 or t − 3 = 0

The "t = −0.2" is a negative time, impossible in our case.

The "t = 3" is the answer we want:

The ball hits the ground after 3 seconds!

Here is the graph of the Parabola h = −5t2 + 14t + 3

It shows you the height of the ball vs time

Some interesting points:

(0,3) When t=0 (at the start) the ball is at 3 m

(−0.2,0) says that −0.2 seconds BEFORE we threw the ball it was at ground level. This never happened! So our common sense says to ignore it.

(3,0) says that at 3 seconds the ball is at ground level.

Also notice that the ball goes nearly 13 meters high.

Note: You can find exactly where the top point is!

The method is explained in Graphing Quadratic Equations, and has two steps:

Find where (along the horizontal axis) the top occurs using −b/2a:

• t = −b/2a = −(−14)/(2 × 5) = 14/10 = 1.4 seconds

Then find the height using that value (1.4)

• h = −5t2 + 14t + 3 = −5(1.4)2 + 14 × 1.4 + 3 = 12.8 meters

So the ball reaches the highest point of 12.8 meters after 1.4 seconds.

Example: New Sports Bike

You have designed a new style of sports bicycle!

Now you want to make lots of them and sell them for profit.

Your costs are going to be:

• $700,000 for manufacturing set-up costs, advertising, etc
• $110 to make each bike

Based on similar bikes, you can expect sales to follow this "Demand Curve":

• Unit Sales = 70,000 − 200P

Where "P" is the price.

For example, if you set the price:

• at $0, you just give away 70,000 bikes
• at $350, you won't sell any bikes at all
• at $300 you might sell 70,000 − 200×300 = 10,000 bikes

So ... what is the best price? And how many should you make?

Let us make some equations!

How many you sell depends on price, so use "P" for Price as the variable

• Unit Sales = 70,000 − 200P
• Sales in Dollars = Units × Price = (70,000 − 200P) × P = 70,000P − 200P2
• Costs = 700,000 + 110 x (70,000 − 200P) = 700,000 + 7,700,000 − 22,000P = 8,400,000 − 22,000P
• Profit = Sales-Costs = 70,000P − 200P2 − (8,400,000 − 22,000P) = −200P2 + 92,000P − 8,400,000

Profit = −200P2 + 92,000P − 8,400,000

Yes, a Quadratic Equation. Let us solve this one by Completing the Square.

Solve: −200P2 + 92,000P − 8,400,000 = 0

Step 1 Divide all terms by -200

Step 2 Move the number term to the right side of the equation:

Step 3 Complete the square on the left side of the equation and balance this by adding the same number to the right side of the equation:

(b/2)2 = (−460/2)2 = (−230)2 = 52900

P2 – 460P + 52900 = −42000 + 52900

Step 4 Take the square root on both sides of the equation:

P – 230 = ±√10900 = ±104 (to nearest whole number)

Step 5 Subtract (-230) from both sides (in other words, add 230):

P = 230 ± 104 = 126 or 334

What does that tell us? It says that the profit is ZERO when the Price is $126 or $334

But we want to know the maximum profit, don't we?

It is exactly half way in-between! At $230

And here is the graph:

The best sale price is $230, and you can expect:

• Unit Sales = 70,000 − 200 x 230 = 24,000
• Sales in Dollars = $230 x 24,000 = $5,520,000
• Costs = 700,000 + $110 x 24,000 = $3,340,000
• Profit = $5,520,000 − $3,340,000 = $2,180,000

A very profitable venture.

Example: Small Steel Frame

Your company is going to make frames as part of a new product they are launching.

The frame will be cut out of a piece of steel, and to keep the weight down, the final area should be 28 cm2

The inside of the frame has to be 11 cm by 6 cm

What should the width x of the metal be?

Area of steel before cutting:

Area = (11 + 2x) × (6 + 2x) cm2

Area = 66 + 22x + 12x + 4x2

Area of steel after cutting out the 11 × 6 middle:

Area = 4x2 + 34x + 66 − 66

Here is the graph of 4x2 + 34x :

The desired area of 28 is shown as a horizontal line.

The area equals 28 cm2 when:

x is about −9.3 or 0.8

The negative value of x make no sense, so the answer is:

x = 0.8 cm (approx.)

Example: River Cruise

A 3 hour river cruise goes 15 km upstream and then back again. The river has a current of 2 km an hour. What is the boat's speed and how long was the upstream journey?

There are two speeds to think about: the speed the boat makes in the water, and the speed relative to the land:

• Let x = the boat's speed in the water (km/h)
• Let v = the speed relative to the land (km/h)

Because the river flows downstream at 2 km/h:

• when going upstream, v = x−2 (its speed is reduced by 2 km/h)
• when going downstream, v = x+2 (its speed is increased by 2 km/h)

We can turn those speeds into times using:

time = distance / speed

(to travel 8 km at 4 km/h takes 8/4 = 2 hours, right?)

And we know the total time is 3 hours:

total time = time upstream + time downstream = 3 hours

Put all that together:

total time = 15/(x−2) + 15/(x+2) = 3 hours

Now we use our algebra skills to solve for "x".

First, get rid of the fractions by multiplying through by (x-2)(x+2):

3(x-2)(x+2) = 15(x+2) + 15(x-2)

Expand everything:

3(x2−4) = 15x+30 + 15x−30

Bring everything to the left and simplify:

3x2 − 30x − 12 = 0

It is a Quadratic Equation! Let us solve it using the Quadratic Formula:

Where a, b and c are from the
Quadratic Equation in "Standard Form": ax2 + bx + c = 0

Solve 3x2 - 30x - 12 = 0

Coefficients are:a = 3, b = −30 and c = −12

Quadratic Formula:x = [ −b ± √(b2−4ac) ] / 2a

Put in a, b and c:x = [ −(−30) ± √((−30)2−4×3×(−12)) ] / (2×3)

Solve:x = [ 30 ± √(900+144) ] / 6

Answer: x = −0.39 or 10.39 (to 2 decimal places)

x = −0.39 makes no sense for this real world question, but x = 10.39 is just perfect!

Answer: Boat's Speed = 10.39 km/h (to 2 decimal places)

And so the upstream journey = 15 / (10.39−2) = 1.79 hours = 1 hour 47min

And the downstream journey = 15 / (10.39+2) = 1.21 hours = 1 hour 13min

Example: Resistors In Parallel

Two resistors are in parallel, like in this diagram:

The total resistance has been measured at 2 Ohms, and one of the resistors is known to be 3 ohms more than the other.

What are the values of the two resistors?

The formula to work out total resistance "RT" is:

1RT   =   1R1 + 1R2

In this case, we have RT = 2 and R2 = R1 + 3

1 2   =   1 R1 + 1 R1+3

To get rid of the fractions we can multiply all terms by 2R1(R1 + 3) and then simplify:

Multiply all terms by 2R1(R1 + 3):2R1(R1+3)2 = 2R1(R1+3)R1 + 2R1(R1+3)R1+3

Then simplify:R1(R1 + 3) = 2(R1 + 3) + 2R1

Expand: R12 + 3R1 = 2R1 + 6 + 2R1

Bring all terms to the left:R12 + 3R1 − 2R1 − 6 − 2R1 = 0

Yes! A Quadratic Equation !

Let us solve it using our Quadratic Equation Solver.

• Enter 1, −1 and −6
• And you should get the answers −2 and 3

R1 cannot be negative, so R1 = 3 Ohms is the answer.

The two resistors are 3 ohms and 6 ohms.

Others

Quadratic Equations are useful in many other areas:

For a parabolic mirror, a reflecting telescope or a satellite dish, the shape is defined by a quadratic equation.

Which lines indicate no solution?

Which equation has no solution?

What are solutions on a parabola?

Does 0 have no solution?