Two unknown tangents to a parabola

[Jakotheshadows]

"Draw a diagram to show that there are two tangent lines to the parabola y = x^2 that pass through the point (0, -4). Find the coordinates of the points where these tangent lines intersect the parabola."

I know that y'= 2x by the power rule, but I'm not sure how to find the points where these specific undetermined tangents intersect the parabola. There's something I'm just not seeing, and I really need to know how to start working on this. Is there some way to determine what these tangents are with the knowledge that they both pass through (0, -4) and that they are tangent to y=x^2 at some unknown point? (mspaint ftw )

 

[wjm11]

"Draw a diagram to show that there are two tangent lines to the parabola y = x^2 that pass through the point (0, -4). Find the coordinates of the points where these tangent lines intersect the parabola."

I know that y'= 2x by the power rule, but I'm not sure how to find the points where these specific undetermined tangents intersect the parabola. There's something I'm just not seeing, and I really need to know how to start working on this. Is there some way to determine what these tangents are with the knowledge that they both pass through (0, -4) and that they are tangent to y=x^2 at some unknown point?

You have a point and a slope (from the derivative); write an equation:

y – (-4) = (2x)(x-0) or

y + 4 = 2x^2 or

y = 2x^2 – 4

This forms a system of equations with the original function, y = x^2.

Set them equal and solve for x:

2x^2 – 4 = x^2

x^2 = 4

x = +/- 2

These are the x values of the tangent points on the original function. Make sense?

 

[Jakotheshadows]

This forms a system of equations with the original function, y = x^2.

How does that work? Could you elaborate on how that works? Is there an alternative method?

 

[stapel]

This forms a system of equations with the original function, y = x^2....

How does that work? Could you elaborate on how that works?

It works in the way shown in the solution that you were provided. :shock:

It would appear that you are unfamiliar with how to solve non-linear systems of equations. This was supposed to have been covered back in algebra, and you will need this knowledge, so please take a break to learn how. :wink:

You have the parabola y = x[sup:1wffwzr5]2[/sup:1wffwzr5], so the slope at any point along this curve will be dy/dx = 2x = m. For a tangent line touching the curve (x, y) = (x, x[sup:1wffwzr5]2[/sup:1wffwzr5]) and passing through the point (x, y) = (0, -4), the slope relation will be:

. . . . .\(\displaystyle 2x\, =\, \frac{x^2\, +\, 4}{x}\)

. . . . .\(\displaystyle 2x^2\, =\, x^2\, +\, 4\)

. . . . .\(\displaystyle x^2\, -\, 4\, =\, 0\)

The above would be an alternative method of solution.

 

[Jakotheshadows]

I am familiar with solving systems of equations. What I am not familiar with is why that is a system of equations, or how to recognize what will become systems of equations in future problems.

edit: nevermind.. I needed to think about the problem visually to understand that it was a system. Thanks guys

 

[alexe]

There's a different way that might be a little easier:

You have point (0, -4) and you create (x, y).

Using the slope formula you get, y-(-4) / x - 0 = (y+4)/x

y' which is a slope of the tangent line is 2x. Set both equations equal to each other:

2x= (y+4)/x

y+4=2x^2

y=2x^2 - 4

Since y = x^2

x^2=2x^2 - 4

x^2-4=0

x^2=4

x= + / - 2

Solving for the initial equation you get points (2, 4) and (-2, 4).