The parabola \(y = x^2+2 \) and the hyperbola \(y^2 - mx^2 = 1\) are tangent. Find \(m\)

HelpBot Oct 21, 2020

#1**+2 **

I do not have time to present this properly and maybe getting an outline is better anyway..

First, draw a rough sketch, this will be enough to tell you that y must be greater than 2. It will also tell you that there is only one y value (but 2 x vales for the solution.)

second, Solve the 2 equations simultaneously, (for y)

y will be in terms of m but if you know roughly what it looks like you will know that there is only one solution.

So the discriminant must be 0.

From this, the value of m can be determined.

You are welcome to ask questions (tell me what you have done though)

**Please no one else answer, unless answering a specific further question from Helpbot**

Melody Oct 22, 2020

#2**+1 **

Thx. I graphed the parabola and plugged in values for m.

y=x^2+2

y=-3x^2=1

And solved.

Hope im on the right track thx

HelpBot Oct 22, 2020

#4

#5**+1 **

Um I thought it was, but I didn't look properly and you did not expressly ask me to look. Sorry.

I will give you a more comprehensive answer.

First, draw a rough sketch, this will be enough to tell you that y must be greater than 2. It will also tell you that there is only one y value (but 2 x vales for the solution.)

\(y = x^2+2\) is a concave up parabola. The axis of symmetry is the y axis and the vertex is (0,2)

You should have enough knowledge to know this.

--------------------

\( y^2 - mx^2 = 1\) is a hyperbola so long as **m is positive** real number

Hypperbolas of the form \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\) have a vertical axis, the y axis is the axis of symmetry and a ist the y intercept

So in our case, a=+/-1 So the vertices are (0,1) and (0,-1)

So this is our initial sketch. The red is the parabola and the blue is the hyperbola.

m will change the steepness of the hyperbola so you want to find the m that will make the graph look like this.

So by now you should at least understand what you are being asked, as far as the outcome goes.

-------------------------------

So now you have to solve

\(y=x^2+2\qquad and \qquad y^2-mx^2=1\qquad \text{simultaneously}\\ \text{we know already}\\ \text{ m is positive and y is positive and there is only one value of y}\)

\(y=x^2+2\qquad and \qquad y^2-mx^2=1\\ x^2=y-2\qquad and \qquad x^2=\frac{1-y^2}{-m}\\ so\\ y-2=\frac{1-y^2}{-m}\\ -my+2m=1-y^2\\ y^2-my+(2m-1)=0\\\)

now solve using the quadratic formula

y will be in terms of m but if you know roughly what it looks like you will know that there is only one solution.

So the discriminant must be 0.

From this, the value of m can be determined.

You are welcome to ask questions (tell me what you have done though)

LaTex:

y=x^2+2\qquad and \qquad y^2-mx^2=1\\

x^2=y-2\qquad and \qquad x^2=\frac{1-y^2}{-m}\\

so\\

y-2=\frac{1-y^2}{-m}\\

-my+2m=1-y^2\\

y^2-my+(2m-1)=0\\

Melody
Oct 23, 2020

#8

#11**+2 **

I solved for y and got

y^2=my-2m+1

y=sqrt(my-2m+1)

Hmmm how do u remove the y from my?

HelpBot Oct 26, 2020

#12**+1 **

Thanks for answering Helpbot,

I am sorry to say that you have not solved it.

You see you have a y on both sides.

You have to have y on one side and en expression with m on the other side.

Think of m just as a constant. I mean treat it like it is a number that stays the same all the time.

\(y^2-my+(2m-1)=0\\ \text{This is a quadatic equation}\\ \text{It is of the form}\\ ay^2+by+c=0\\ \)

NOW to solve this you use the quadratic formula.

\(y = {-b \pm \sqrt{b^2-4ac} \over 2a} \)

So now you solve yours.

Hint: a=1 b=-m c=(2m-1)

Your answer will be will have an m in it.

--------------

AFTER you have done this, think about how many answers there can be for y.

How can you make this happen?

Melody
Oct 26, 2020