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# halp thx

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

Oct 21, 2020

#1
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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)

Oct 22, 2020
#2
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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

Oct 22, 2020
#3
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That is great Helpbot Melody  Oct 23, 2020
#4
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Thanks :) Is that correct?

Oct 23, 2020
#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.

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$$y^2 - mx^2 = 1$$   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:

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

Melody  Oct 23, 2020
#6
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Thank you so much Oct 23, 2020
#7
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What final answer did you get?

Melody  Oct 23, 2020
#8
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I'm still trying to solve the equation, I didn't get much time before rip f Oct 24, 2020
#9
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Well get back to me when you do.

You will have to private message me with the address, otherwise, I will not see it.

Melody  Oct 24, 2020
#10
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OK thanks HelpBot  Oct 24, 2020
#11
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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?

Oct 26, 2020
#12
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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)

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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
edited by Melody  Oct 26, 2020
edited by Melody  Oct 26, 2020