y = 2x + m is Tangent to Ellipse (x^2)/39+(y^2)/13=1

Nikolas111

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Hello everybody

I would like to ask you whether you could help me with this problem. The problem is I have a straight line with an equation y=2x+m where m is a real parameter and this equation is a tangent to ellipse which has an equation (x^2)/39+(y^2)/13=1 the question is what are those two m parameters ;) .

Thank you very much in advance ;)
 
What have you tried? If you are in calculus, you may wish to examine the first derivative and figure out where the ellipse has a slope of 2. Where did "2" come from? Is there more than one solution?

Let's see your best efforts.
 
Hello,
Well If I'm not wrong the first derivative of x^2/(39)+y^2/13=1 is y'=-x/y and 2 is the slope that comes from the formula(y2-y1)/(x2-x1) but I don't know how to continue or where to plug in that slope .
Thank you
 
Hello,
Well If I'm not wrong the first derivative of x^2/(39)+y^2/13=1 is y'=-x/y and 2 is the slope that comes from the formula(y2-y1)/(x2-x1) but I don't know how to continue or where to plug in that slope .
Thank you

x^2/(39)+y^2/13=1

2*x/(39)+2*y*y'/13=0 ...... *39

x + 3*y*y' = 0

y' = -(1/3) * (x/y) ................ continue
 
Hello,
Well If I'm not wrong the first derivative of x^2/(39)+y^2/13=1 is y'=-x/(3y) and 2 is the slope that comes from the formula(y2-y1)/(x2-x1) but I don't know how to continue or where to plug in that slope .
Thank you
How did you get these co-ordinates?

Please post the complete problem.
 
If you mean that first derivative it was just a misprint because I was counting a similair problem so I accidentally wrote different result here and what concerns of the second step I don't know how to plug that slope to the equation .
 
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If you mean that first derivative it was just a misprint because I was counting a similair problem so I accidentally wrote different result here and what concerns of the second step I don't know how to plug that slope to the equation .
Please post the complete problem!
 
It's not in English language...

Problem : A straight line y=2x+m where m is a real parameter and is a tangent of an ellipse which is given by an equation x^2/39+y^2/13=1 for two values of real parameters m1 and m2 .The product of m1 and m2 equals?

That's everything what it says.
 
It's not in English language...

Problem : A straight line y=2x+m where m is a real parameter and is a tangent of an ellipse which is given by an equation x^2/39+y^2/13=1 for two values of real parameters m1 and m2 .The product of m1 and m2 equals?

That's everything what it says.

Okay, well, here's how I'd solve this problem. No Calculus required. It may not be most elegant, or even the easiest, solution, but it does work. We know that the line y = 2x + m must be tangent to the ellipse. We need to find two values m = m1 and m = m2, for which this property holds. Recall that for a line to be tangent to a function, the two must intersect exactly once. But what does that mean in this case?

If we plug the value we have for y into the equation for the ellipse, we'll get \(\displaystyle \dfrac{x^2}{39}+\dfrac{(2x+m)^2}{13}=1\). This can be rearranged into a quadratic in terms of m and x. Solving this quadratic for x will give either 0, 1, or 2 solutions, depending on the value of m. How does this result relate to how many times the line and the ellipse intersect? Given that the line needs to be tangent to the ellipse? how many solutions should the quadratic have? Can you find the only two values of m that produce the desired number of solutions?

This method requires a lot of Algebra and rearranging terms and what not, but depending on your proficiencies with Algebra and Calculus respectively, it may be easier for you.
 
Hello

I tried to solve it x^2/39 + y^2/13 =1
1. x^2+3y^2=39
2. x^2+3y^2-39=0
3. x^2+3*(2x+m)^2 -39=0
4. x^2+3*(4x^2+4xm+m^2)-39=0
5. x^2+12x^2+12xm+3m^2-39=0
6. 13x^2+12xm+3m^2-39=0

D=b^2-4ac
1. D=(12m)^2 - 4*(13*3m^2 *(-39))
2. D=144m^2+6084m^2=0
3. D=sqrt(6228m^2)
4. (-12m+-sqrt(6228m^2))/26
5.(-6+-3m*sqrt(173))/13



but this way of solving is not correct because those parameters have to be real
 
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Hello

I tried to solve it x^2/39 + y^2/13 =1
1. x^2+3y^2=39
2. x^2+3y^2-39=0
3. x^2+3*(2x+m)^2 -39=0
4. x^2+3*(4x^2+4xm+m^2)-39=0
5. x^2+12x^2+12xm+3m^2-39=0
6. 13x^2+12xm+3m^2-39=0

D=b^2-4ac
1. D=(12m)^2 - 4*(13*3m^2 *(-39))
2. D=144m^2+6084m^2=0
3. D=sqrt(6228m^2)
4. (-12m+-sqrt(6228m^2))/26
5.(-6+-3m*sqrt(173))/13

but this way of solving is not correct because those parameters have to be real

This is a very good attempt. You're almost there, you just made one small error. Once you've found your quadratic, you're correct that you want to set the discriminant to zero in order to find values of m that produce one real solution. However, you've plugged in the wrong values for a, b, and c in the discriminant. The general form of a quadratic is ax^2 + bx + c = 0. Let's rewrite your quadratic and see what we can come up with: (13)x^2 + (12m)x + (3m^2 - 39) = 0. I think the grouping symbols make it easier to see what's going on. You'll have:

a = 13
b = 12m
c = 3m^2 - 39

Try continuing from here and see what you get. You'll end up with the discriminant itself being a quadratic that gives two real solutions for m. Compare that to what you did before and see if you can figure out why what it was wrong. :)
 
Well I got it the problem was in the way of distribution :) thank you very much for helping me and being patient :)
 
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