Ellipse analyticaly geometry problem

[Government$]

Homework Statement

Find parameter a so that line y=ax + 11 touches ellipse 3x^2 + 2y^2 = 11

The Attempt at a Solution

|

I can rewrite ellipse equation like \frac{x^2}{\frac{11}{3}} + \frac{y^2}{\frac{11}{2}} = 1

And i know that line y=kx + n touches ellipse when a^2k^2 + b^2 = n^2

So in essence i am looking for a slope of a line.
({\frac{11}{3}})^2k^2 + ({\frac{11}{2}})^2 = 11^2

({\frac{121}{9}})k^2 + ({\frac{121}{4}}) = 121

When i solve for k i get k^2 = 6.75

Problem is that this is not a solution. Here is what my textbook says:

Line that touches ellpise if and only if system y=ax + 11, 3x^2 + 2y^2 = 11 has one solution i.e. when discriminant of quadratic equation 3x^2 + 2(ax+ 11)^2 = 11 is equal to 0, and for that a = \pm \sqrt{\frac{63}{2}}

I tried graphing this problem with both solutions and line doesn't touches ellipse.

 

[mfb]

I tried graphing this problem with both solutions and line doesn't touches ellipse.

Looks touching (using sqrt(63/2))
This one does not (using sqrt(6.75))

You used two different equations to plug in your numbers in a^2k^2+b^2=n^2 - either use the given equation (with a^2=3, b^2=2) or the modified equation (with n^2=1).

 

[SteamKing]

mfb: shouldn't the ellipse plot the same regardless of the equation of the line?

If you look at both plots, you will see that the major axis of the second ellipse is at right angles to the major axis of the first ellipse. Warum das?

 

[Saitama]

Homework Statement

Find parameter a so that line y=ax + 11 touches ellipse 3x^2 + 2y^2 = 11

The Attempt at a Solution

|

I can rewrite ellipse equation like \frac{x^2}{\frac{11}{3}} + \frac{y^2}{\frac{11}{2}} = 1

And i know that line y=kx + n touches ellipse when a^2k^2 + b^2 = n^2

So in essence i am looking for a slope of a line.
({\frac{11}{3}})^2k^2 + ({\frac{11}{2}})^2 = 11^2

({\frac{121}{9}})k^2 + ({\frac{121}{4}}) = 121

When i solve for k i get k^2 = 6.75

Problem is that this is not a solution. Here is what my textbook says:

Line that touches ellpise if and only if system y=ax + 11, 3x^2 + 2y^2 = 11 has one solution i.e. when discriminant of quadratic equation 3x^2 + 2(ax+ 11)^2 = 11 is equal to 0, and for that a = \pm \sqrt{\frac{63}{2}}

I tried graphing this problem with both solutions and line doesn't touches ellipse.

What are the values of a^2 and b^2 to be used in the equation: $a^2k^2+b^2=n^2$. You squared both a^2 and b^2 while substituting them in the equation.

 

[SammyS]

Homework Statement

Find parameter a so that line y=ax + 11 touches ellipse 3x^2 + 2y^2 = 11

The Attempt at a Solution

|

I can rewrite ellipse equation like \frac{x^2}{\frac{11}{3}} + \frac{y^2}{\frac{11}{2}} = 1

And i know that line y=kx + n touches ellipse when a^2k^2 + b^2 = n^2

So in essence i am looking for a slope of a line.
({\frac{11}{3}})^2k^2 + ({\frac{11}{2}})^2 = 11^2

({\frac{121}{9}})k^2 + ({\frac{121}{4}}) = 121

When i solve for k i get k^2 = 6.75

Problem is that this is not a solution. Here is what my textbook says:

Line that touches ellipse if and only if system y=ax + 11, 3x^2 + 2y^2 = 11 has one solution i.e. when discriminant of quadratic equation 3x^2 + 2(ax+ 11)^2 = 11 is equal to 0, and for that a = \pm \sqrt{\frac{63}{2}}

I tried graphing this problem with both solutions and line doesn't touches ellipse.

I see a couple of major problems.

1. 11/2 > 11/3, so the major axis of the ellipse is along the y-axis .

2. The standard form for the equation of such an ellipse is often written as follows.

\displaystyle \frac{x^2}{b^2}+\frac{y^2}{a^2}=1
​

So that a² = 11/2 , not a = 11/2 , etc.

 

[mfb]

mfb: shouldn't the ellipse plot the same regardless of the equation of the line?

It is, but the scales used in the plots are different.

 

[SteamKing]

mfb: I appreciate your efforts at illustration, but the Mathematica plots of the tangent line candidates still make no sense. Both lines have y-intercepts of (0, 11), and neither line passes thru this point.

 

[mfb]

Do we see different graphs? Both lines hit (0,11).
Here are the raw inputs to WolframAlpha:
y=sqrt(63/2)*x+11 and 3x^2+2y^2=11
y=sqrt(6.75)*x+11 and 3x^2+2y^2=11
And here a direct comparison:
y=sqrt(63/2)*x+11 and y=sqrt(6.75)*x+11 and 3x^2+2y^2=11

 

[SteamKing]

mfb: I apologize. Overlooked scale for x-axis.