Ellipse analyticaly geometry problem

Thank you for your patience.In summary, we are trying to find the parameter a so that the line y=ax+11 touches the ellipse 3x^2+2y^2=11. However, there were some errors in the attempt at a solution, such as using the wrong values for a^2 and b^2 and not taking into account the orientation of the ellipse. After correcting these errors, the textbook provided a solution stating that the discriminant of the quadratic equation 3x^2+2(ax+11)^2=11 should be equal to 0, giving the solution of a = \pm \sqrt{\frac{63}{2}}. Graphing both solutions confirms that only the positive value satisfies
  • #1
Government$
87
1

Homework Statement


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


The Attempt at a Solution

|

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

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

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

[itex]({\frac{121}{9}})k^2 + ({\frac{121}{4}}) = 121[/itex]

When i solve for k i get [itex]k^2 = 6.75[/itex]

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 [itex]a = \pm \sqrt{\frac{63}{2}}[/itex]

I tried graphing this problem with both solutions and line doesn't touches ellipse.
 
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  • #2
Government$ said:
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).
 
  • #3
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?
 
  • #4
Government$ said:

Homework Statement


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


The Attempt at a Solution

|

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

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

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

[itex]({\frac{121}{9}})k^2 + ({\frac{121}{4}}) = 121[/itex]

When i solve for k i get [itex]k^2 = 6.75[/itex]

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 [itex]a = \pm \sqrt{\frac{63}{2}}[/itex]

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.
 
  • #5
Government$ said:

Homework Statement


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

The Attempt at a Solution

|

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

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

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

[itex]({\frac{121}{9}})k^2 + ({\frac{121}{4}}) = 121[/itex]

When i solve for k i get [itex]k^2 = 6.75[/itex]

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 [itex]a = \pm \sqrt{\frac{63}{2}}[/itex]

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.
[itex] \displaystyle \frac{x^2}{b^2}+\frac{y^2}{a^2}=1[/itex]
So that a2 = 11/2 , not a = 11/2 , etc.
 
  • #6
SteamKing said:
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.
 
  • #7
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.
 
  • #8
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
 
  • #9
mfb: I apologize. Overlooked scale for x-axis.
 

1. What is an ellipse?

An ellipse is a geometric shape that resembles a flattened circle. It is defined as the set of all points in a plane where the sum of the distances from two fixed points, called the foci, is constant.

2. How do you find the length of the major and minor axes of an ellipse?

The major and minor axes of an ellipse can be found by using the distance formula to find the distance between the foci, and then dividing that distance by 2 to get the length of the major axis. The minor axis can be found by using the Pythagorean theorem to calculate the length of the semi-minor axis, and then doubling it to get the length of the minor axis.

3. What is the standard form of an ellipse equation?

The standard form of an ellipse equation is (x-h)^2/a^2 + (y-k)^2/b^2 = 1, where (h,k) is the center of the ellipse and a and b are the lengths of the semi-major and semi-minor axes, respectively.

4. How do you graph an ellipse on a coordinate plane?

To graph an ellipse, first find the center of the ellipse and plot it on the coordinate plane. Then, use the lengths of the major and minor axes to plot the foci and the vertices. Finally, use a compass to draw the curve of the ellipse connecting the vertices.

5. How do you determine if a point is inside, outside, or on an ellipse?

To determine if a point is inside, outside, or on an ellipse, plug the coordinates of the point into the ellipse equation. If the resulting value is less than 1, the point is inside the ellipse. If the resulting value is greater than 1, the point is outside the ellipse. If the resulting value is equal to 1, the point is on the ellipse.

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