Help! Ellipse/tangent question driving me crazy

[GilbertKeithChesterton]

Question is from Schaum's Guide to Calculus, p.97 q.18:

For the ellipse [MATH][/MATH][MATH]b^2x^2+a^2y^2=a^2b^2[I][/I][/MATH] show that the equations of its tangent lines of slope m are [MATH]y=mx \pm \sqrt{a^2m^2+b^2}[/MATH]
Question is in chapter on tangent lines and is mostly based on taking implicit derivatives and plugging into point-slope format for the tangent lines. I've seen complicated derivations based on substituting mx+c into the ellipse question and solving for c (by setting the discriminant of the quadratic to zero) but I'm 99.999% certain the book isn't asking for this as that would be far more complex than anything yet covered up to this point.

Really appreciate any help anyone can offer here!

Andrew

 

[Steven G]

First you need to find all the points on the ellipse whose slopes are m. Then for each of those points you need to find the equation of the tangent line. This part is easy since you have the slope (m) and the points.

Therefore the main part of this problem is finding all the points on the ellipses whose tangent lines have a slope of m.

Please find these points, post back and we can go from there.

 

[GilbertKeithChesterton]

Hello, with the slope being the(from implicit differentiation) [MATH]y'=-\frac{b^2x}{a^2y}[/MATH], this would be the slope [MATH]m[/MATH] at all points [MATH](x_1,y_1)[/MATH], correct?

 

[Steven G]

Hello, with the slope being the(from implicit differentiation) [MATH]y'=-\frac{b^2x}{a^2y}[/MATH], this would be the slope [MATH]m[/MATH] at all points [MATH](x_1,y_1)[/MATH], correct?

No, that is not correct. You found, if I remember correctly, the correct y'. Now what exactly do you mean y' would be the slope m at all points (x1,y1)?? Do these points at least have to be on the ellipse?? Image m is instead 3. What would you do then. In fact try to do the problem if you were told that m = 3. Then redo it the same exact way exact use m. I do not think that you realize that m is a constant. It does NOT CHANGE.

Please try to do the problem with 3 instead of m (or use m but think 3!) and see what you get. NOT every point on the ellipse have a slope of m, just like every point on the ellipse does not have a slope of 3.

 

[HallsofIvy]

Hello, with the slope being the(from implicit differentiation) [MATH]y'=-\frac{b^2x}{a^2y}[/MATH], this would be the slope [MATH]m[/MATH] at all points [MATH](x_1,y_1)[/MATH], correct?

Yes, and so [math]-\frac{b^2x_1}{a^2y_1}= m[/math] leading to [math]y_1= -\frac{b^2x_1}{ma^2}[/math]. Also, [math](x_1, y_1)[/math] is a point on the ellipse [math]\frac{x^2}{a^2}+ \frac{y^2}{b^2}= 1[/math] so also [math]y_1= \pm b\sqrt{1- \frac{x_1^2}{a^2}}[/math].