Andrew J.

asked • 06/15/20# Calculus question - tangent line equation of ellipse

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

*For the ellipse [MATH]*b^2x^2+a^2y^2=a^2b^2*[/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

## 1 Expert Answer

Hi Andrew,

There is another way. Let (x_{1}, y_{1}) be the point of intersection of the ellipse and a tangent line of slope m. Note that y_{1} cannot be zero since otherwise the tangent lines are vertical (x = ± a). By implicit differentiation, show that m = - b^{2}x_{1}/(a^{2}y_{1}). In your notation, c = y_{1} - mx_{1}, which I’ll now compute.

y_{1 }- mx_{1} = y_{1} + (mb^{2}x_{1}^{2})/(a^{2}y_{1})

..............= (a^{2}y_{1}^{2} + b^{2}x_{1}^{2})/(a^{2}y_{1})

..............= a^{2}b^{2}/(a^{2}y_{1}) (by the ellipse equation)

..............= b^{2}/y_{1 }

Now

m^{2} = (b^{4}x_{1}^{2})/(a^{4}y_{1}^{2}) = b^{2}/a^{2} • (b^{2}x_{1}^{2})/(a^{2}y_{1}^{2})

......= b^{2}/a^{2} • (a^{2}b^{2} - a^{2}y_{1}^{2})/(a^{2}y_{1}^{2}) (by the ellipse equation)

......= b^{2}/a^{2} • (b^{2}/y_{1}^{2} - 1)

......= 1/a^{2} • (b^{4}/y_{1}^{2} - b^{2})

and so a^{2}m^{2} = b^{4}/y_{1}^{2} - b^{2}, or, a^{2}m^{2} + b^{2} = b^{4}/y_{1}^{2}. Taking square roots results in b^{2}/y_{1} = ± √(a^{2}m^{2} + b^{2}). Hence, the tangent lines are

y = mx + c = mx ± √(a^{2}m^{2} + b^{2})

as desired.

If you have more calculus questions, feel free to message me and schedule a lesson.

Andrew J.

Thank very much Eugene. You are a genius - I wouldn't have figured this out in a million years. Is there a more intuitive way to get there? I had gotten to c=b^2/y_1 but I don't think I every would have known to start working on M^2 from there. Thanks Again!06/15/20

Eugene E.

06/15/20

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Eugene E.

06/15/20