Calculus question - tangent line equation of ellipse

Question

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

Answer 1

Hi Andrew,

There is another way. Let (x₁, y₁) be the point of intersection of the ellipse and a tangent line of slope m. Note that y₁ cannot be zero since otherwise the tangent lines are vertical (x = ± a). By implicit differentiation, show that m = - b²x₁/(a²y₁). In your notation, c = y₁ - mx₁, which I’ll now compute.

y₁ - mx₁ = y₁ + (mb²x₁²)/(a²y₁)

..............= (a²y₁² + b²x₁²)/(a²y₁)

..............= a²b²/(a²y₁) (by the ellipse equation)

..............= b²/y₁

Now

m² = (b⁴x₁²)/(a⁴y₁²) = b²/a² • (b²x₁²)/(a²y₁²)

......= b²/a² • (a²b² - a²y₁²)/(a²y₁²) (by the ellipse equation)

......= b²/a² • (b²/y₁² - 1)

......= 1/a² • (b⁴/y₁² - b²)

and so a²m² = b⁴/y₁² - b², or, a²m² + b² = b⁴/y₁². Taking square roots results in b²/y₁ = ± √(a²m² + b²). Hence, the tangent lines are

y = mx + c = mx ± √(a²m² + b²)

as desired.

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