For the ellipse $b^2x^2+a^2y^2=a^2b^2$, show that the equations of the tangent lines of slope $m$ are $y=mx\pm \sqrt{a^2m^2+b^2}.$

Question

I am trying to answer this question from Schaum's Calculus. Can someone please let me know if my answer looks kosher?

If we substitute $y=mx\pm \sqrt{a^2m^2+b^2}$ into the equation $b^2x^2+a^2y^2=a^2b^2,$ we obtain

$$b^2x^2+a^2(mx\pm \sqrt{a^2m^2+b^2})^2=a^2b^2,$$

which can be algebraically manipulated to arrive at

$$mx\pm \sqrt{a^2m^2+b^2}=\pm \sqrt{\frac{a^2b^2-b^2x^2}{a^2}}.$$

Solving the ellipse equation for $y$ yields

$$y=\pm \sqrt{\frac{a^2b^2-b^2x^2}{a^2}}.$$

Thus, I get that

$$y=mx\pm \sqrt{a^2m^2+b^2}.$$

Have I used any tautologies here, or does it look good? Thanks for your help!

Answer 1

Here is a way:

Let the tangent be $y=mx+c$. Put this in the ellipse's equation and then set the discriminant of a quadratic in $x$ equal to zero (Since the tangent and ellipse intersect at only one point).

Another way to go forward would be using the derivative. I think you did a circular reasoning to prove the statement...Nowhere you used the formula of the tangent ...You might as well use $y=\textit{Anything}$ and still arrive at the same result...

Edit:

Putting $y=mx+c$ into the ellipse's equation, we get $$b^2x^2+a^2(m^2x^2+2mxc+c^2)-a^2b^2=0$$ Rearranging for the quadratic in $x$

$$x^2(a^2m^2+b^2)+2mca^2x+c^2a^2-a^2b^2=0$$ Setting the discriminant zero and solving for $c$ we get

$c^2=a^2m^2+b^2$ And we are done.

Note that we are getting two values of $c$ which are both valid as there could be $2$ tangents with a given slope $m$ to a standard ellipse which can be seen from the graph.

Answer 2

Someone on a math tutors' board was able to show how me how to solve this one without resorting to the quadratic equation (which I think is what the text was looking for). I've posted it below. I don't think I would have gotten this in a million years on my own. Genuis answer.

Let $(x_1, y_1)$ 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, $m = - \frac{b^2x_1}{a^2y_1}$. Given y$=mx+c$, then $c = y_1-mx_1$.

$y_1-mx_1 = y_1+\frac{b^2x_1^2}{a^2y_1}$

..............$= \frac{a^2y_1^2 + b^2x_1^2}{a^2y_1}$

..............$= \frac{a^2b^2}{a^2y_1}$ (by the ellipse equation)

..............$= \frac{b^2}{y_1}$

Now

$m^2 = \frac{b^4x_1^2}{a^4y_1^2} = \frac{b^2}{a^2}•\frac{b^2x_1^2}{a^2y_1^2}$

......$= \frac{b^2}{a^2}•\frac{a^2b^2-a^2y_1^2}{a^2y_1^2}$ (by the ellipse equation)

......$= \frac{b^2}{a^2}•\frac{b^2}{y_1^2-1}$

......$= \frac{1}{a^2}•\frac{b^4}{y_1^2-b^2}$

and so $a^2m^2=\frac{b^4}{y_1^2}-b^2$, or, $a^2m^2+b^2=\frac{b^4}{y_1^2}$. Taking square roots results in $\frac{b^2}{y_1}=±\sqrt{a^2m^2+b^2}$. Hence, the tangent lines are

$y=mx+c=mx±\sqrt{a^2m^2+b^2}$

as desired.

Answer 3

Here is another sequence of steps that avoids using the quadratic formula and holds off writing the line equation until near the end. Once we have the slope of the tangent line at a point $ \ (x_0 \ , \ y_0) \ $ by implicit differentiation as $$ b^2·x^2 \ + \ a^2·y^2 \ \ = \ \ a^2·b^2 \ \ \rightarrow \ \ \left. \frac{dy}{dx} \right\vert_{(x_0 \ , \ y_0)} \ \ = \ \ -\frac{b^2·x_0}{a^2·y_0} \ \ = \ \ m \ \ , \ \ y_0 \ \neq \ 0 \ \ , $$ we may write $ \ b^4·x_0^2 \ = \ m^2·a^4·y_0^2 \ \Rightarrow \ a^2·y_0^2 \ = \ \frac{b^4}{m^2·a^2}·x_0^2 \ \ $ (the "trickiest" part of the derivation is "eliminating" the coordinates of the point at the best place in its development). [Incidentally, locating the "vertical tangents" at $ \ y_0 \ = \ 0 \ \Rightarrow \ x_0 \ = \ \pm a \ \ $ and the "horizontal tangents" at $ \ x_0 \ = \ 0 \ \Rightarrow \ y_0 \ = \ \pm b \ \ $ shows that the ellipse is contained with the "box" $ \ [ \ -a \ , \ +a \ ] \ \times \ [ \ -b \ , \ +b \ ] \ \ , $ as we would expect for the given major and minor axes.]

Inserting this into the ellipse equation produces $$ b^2·x_0^2 \ + \ \frac{b^4}{m^2·a^2}·x_0^2 \ \ = \ a^2·b^2 \ \ \Rightarrow \ \ x_0^2 \ \ = \ \ \frac{a^2}{1 \ + \ \frac{b^2}{m^2·a^2}} \ \ = \ \ \frac{m^2·a^4}{m^2·a^2 \ + \ b^2 } $$ $$ \Rightarrow \ \ y_0^2 \ \ = \ \ \frac{b^4}{m^2·a^4}·x_0^2 \ \ = \ \ \frac{b^4}{m^2·a^2 \ + \ b^2 } \ \ . $$

The tangent lines to the ellipse with slope $ \ m \ $ are then $$ y \ - \ y_0 \ \ = \ \ m·(x \ - \ x_0) \ \ \Rightarrow \ \ y \ \ = \ \ mx \ - \ m·x_0 \ + \ y_0 $$ $$ \Rightarrow \ \ y \ \ = \ \ mx \ - \ m·\left(\pm \ \frac{m ·a^2}{\sqrt{m^2·a^2 \ + \ b^2 }} \right) \ + \ \left(\pm \ \frac{b^2}{\sqrt{m^2·a^2 \ + \ b^2 }} \right) $$ (so there are two such tangent lines) $$ = \ \ mx \ - \ \ \frac{\pm \ (m^2 ·a^2 \ + \ b^2)}{\sqrt{m^2·a^2 \ + \ b^2 }} \ \ = \ \ mx \ \mp \ \sqrt{m^2·a^2 \ + \ b^2 } \ \ . $$

Because the ellipse is symmetrical about the origin, the tangent points with identical slope are found at $ \ ( \pm x_0 \ , \ \pm y_0) \ $ and the intercepts of the tangent lines are likewise symmetrically located.