I have two ellipses: $x^2/25+y^2/9=1$ and $x^2/16+y^2/25=1$. I have to find the equations of common tangents.
I understand that $xx_0/25+yy_0/9=1$ could be the equation of common tangents, ($x_0,y_0$) being the point of intersection. We will have 4 points of intersection thus 4 tangent equations.
I tried to find the points of intersection by writing
$x^2/25+y^2/9=x^2/16+y^2/25$
$9|x|=16|y|$
$x=16/9y$ and $x=-16/9y$
By inserting these into one of the ellipses equations, I got
$y=45/sqrt(481)$ and $y=45-/sqrt(481)$
I inserted ($x,y$) instead of ($x_0,y_0$) but this doesn't seem to be the right answer. What am I doing wrong?
[Math] Common tangents of two ellipses
analytic geometryconic sectionsgeometrytangent line
Related Solutions
This probably isn't the kind of answer you're looking for, but it's one I like because it doesn't use calculus.
The ellipse is the image of the unit circle under the linear mapping $T(x,y) = (ax,by)$. The tangent line to the ellipse is the image under $T$ of the tangent line to the circle.
We have $n = T^{-1}(x_0,y_0) = (x_0/a,y_0/b)$, the corresponding point on the unit circle. The equation of the tangent line to the circle at $n$ is $1 = n \cdot (x,y) = xx_0/a + yy_0/b$.
To get the equation of the tangent line to the ellipse at $T(n) = (x_0,y_0)$, we substitute $T^{-1}(x,y) = (x/a,y/b)$ into this equation to obtain $$1 = xx_0/a^2 + yy_0/b^2.$$
Take the reflections of $X$ in both the common tangents and mark them as $X'$ and $X''$. Join $X_1$ and $X_2$ to $X'$ and $X''$. $X_1X' = X_1X'' = 15$ and $X_2X' = X_2X'' = 17$. This can be shown using two properties of ellipse:
- Sum of distances of any point on ellipse to foci is a constant.
- A ray passing through one of the foci bounces back to the other focus after reflecting off the surface of the ellipse. (Can be proven with ease using property 1).
Now $\bigtriangleup X_1X_2X'$ and $\bigtriangleup X_1X_2X''$ are right angled using Pythagoras' theorem. This implies $X'$, $X_1$ and $X''$ are collinear and $X_1X_2$ is the perpendicular bisector of $X'X''$.
Also notice $Y$ has to be the circumcenter of the $\bigtriangleup{XX''X'}$ since $Y$ lies on the perpendicular bisectors of $XX''$ and $XX'$. $\implies X''Y = XY$
Imagine $X_2$ at $(0,0)$, $X_1$ at $(8,0)$. Can be shown that $X$ lies at $(2, \pm\sqrt{45})$. Let $Y$ be at $(8+x, 0)$.
$X''Y^2 = x^2 + 225 = XY^2 = (x+6)^2 + 45 \implies x = 12 \implies XY^2 = 369$.
Best Answer
There exists a line $y = mx + b$
$\frac {x^2}{25} + \frac {(mx+b)^2}{9} = 1\\ (9+ 25m^2) x^2 + 50mbx +25b^2- 225 = 0$
Since the line is tangent
$x = \frac {-50mb \pm \sqrt {(50mb)^2 - 4(9+25m^2)(25b^2 + 225)}}{2(9+25m^2)}$
Since the line is tangent (and not intersecting) the discriminant is $0$
$(50mb)^2 - 4(9+25m^2)(25b^2 - 225) = 0$
Applying the line to the other equation
$(32mb)^2 - 4(25+16m^2)(16b^2 - 400) = 0$
Which simplifies to
$b^2 - 25m^2 - 9 = 0\\ b^2 - 16m^2 + 25$
Respectively
$m^2 = \frac {16}{9}\\ b^2 = \frac {481}{9}$
$y = \pm \frac {4}{3} x \pm \frac {\sqrt {481}}{3} $