Find the equations of the lines that are tangents to the elipse
x^2 + 4y^2 = 16 and that also passes through the point (4,6).
Hi
First the line $\displaystyle x=4$ is tangent to the ellipse $\displaystyle x^2 + 4y^2 = 16$ and passes through the point (4,6).
If there is another one its equation is $\displaystyle y=mx+p$
It is tangent to the ellipse. Therefore there is only one contact point between the ellipse and the line. In other words the system
$\displaystyle x^2 + 4y^2 = 16$
$\displaystyle y=mx+p$
has one single solution
It means that the equation $\displaystyle x^2 + 4(mx+p)^2 = 16$ or $\displaystyle (1+4m^2)x^2+8mpx+4p^2-16=0$ has one solution. Its discriminant is thus equal to 0.
$\displaystyle (4mp)^2-(1+4m^2)(4p^2-16)=0$
$\displaystyle -p^2+4+16m^2=0$
The line passes through the point (4,6). Therefore $\displaystyle 6=4m+p$ or $\displaystyle p=6-4m$.
Then $\displaystyle -(6-4m)^2+4+16m^2=0$ giving $\displaystyle m=\frac{2}{3}$ and $\displaystyle p=\frac{10}{3}$
Here is the idea. Lets suppose that the point of tangency is (a,b)
Plugging this point in we get our 1st equation
$\displaystyle E_1: a^2+4b^2=16$
Now lets take the derivative of the equation(implicitly)
$\displaystyle 2x+8y\frac{dy}{dx}=0 \iff \frac{dy}{dx}=\frac{-x}{4y}$
So now we know the slope is $\displaystyle \frac{dy}{dx}=\frac{-a}{4b}$
We also know that the tangent line passes through (4,6) so we get the slope is
$\displaystyle m=\frac{6-b}{4-a}$ and we know that the slopes have to be equal.... so setting them equal yields our 2nd equation.
$\displaystyle E_2: \frac{6-b}{4-a}=\frac{-a}{4b}$
This is a system of non-linear equations...
Solveing yeilds the solution $\displaystyle \left(\frac{-16}{5},\frac{6}{5}\right)$
So plugging into the equation for the derivative we get
$\displaystyle \frac{dy}{dx}=\frac{\frac{16}{5}}{(4)\frac{6}{5}}= \frac{2}{3}$
$\displaystyle y=\frac{2}{3}x+b \implies 6=\frac{2}{3}(4)+b \iff \frac{10}{3}=b$
So we end up with $\displaystyle y=\frac{2}{3}x+\frac{10}{3}$
Good luck.
TES