Math Help - differentiation question 2

1. differentiation question 2

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).

2. Hi

First the line $x=4$ is tangent to the ellipse $x^2 + 4y^2 = 16$ and passes through the point (4,6).

If there is another one its equation is $y=mx+p$
It is tangent to the ellipse. Therefore there is only one contact point between the ellipse and the line. In other words the system
$x^2 + 4y^2 = 16$
$y=mx+p$
has one single solution

It means that the equation $x^2 + 4(mx+p)^2 = 16$ or $(1+4m^2)x^2+8mpx+4p^2-16=0$ has one solution. Its discriminant is thus equal to 0.
$(4mp)^2-(1+4m^2)(4p^2-16)=0$
$-p^2+4+16m^2=0$

The line passes through the point (4,6). Therefore $6=4m+p$ or $p=6-4m$.

Then $-(6-4m)^2+4+16m^2=0$ giving $m=\frac{2}{3}$ and $p=\frac{10}{3}$

3. Originally Posted by math123456
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).
Here is the idea. Lets suppose that the point of tangency is (a,b)

Plugging this point in we get our 1st equation

$E_1: a^2+4b^2=16$

Now lets take the derivative of the equation(implicitly)

$2x+8y\frac{dy}{dx}=0 \iff \frac{dy}{dx}=\frac{-x}{4y}$

So now we know the slope is $\frac{dy}{dx}=\frac{-a}{4b}$

We also know that the tangent line passes through (4,6) so we get the slope is

$m=\frac{6-b}{4-a}$ and we know that the slopes have to be equal.... so setting them equal yields our 2nd equation.

$E_2: \frac{6-b}{4-a}=\frac{-a}{4b}$

This is a system of non-linear equations...

Solveing yeilds the solution $\left(\frac{-16}{5},\frac{6}{5}\right)$

So plugging into the equation for the derivative we get

$\frac{dy}{dx}=\frac{\frac{16}{5}}{(4)\frac{6}{5}}= \frac{2}{3}$

$y=\frac{2}{3}x+b \implies 6=\frac{2}{3}(4)+b \iff \frac{10}{3}=b$

So we end up with $y=\frac{2}{3}x+\frac{10}{3}$
Good luck.

TES