Find the equations of all tangent lines to the curve $x^2 + 4y^2 = 8$ that pass through the point (−4, 0).

Question

So I stared off by implicitly differentiating $x^2 + 4y^2 = 8$, and I got $\frac{dy}{dx }= \frac{-x}{4y}$.

I then plugged this into the point-slope formula$ (y - y1 = m (x - x1))$ along with my point $(-4,0)$ to get $y = \frac{-x }{ 4y}(x + 4)$.

Now I know I can plug this back into the formula I started with, $x^2 + 4y^2 = 8$, to get $x^2 + 4((-x^2 - 4x) / 4y)^2 = 8$ and solve for $x$ and $y$ to get my answer but this is where I got stuck.

How would I start solving for $X$ or $Y$?

Answer 1

If $(x,y)\neq(-4,0)$, then the slope of the line defined by $(-4,0)$ and by $(x,y)$ is $\frac y{x+4}$. So, you're after the points of the curve such that $\frac y{x+4}=-\frac x{4y}$. This means that you should solve the system$$\left\{\begin{array}{l}\frac y{x+4}=-\frac x{4y}\\x^2+4y^2=8.\end{array}\right.$$That's not hard. The solutions are $(-2,\pm1)$; see the picture below.

Answer 2

We have the tangent line given in the form $$y=mx+n$$ since $$P(-4;0)$$ is situated on the line we get $$y=mx+4m$$ so we have to solve the equation $$x^2+4(mx+4m)^2=8$$ Solve this equation and set the discriminant equal to zero to get $m$. For your work: It is $$1-4m^2=0$$