Equations of lines tangent to an ellipse Determine the equations of the lines that are tangent to the ellipse $\displaystyle{\frac{1}{16}x^2 + \frac{1}{4}y^2 = 1}$
and pass through $(4,6)$.
I know one tangent should be $x = 4$ because it goes through $(4,6)$ and is tangent to the ellipse but I don't know how to find the other tangents. Any help is appreciated.
 A: You need a line that passes through the point $(4,6)$ and that touches the ellipse at just one point.  The vertical line does that and you've already found it.  Obviously there is exactly one other tangent line (and if that's not obvious to you, then draw the picture and look at it!).
Nonvertical lines passing through the point $(4,6)$ have equation $y-6=m(x-4)$.
That implies $y=mx-4m+6$, so we can put $mx-4m+6$ in place of $y$:
$$
\frac{x^2}{16} + \frac{(mx-4m+6)^2}{4} = 1.
$$
This is equivalent to
$$
\underbrace{(1+4m)}x^2 + \underbrace{-8m(4m-6)}\ x + \underbrace{4(4m-6)^2 -16} = 0.
$$
This equation is quadratic in $x$.  We therefore want a quadratic equation with exactly one solution.  A quadratic equation $ax^2+bx+c=0$ has exactly one solution precisely if its discriminant $b^2-4ac$ is $0$.  So we have
$$
b^2-4ac = \underbrace{64m^2(4m-6)^2 - 4(1+4m)(4(4m-6)^2-16) = 0}.
$$
Now we only need to solve this last equation for $m$.
A: Try changing the variables, linear transf, x=4X, y=2Y.   This turns the ellipse into the unit circle, and (4,6) goes to (1,3).
Now draw it and use elementary trigonometry to get the slope and the equation, in X and Y. Convert back to x and y.
A: We have that $x^2+4y^2=16$, so using implicit differentiation gives $2x+8yy^{\prime}=0$ and therefore $\;\;\;\displaystyle y ^{\prime}=-\frac{x}{4y}$. 
Since the slope of the line between $(x,y)$ and $(4,6)$ is given by $\frac{y-6}{x-4}$, you have that $\displaystyle -\frac{x}{4y}=\frac{y-6}{x-4}$.
This gives $-x^2+4x=4y^2-24y$, so now you can use this equation and the equation
$x^2+4y^2=16$ to find the other point of tangency.
A: The equation of the tangent line to ellipse at the point $(x_0,y_0)$ is $y-y_0=m(x-x_0)$ where $m$ is the slope of the tangent. This is given by $m=\frac{dy}{dx}|_{x=x_0}.$ (Note that at $x=\pm 4$ this doesn't work, because at such points the tangent is given by $x=\pm 4.$)  Taking derivatives we get $\frac{x_0}{8}+\frac{y_0}{2}\frac{dy}{dx}|_{x=x_0}=0,$ that is, $\frac{dy}{dx}|_{x=x_0}=-\frac{x_0}{4y_0}.$  So the equation of the tangent line is 
$$y-y_0=-\frac{x_0}{4y_0}(x-x_0).$$
If we assume that the point $(4,6)$ belongs to this line then 
$$6-y_0=-\frac{x_0}{4y_0}(4-x_0) \implies 24y_0-4y^2_0=x^2_0-4x_0. $$ Since $(x_0,y_0)$ belongs to the ellipse, we have to solve the system
$$\left\{\begin{array}{rcl} \displaystyle x_0^2+4y^2_0 & = & 16\\ x_0^2+4y_0^2-4x_0-24y_0&=&0\end{array}\right.$$ This system has two solutions. One solution is $(4,0)$ and we know that the tangent at this point pass through $(4,6).$ The other solution is $\left(-\frac{16}{5},\frac{6}{5}\right).$ So this is the other point that satisfies the given condition.
A: Use parametric equation of tangent to ellipse xcosø/a + ysinø/b =1 
Since this satisfies (4,6) put x and y as 4 and 6 and you get a trigonometric equation. Square on both sides and put sin^2ø as 1-cos^2ø you get two values of cos ø. Find corresponding values of sinø and substitute in the original equation of tangent!
Hope it helps. Cheers!
