conic sections, ellipse A particle is travelling clockwise on the elliptical orbit given by
$$\displaystyle \frac{x^2}{100} + \frac{y^2}{25} = 1$$
The particle
leaves the orbit at the point $(-8, 3)$ and travels in a straight line tangent to the ellipse.
At which point will the particle cross the $y$-axis?
 A: HINT:
From Article $262$ of this, the equation of the tangent at $P(h,k)$ of $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\text{   is      }\frac{x\cdot h}{a^2}+\frac{y\cdot k}{b^2}=1$$
So, the equation of the tangent here will be $$\frac{x\cdot (-8)}{100}+\frac{y\cdot 3}{25}=1$$
Now to cross the $y$ axis, $x=0$
A: We have $x^2/100+y^2/25=1$ so if we set $F(x,y)=x^2/100+y^2/25-1=0$ then $$y'=m_{\text{tangant} }=\frac{-F_x}{F_y}=\frac{-x}{4y}$$ and so $$m_{(-8,3)}=\frac{8}{12}$$ and the equation of the tangent line as @lab noted is $$y=\frac{8}{12}(x+8)+3$$

A: I like to work with homogeneous coordinates. 
The ellipse is represented by 
$$C = \begin{bmatrix} \frac{1}{100} & 0 & 0 \\
0 & \frac{1}{25} & 0 \\
0 & 0 & -1 \end{bmatrix}$$
the 3×3 matrix such that 
$$ \left. \begin{pmatrix} x & y & 1 \end{pmatrix} \begin{bmatrix} \frac{1}{100} & 0 & 0 \\
0 & \frac{1}{25} & 0 \\
0 & 0 & -1 \end{bmatrix} \begin{pmatrix} x \\ y \\ 1 \end{pmatrix} =0 \right\} \frac{x^2}{100} + \frac{y^2}{25} -1 =0 $$
The tangent point is $P=(-8,3,1)$ (you verify tangency by $P\cdot C P=0$)
The tangent line is $L=C P = (-\frac{2}{25}, \frac{3}{25}, -1)$ with equation
$$ \left. (x,y,1) (-\frac{2}{25}, \frac{3}{25}, -1)=0 \right\} -2x+3y-25=0 $$
You can work out the rest.
