conic sections, ellipse

Question

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?

Answer 1

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, $x=0$

Answer 2

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 $$y=\frac{8}{12}(x+8)+3$$