# Ellipse Discriminant Eccentricity

Discriminant can be used to ellipses for identifying the status of their intersections in conjunction with eccentricity.

The eccentricity of $\displaystyle \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is $\displaystyle e^2 = 1-\frac{b^2}{a^2}$.

### Worked Example of Ellipse Discriminant Eccentricity

(a)    The tangent $\ell$ has equation $y=mx+k$. Show that $a^2m^2 + b^2 = k^2$.

\begin{aligned} \displaystyle \require{AMSsymbols} \require{color} \frac{x^2}{a^2} + \frac{(mx+k)^2}{b^2} &= 1 &\color{red} \text{substitute } y=mx+k \text{ into } \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \\ \frac{x^2}{a^2} + \frac{m^2x^2 + 2mkx + k^2}{b^2} &= 1 &\color{red} \text{expand} \\ b^2x^2 + a^2m^2x^2 + 2a^2mkx + a^2k^2 &= a^2 b^2 &\color{red} \text{multiply both sides by } a^2b^2 \\ (b^2 + a^2m^2)x^2 + 2a^2mkx + (a^2k^2-a^2 b^2) &= 0 &\color{red} \text{rearrange in terms of } x \\ (2a^2mk)^2-4 \times (b^2 + a^2m^2) \times (a^2k^2-a^2 b^2) &= 0 &\color{red} \triangle = 0 \text{ for a tangent} \\ 4a^4m^2k^2-4(a^2b^2k^2-a^2b^4 + a^4m^2k^2-a^4b^2m^2) &= 0 \\ a^4m^2k^2-a^2b^2k^2 + a^2b^4-a^4m^2k^2 + a^4b^2m^2 &= 0 \\ -a^2b^2k^2 + a^2b^4 + a^4b^2m^2 &= 0 \\ -b^2k^2 + b^4 + a^2b^2m^2 &= 0 \\ -k^2 + b^2 + a^2m^2 &= 0 \\ \therefore a^2m^2 + b^2 &= k^2 \end{aligned}

(b)    Show that the shortest distance from $S$ to $\ell$ is $\displaystyle SQ=\frac{|mae + k|}{\sqrt{1+m^2}}$.

\begin{aligned} \displaystyle \require{AMSsymbols} \require{color} mx-y+k &= 0 &\color{red} \text{equation of } \ell \text{in general form} \\ SQ &= \frac{|m \times ae-0 + k|}{\sqrt{1+m^2}} &\color{red} \text{perpendicular distance from } S(ae,0) \\ \therefore SQ &= \frac{|mae + k|}{\sqrt{1+m^2}} \end{aligned}

(c)    Prove that $SQ \times S’Q’ = b^2$.

\begin{aligned} \displaystyle \require{AMSsymbols} \require{color} S’Q’ &= \frac{|m \times -ae-0 + k|}{\sqrt{1+m^2}} &\color{red} \text{perpendicular distance from } S'(-ae,0) \\ S’Q’ &= \frac{|-mae + k|}{\sqrt{1+m^2}} \\ SQ \times \ S’Q’ &= \frac{|mae + k|}{\sqrt{1+m^2}} \times \frac{|-mae + k|}{\sqrt{1+m^2}} \\ &= \frac{|(mae + k)(-mae+k)|}{1+m^2} \\ &= \frac{|-m^2a^2e^2 + k^2|}{1+m^2} \\ &= \frac{|-m^2a^2e^2 + a^2m^2 + b^2|}{1+m^2} \\ &= \frac{|-m^2(1-b^2) + a^2m^2 + b^2|}{1+m^2} &\color{red} a^2e^2=1-b^2 \\ &= \frac{|-m^2a^2 + m^2b^2 + a^2m^2 + b^2|}{1+m^2} \\ &= \frac{|m^2b^2 + b^2|}{1+m^2} \\ &= \frac{m^2b^2 + b^2}{1+m^2} &\color{red} m^2b^2 + b^2 \ge 0 \\ &= \frac{(m^2 + 1)b^2}{1+m^2} \\ &= b^2 \end{aligned}