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 \( a^2m^2 + b^2 = k^2 \).

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

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

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

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

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

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