Derivation of Hyperbolae

Let the coordinates of a point on a hyperbola be (x,y)(x,y), the distance from the centre of the hyperbola to either vertex aa and the distance from the centre of the hyperbola to either focus cc.

The difference of the distances from a point on the hyperbola to the two foci is equivalent to both (x+c)2+y2−(x−c)2+y2\sqrt{(x+c)^2+y^2}-\sqrt{(x-c)^2+y^2} and 2a2a, thus (x+c)2+y2−(x−c)2+y2=2a\sqrt{(x+c)^2+y^2}-\sqrt{(x-c)^2+y^2}=2a. The equation is true for all points (x,y)(x,y) on the hyperbola because the difference is invariant of (x,y)(x,y).

Derivation:

(x+c)2+y2−(x−c)2+y2=2a(x+c)2+y2=2a+(x−c)2+y2(x+c)2+y2=4a2+4a(x−c)2+y2+(x−c)2+y2x2+2cx+c2+y2=4a2+4a(x−c)2+y2+x2−2cx+c2+y22cx=4a2+4a(x−c)2+y2−2cx4cx−4a2=4a(x−c)2+y2cx−a2=a(x−c)2+y2(cx−a2)2=a2(x2−2cx+c2+y2)c2x2−2a2cx+a4=a2x2−2a2cx+a2c2+a2y2c2x2+a4=a2x2+a2c2+a2y2c2x2−a2x2−a2y2=a2c2−a4x2(c2−a2)−a2y2=a2(c2−a2)Let a2+b2=c2⇒b2=c2−a2x2b2−a2y2=a2b2\begin{aligned} \sqrt{(x+c)^2+y^2}-\sqrt{(x-c)^2+y^2}&=2a\\ \sqrt{(x+c)^2+y^2}&=2a+\sqrt{(x-c)^2+y^2}\\ (x+c)^2+y^2&=4a^2+4a\sqrt{(x-c)^2+y^2}+(x-c)^2+y^2\\ x^2+2cx+c^2+y^2&=4a^2+4a\sqrt{(x-c)^2+y^2}+x^2-2cx+c^2+y^2\\ 2cx&=4a^2+4a\sqrt{(x-c)^2+y^2}-2cx\\ 4cx-4a^2&=4a\sqrt{(x-c)^2+y^2}\\ cx-a^2&=a\sqrt{(x-c)^2+y^2}\\ \left(cx-a^2\right)^2&=a^2\left(x^2-2cx+c^2+y^2\right)\\ c^2x^2-2a^2cx+a^4&=a^2x^2-2a^2cx+a^2c^2+a^2y^2\\ c^2x^2+a^4&=a^2x^2+a^2c^2+a^2y^2\\ c^2x^2-a^2x^2-a^2y^2&=a^2c^2-a^4\\ x^2\left(c^2-a^2\right)-a^2y^2&=a^2\left(c^2-a^2\right)\quad\text{Let }a^2+b^2=c^2\Rightarrow b^2=c^2-a^2\\ x^2b^2-a^2y^2&=a^2b^2\\ \end{aligned}

∴x2a2−y2b2=1\therefore\boxed{\frac{x^2}{a^2}-\frac{y^2}{b^2}=1}

#Algebra

Note by Gandoff Tan
1 year, 6 months ago

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