Hyperbola Examples

Example 1 : If the foci of a hyperbola are foci of the ellipse $x^2\over 25$ + $y^2\over 9$ = 1. If the eccentricity of the hyperbola be 2, then its equation is :

Solution : For ellipse e = $4\over 5$, so foci = ($\pm$4, 0)

for hyperbola e = 2, so a = $ae\over e$ = $4\over 2$ = 2, b = $2\sqrt{4-1}$ = 2$\sqrt{3}$

Hence the equation of the hyperbola is $x^2\over 4$ – $y^2\over 12$ = 1

---

Example 2 : The eccentricity of the conjugate hyperbola to the hyperbola $x^2-3y^2$ = 1 is-

Solution : Equation of the conjugate hyperbola to the hyperbola $x^2-3y^2$ = 1 is

$-x^2-3y^2$ = 1 $\implies$ $-x^2\over 1$ + $y^2\over {1/3}$ = 1

Here $a^2$ = 1, $b^2$ = $1\over 3$

$\therefore$   eccentricity e = $\sqrt{1 + a^2/b^2}$ = $\sqrt{1+3}$ = 2

---

Example 3 : Find the equation of the tangent to the hyperbola $x^2 – 4y^2$ = 36 which is perpendicular to the line x – y + 4 = 0

Solution : Let m be the slope of the tangent, since the tangent is perpendicular to the line x – y = 0

$\therefore$   m$\times$1 = -1 $\implies$ m = -1

Since $x^2-4y^2$ = 36 or $x^2\over 36$ – $y^2\over 9$ = 1

Comparing this with $x^2\over a^2$ – $y^2\over b^2$ = 1

$\therefore$   $a^2$ = 36 and $b^2$ = 9

So the equation of the tangent are y = -1x $\pm$ $\sqrt{36\times {-1}^2 – 9}$

$\implies$ y = x $\pm$ $\sqrt{27}$ $\implies$ x + y $\pm$ 3$\sqrt{3}$ = 0

---

Example 4 : Find the asymptotes of the hyperbola $2x^2 + 5xy + 2y^2 + 4x + 5y$ = 0. Find also the general equation of all the hyperbolas having the same set of asymptotes.

Solution : Let $2x^2 + 5xy + 2y^2 + 4x + 5y + k$ = 0 be asymptotes. This will represent two straight line

so $abc + 2fgh – af^2 – bg^2 – ch^2$ = 0 $\implies$ 4k + 25 – $25\over 2$ – 8 – $25\over 4$k = 0

$\implies$ k = 2

$\implies$ $2x^2 + 5xy + 2y^2 + 4x + 5y + 2$ = 0 are asymptotes

$\implies$ (2x+y+2) = 0 and (x+2y+1) = 0 are asymptotes

and   $2x^2 + 5xy + 2y^2 + 4x + 5y + c$ = 0 is general equation of hyperbola.

Practice these given hyperbola examples to test your knowledge on concepts of hyperbola.