# Hyperbola Questions

## What is the Equation of Director Circle of Hyperbola ?

Solution : The locus of the intersection of tangents which are at right angles is known as director circle of the hyperbola. The equation to the director circle is : $$x^2+y^2$$ = $$a^2-b^2$$ If $$b^2$$ < $$a^2$$, this circle is real ; If $$b^2$$ = $$a^2$$ the radius of the circle is zero & it …

## Find the normal to the hyperbola $$x^2\over 16$$ – $$y^2\over 9$$ = 1 whose slope is 1.

Solution : We have, $$x^2\over 16$$ – $$y^2\over 9$$ = 1 Compare given equation with $$x^2\over a^2$$ – $$y^2\over b^2$$ = 1 a = 4 and b = 3 Since the normal to the given hyperbola whose slope is ‘m’, is  y = mx $$\mp$$ $${m(a^2+b^2)}\over \sqrt{a^2 – m^2b^2}$$ Hence, required equation of normal is …

## Angle between asymptotes of hyperbola xy=8 is

Solution : Since given hyperbola xy = 8 is rectangular hyperbola. And eccentricity of rectangular hyperbola is $$\sqrt{2}$$ Angle between asymptotes of hyperbola is $$2sec^{-1}(e)$$ $$\implies$$ $$\theta$$ = $$2sec^{-1}(\sqrt{2})$$ $$\implies$$ $$\theta$$ = $$2sec^{-1}(sec 45)$$ $$\implies$$ $$\theta$$ = 2(45) = 90 Similar Questions Find the normal to the hyperbola $$x^2\over 16$$ – $$y^2\over 9$$ = 1 …

## 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 $$\implies$$ …

## 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 $$\therefore$$; $$a^2$$ …

## 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 Similar Questions Angle between asymptotes of hyperbola xy=8 is Find the …

## 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 Similar Questions Find the equation of the ellipse …