Hyperbola Questions

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 […]

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

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

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 $\implies$

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 $\therefore$; $a^2$

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