How do you find an equation of hyperbola with given endpoints of the transverse axis: \[y = \dfrac{3}{{10}}x\] Asymptote: \[y = \dfrac{3}{{10}}x\] ?

Question

How do you find an equation of hyperbola with given endpoints of the transverse axis:  $$y = \dfrac{3}{{10}}x$$  Asymptote:  $$y = \dfrac{3}{{10}}x$$ ?

Vedantu Content Team · Accepted Answer

Hint: We need to find the equation of hyperbola with the given points foci, $$F(0, \pm a)$$ of the transverse axis and asymptote equation of the transverse axis.The equation of asymptote is  $$y = \dfrac{a}{b}x$$  ………………. $$(A)$$  The equation of hyperbola is  $$\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1$$  ………………… $$(B)$$ To plot a graph by the given points mentioned below\n    \n    \n    \n    \n  Complete step-by-step answer:Given,Focus,  $$F = (0, \pm 6)$$ Where,  $$a = 6$$ .The given asymptote equation, we have $$y = \dfrac{3}{{10}}x$$  …………… $$(1)$$ By substitute the equation  $$(A)$$  in  $$(1)$$ , we get $$\dfrac{a}{b}x = \dfrac{3}{{10}}x$$ By remove $$x$$  on both sides, we get $$\dfrac{a}{b} = \dfrac{3}{{10}}$$ By substitute the value,  $$a = 6$$  $$\dfrac{6}{b} = \dfrac{3}{{10}}$$  By simplify the above to find the value, $$  6 \times 10 = 3b   \\  b = \dfrac{{60}}{3} = 20 \; $$ Now, we get $$b = 20$$ To find the equation of hyperbola by substitute the value to the formulaWe know that, The equation of hyperbola is  $$\dfrac{{{x^2}}}{{{a^2}}} - \dfrac{{{y^2}}}{{{b^2}}} = 1$$ Here, we have the value  $$a = 6,b = 20$$ To substitute the values, we get $$\dfrac{{{x^2}}}{{{6^2}}} - \dfrac{{{y^2}}}{{{{20}^2}}} = 1$$ By simplify the square of denominator, we get $$\dfrac{{{x^2}}}{{36}} - \dfrac{{{y^2}}}{{400}} = 1$$  Hence, the equation of hyperbola is  $$\dfrac{{{x^2}}}{{36}} - \dfrac{{{y^2}}}{{400}} = 1$$ So, the correct answer is “$$\dfrac{{{x^2}}}{{36}} - \dfrac{{{y^2}}}{{400}} = 1$$ ”.Note: We need to find the equation of hyperbola with the given focus and equation of asymptote of the transverse axis. To solve the equation of hyperbola by finding the value of  $$a$$ and $$b$$ by the asymptote equation and the focus value,  $$F(0, \pm a)$$ . We should remember the equation of parabola, hyperbola and asymptote to solve the similar problem with different values like  $$y = \dfrac{1}{2}x + 4$$ etc.….