Question

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

Answer 1

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

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 formula
We 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.….