Question

How do you find the standard form of the equation of the hyperbola given the properties vertices (-10, 5), asymptotes $y=\pm \dfrac{1}{2}\left( x-6 \right)+5$?

Answer 1

Assume the equation $\dfrac{{{\left( x-h \right)}^{2}}}{{{a}^{2}}}-\dfrac{{{\left( y-k \right)}^{2}}}{{{b}^{2}}}=1$ as the equation of the hyperbola in its standard form. Here, (h, k) denotes the centre of the hyperbola. Now, consider the general equation of asymptote with a horizontal transverse axis given as: - $y=\pm \dfrac{b}{a}\left( x-h \right)+k$. Compare this equation with the equation of given asymptote and find the values of h, k and also find a relation between ‘a’ and ‘b’. Substitute the given properties vertices (-10, 5) in the equation of hyperbola and find another relation between ‘a’ and ‘b’. Solve the two equations and find the values of ‘a’ and ‘b’. In the end, substitute these values of ‘a’ and ‘b’ in the equation of hyperbola to get the answer.

Complete step-by-step solution:
Here, we have been provided with the properties vertices, (-10, 5) and equation of the asymptotes, $y=\pm \dfrac{1}{2}\left( x-6 \right)+5$, of the hyperbola and we are asked to find the standard form of the equation of this hyperbola.
Now, we know that equation of a hyperbola with centre (h, k) is given by the equation: - $\dfrac{{{\left( x-h \right)}^{2}}}{{{a}^{2}}}-\dfrac{{{\left( y-k \right)}^{2}}}{{{b}^{2}}}=1$. Also, the general equation of the asymptote with a horizontal transverse axis is given as: - $y=\pm \dfrac{b}{a}\left( x-h \right)+k$. So, on comparing it with the given equation of asymptote of the hyperbola $y=\pm \dfrac{1}{2}\left( x-6 \right)+k$, we get,
$\Rightarrow h=6$ and k = 5
Also, we have $\dfrac{b}{a}=\dfrac{1}{2}$ - (1)
So, the equation of hyperbola becomes: -
$\Rightarrow \dfrac{{{\left( x-6 \right)}^{2}}}{{{a}^{2}}}-\dfrac{{{\left( y-5 \right)}^{2}}}{{{b}^{2}}}=1$
Now, it is given that the property vertices of the hyperbola is (-10, 5). Here, property vertices means that the hyperbola passes through this point. That means this point will satisfy the equation of the hyperbola, so we have on substituting this point in the equation of hyperbola,

& \Rightarrow \dfrac{{{\left( -10-6 \right)}^{2}}}{{{a}^{2}}}-\dfrac{{{\left( 5-5 \right)}^{2}}}{{{b}^{2}}}=1 \\\ & \Rightarrow \dfrac{{{\left( -16 \right)}^{2}}}{{{a}^{2}}}=1 \\\ & \Rightarrow {{a}^{2}}=256 \\\ & \Rightarrow a=\pm 16 \\\ \end{aligned}$$ Now, substituting the value of a in equation (1), we get, $$\begin{aligned} & \Rightarrow \dfrac{b}{\pm 16}=\dfrac{1}{2} \\\ & \Rightarrow b=\pm \left( \dfrac{16}{2} \right) \\\ & \Rightarrow b=\pm 8 \\\ \end{aligned}$$ **Finally, substituting the values of ‘a’ and ‘b’ in the equation of hyperbola, we get, $$\Rightarrow \dfrac{{{\left( x-6 \right)}^{2}}}{256}-\dfrac{{{\left( y-5 \right)}^{2}}}{64}=1$$ Hence, the above equation is our answer.** **Note:** One must not consider the general equation of hyperbola as $$\dfrac{{{x}^{2}}}{{{a}^{2}}}-\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$$ because this equation is only applicable when the centre of hyperbola is origin. In such case (h, k) = (0, 0). You must remember the general equation of asymptote of the hyperbola otherwise it will be difficult to solve the above question. Remember the meaning of property vertices. Note that here we have obtained two values of ‘a’ and corresponding to values of ‘b’. Both the values are valid because at the end we have to substitute the squared value of both of them.