Question

Show that the lines \(y = \left( {\frac{b}{a}} \right)x\) and \(y = - \left( {\frac{b}{a}} \right)x\) are slant

asymptotes of the hyperbola \(\left( {\frac{{{x^2}}}{{{a^2}}}} \right) - \left( {\frac{{{y^2}}}{{{b^2}}}} \right) = 1\).

Short answer

It is proved thatthe lines \(y = \left( {\frac{b}{a}} \right)x\) and \(y = - \left( {\frac{b}{a}} \right)x\) are slant

asymptotes of the hyperbola \(\left( {\frac{{{x^2}}}{{{a^2}}}} \right) - \left( {\frac{{{y^2}}}{{{b^2}}}} \right) = 1\).

Step-by-step solution

Sketching a Graph of a function

Formula for slant asymptote:

if \(\mathop {\lim }\limits_{x \to \pm \infty } \left( {f\left( x \right) - y\left( x \right)} \right) = 0\), then \(y\left( x \right) = mx + c\) is the slant asymptote.

Determine slant asymptotes

The given equation is \(\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1\).

Simplifying we get,

\(\begin{array}{c}\frac{{{y^2}}}{{{b^2}}} = \frac{{{x^2}}}{{{a^2}}} - 1\\{y^2} = {b^2}\left( {\frac{{{x^2}}}{{{a^2}}} - 1} \right)\\y = \pm \sqrt {{b^2}\left( {\frac{{{x^2}}}{{{a^2}}} - 1} \right)} \end{array}\)

Now,

\(\begin{array}{c}\mathop {\lim }\limits_{x \to \infty } \left( {y - \frac{b}{a}x} \right) = \mathop {\lim }\limits_{x \to \infty } \left( {\sqrt {{b^2}\left( {\frac{{{x^2}}}{{{a^2}}} - 1} \right)} - \frac{b}{a}x} \right)\\ = \mathop {\lim }\limits_{x \to \infty } \frac{{{b^2}\left( {\frac{{{x^2}}}{{{a^2}}} - 1} \right) - \frac{{{b^2}}}{{{a^2}}}{x^2}}}{{\sqrt {{b^2}\left( {\frac{{{x^2}}}{{{a^2}}} - 1} \right)} + \frac{b}{a}x}}\\ = \mathop {\lim }\limits_{x \to \infty } \frac{{ - {b^2}}}{{\sqrt {{b^2}\left( {\frac{{{x^2}}}{{{a^2}}} - 1} \right)} + \frac{b}{a}x}}\\ = 0\end{array}\)

Hence \(y = \frac{b}{a}x\) is a slant asymptote.

Also we have,

\(\begin{array}{c}\mathop {\lim }\limits_{x \to \infty } \left( {y + \frac{b}{a}x} \right) = \mathop {\lim }\limits_{x \to \infty } \left( { - \sqrt {{b^2}\left( {\frac{{{x^2}}}{{{a^2}}} - 1} \right)} + \frac{b}{a}x} \right)\\ = \mathop {\lim }\limits_{x \to \infty } \frac{{{b^2}\left( {\frac{{{x^2}}}{{{a^2}}} - 1} \right) - \frac{{{b^2}}}{{{a^2}}}{x^2}}}{{\sqrt {{b^2}\left( {\frac{{{x^2}}}{{{a^2}}} - 1} \right)} + \frac{b}{a}x}}\\ = \mathop {\lim }\limits_{x \to \infty } \frac{{ - {b^2}}}{{\sqrt {{b^2}\left( {\frac{{{x^2}}}{{{a^2}}} - 1} \right)} + \frac{b}{a}x}}\\ = 0\end{array}\)

Hence \(y = - \left( {\frac{b}{a}} \right)x\) is a slant asymptote.

It is proved thatthe lines \(y = \left( {\frac{b}{a}} \right)x\) and \(y = - \left( {\frac{b}{a}} \right)x\) are slant

asymptotes of the hyperbola \(\left( {\frac{{{x^2}}}{{{a^2}}}} \right) - \left( {\frac{{{y^2}}}{{{b^2}}}} \right) = 1\).