Question

Find the vertices, eccentricity, foci and asymptotes of the hyperbola x^2/8-y^2/4=1.
Also trace it.
Under what conditions on (lamda) the line x - (lamda)y +2 = 0 will be tangent to this hyperbola?
Explain geometrically.

Answer 1

Answer on Question #45089 – Math - Analytic Geometry

Problem.

Find the vertices, eccentricity, foci and asymptotes of the hyperbola $x^{\wedge}2 / 8 - y^{\wedge}2 / 4 = 1$ .

Also trace it.

Under what conditions on (lamda) the line $x - (\text{lamda})y + 2 = 0$ will be tangent to this hyperbola?

Explain geometrically.

Solution.

The equation of the hyperbola is $\frac{x^2}{8} - \frac{y^2}{4} = 1$ , so $a^2 = 8$ and $b^2 = 4$ or $a = 2\sqrt{2}$ and $b = 2$ . Then $c^2 = a^2 + b^3 = 8 + 4 = 12$ , so $c = 2\sqrt{3}$ , the vertices are $(-2\sqrt{2}, 0)$ and $(2\sqrt{2}, 0)$ $((-a, 0)$ and $(a, 0))$ , the eccentricity is $e = \frac{c}{a} = \frac{2\sqrt{3}}{2\sqrt{2}} = \frac{\sqrt{6}}{2}$ , the foci are $(-2\sqrt{3}, 0)$ and $(2\sqrt{3}, 0)$ $((-c, 0)$ and $(c, 0))$ , the asymptotes are $y = \pm \frac{1}{\sqrt{2}} x$ ( $y = \pm \frac{b}{a} x$ ).

The equation of tangent line to this hyperbola that passes through point $(x_0, y_0)$ is $\frac{xx_0}{8} - \frac{yy_0}{4} = 1$ .

We should find such $\lambda$ that $-\frac{x}{2} + \frac{\lambda y}{2} = 1$ ( $x - \lambda y + 2 = 0$ ).

Then

$$- \frac {1}{2} = \frac {x _ {0}}{8} \text { and } - \frac {\lambda}{2} = \frac {y _ {0}}{4}$$

or

$$x _ {0} = - 4 \text { and } y _ {0} = - 2 \lambda ,$$

but $\frac{x_0^2}{8} - \frac{y_0^2}{4} = 1$ , so $2 - \lambda^2 = 1$ . Therefore $\lambda = \pm 1$ .

Hence there are two tangent lines $x + y + 2 = 0$ (blue line) and $x - y + 2 = 0$ (red line). There are two tangent lines because the hyperbola is symmetric with respect to $x$ -axis.

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