Answer to Question #203629 in Analytic Geometry for Agar

Question

Find the vertices, eccentricity, foci and asymptotes of the hyperbola (x²/8) − (y² /4) = 1. Also trace it. Under what conditions on λ the line x+λy = 2 will be tangent to this hyperbola? Explain geometrically.

Answer 1

$$\dfrac{x^2}{8}-\dfrac{y^2}{4}=1$$

$c^2=a^2+b^2=8+4=12$

Vertices: $(-a, 0), (a, 0)$

Vertices: $(-2\sqrt{2}, 0), (2\sqrt{2}, 0)$

Eccentricity $\dfrac{c}{a}=\dfrac{2\sqrt{3}}{2\sqrt{2}}=\sqrt{\dfrac{3}{2}}=\dfrac{\sqrt{6}}{2}>1$

Foci: $(-c, 0), (c, 0)$

Foci: $(-2\sqrt{3}, 0), (2\sqrt{3}, 0)$

$$\dfrac{(-x)^2}{8}-\dfrac{y^2}{4}=\dfrac{x^2}{8}-\dfrac{y^2}{4}$$

The curve is symmetrical about the $y$ -axis.

$$\dfrac{x^2}{8}-\dfrac{(-y)^2}{4}=\dfrac{x^2}{8}-\dfrac{y^2}{4}$$

The curve is symmetrical about the $x$ -axis.

The curve is symmetrical in opposite quadrants.

The curve is not symmetrical about the line $y=x.$

The origin does not lie on the curve.

$y$ - axis: $x=0=>-\dfrac{y^2}{4}=1.$ There is no solution.

There is no $y$ - intercept.

$x$ - axis: $y=0=>\dfrac{x^2}{8}=1, x=\pm2\sqrt{2}.$

$y$ - intercepts: $(-2\sqrt{2}, 0), (2\sqrt{2}, 0).$

Center $: (0, 0)$

First asymptote: $y=-\dfrac{b}{a}x=-\dfrac{\sqrt{2}}{2}x$

Second asymptote: $y=\dfrac{b}{a}x=\dfrac{\sqrt{2}}{2}x$

First directrix: $x=-\dfrac{4\sqrt{3}}{3}$

Second directrix: $x=\dfrac{4\sqrt{3}}{3}$

$$\dfrac{x^2}{8}-\dfrac{y^2}{4}=1$$

If $\lambda =0,$ then the vertical line $x=2$ is not tangent to this hyperbola, since the vertices are $(-2\sqrt{2}, 0), (2\sqrt{2}, 0).$

The equation of a line is represented by

$$y=-\dfrac{1}{\lambda}x+\dfrac{2}{\lambda}, \lambda\not=0$$

The slope of the tangent at the point $(x_1, y_1)$ can be obtained by differentiating the equation of the hyperbola.

$$\dfrac{2x}{8}-\dfrac{2yy'}{4}=0$$

$$x_1-2y_1y'|_{(x_1, y_1)}=0$$

$$y'|_{(x_1, y_1)}=\dfrac{x_1}{2y_1}$$

The equation of the tangent line is

$$y-y_1=\dfrac{x_1}{2y_1}(x-x_1)$$

$$y=\dfrac{x_1}{2y_1}x-\dfrac{x_1^2}{2y_1}+y_1$$

$$y=\dfrac{x_1}{2y_1}x-\dfrac{x_1^2-2y_1^2}{2y_1}$$

$$\dfrac{x_1^2}{8}-\dfrac{y_1^2}{4}=1$$

$$x_1^2-2y_1^2=8$$

The equation of the tangent line is

$$y=\dfrac{x_1}{2y_1}x-\dfrac{8}{2y_1}$$

If $\dfrac{2}{\lambda}=-\dfrac{8}{2y_1},$ then $\lambda=-\dfrac{y_1}{4}.$

$$\dfrac{x_1}{2y_1}=-\dfrac{4}{y_1}$$

$$x_1=-8$$

$$(-8)^2-2y_1^2=8$$

$$y_1=\pm2\sqrt{7}$$

$$Point (-8, -2\sqrt{7}), \lambda=\dfrac{\sqrt{7}}{4}$$

$$Point (-8, 2\sqrt{7}), \lambda=-\dfrac{\sqrt{7}}{4}$$