Answer to Question #202772 in Analytic Geometry for tanya

Question

Find the vertices, eccentricity, foci and asymptotes of the hyperbola (x^2/8) − (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

We compare this equation with $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1.$

Eccentricity is $e=\sqrt{1+\dfrac{b^2}{a^2}}=\sqrt{\dfrac{3}{2}}$

The center is $C=(0,0)$

The vertices are $V'=(−a,0)=(−2\sqrt{2},0)$ and $V=(a,0)=(2\sqrt{2},0)$

To find the foci, we need the distance from the center to the foci $c^2=a^2+b^2=12, \,\,c=\pm 2\sqrt{3}$

The foci are $F'=(-c,0)=(2\sqrt{3},0)$ and $F=(c,0)=(-2\sqrt{3},0)$

The asymptotes are $\dfrac{x^2}{8}-\dfrac{y^2}{4}=0,\,\, y=\pm \dfrac{1}{\sqrt{2}}x$

We compare equation of tangent to hyperbola $x_0\cdot\dfrac{x}{a^2}-y_0\cdot\dfrac{y}{b^2}=1$ with $x+\lambda y=2$

We have $x-\dfrac{a^2}{b^2} \dfrac{y_0}{x_0}y=\dfrac{a^2}{x_0}, \,\, x-2\dfrac{y_0}{x_0}y=\dfrac{8}{x_0}$ and $x_0=4$, so $y_0=\pm2$ and $\lambda=-\dfrac{y_0}{2}=\mp 1$