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

We compare this equation with $x^2/a^2-y^2/b^2=1$

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

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 $x^2/8-y^2/4=0,\,\, y=\pm \frac{1}{\sqrt{2}}x$

We compare equation of tangent to hyperbola $x_0x/a^2-y_0y/b^2=1$ with $x+\lambda y=2$

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