2.3(2): Properties of Hyperbolas (1) | Brief Solutions

Standard form: $x^2/9-y^2/16=1$. Vertices $(\pm3,0)$, foci $(\pm5,0)$, $e=5/3$, asymptotes $y=\pm4x/3$.

$y^2/5-x^2/20=1$. Set $x^2-4y^2=\lambda$ and substitute to get $\lambda=-20$.

$2b=a$. Asymptotes are $y=\pm x/2$ for horizontal foci, or $y=\pm2x$ for vertical foci.

$a=4,b=3,e=5/4$; foci $(\pm5,0)$, vertices $(\pm4,0)$, asymptotes $y=\pm3x/4$.

$a=2,b=\sqrt3,e=\sqrt7/2$; foci $(0,\pm\sqrt7)$, vertices $(0,\pm2)$, asymptotes $y=\pm2x/\sqrt3$.

A

Yes. The points $(\pm3,0),(0,\pm3)$ all lie on $x^2+y^2=9$.

$y^2/18-x^2/18=1$

$(\pm2\sqrt3,0)$;$y=\pm\sqrt2x$

$x^2/9-y^2/16=1$

$2\sqrt{-k}\quad(k<0)$

$x^2/3-y^2/12=1$

A

$x^2-y^2/9=1$ or $y^2/81-x^2/9=1$. Use $b=3$ and consider both focal orientations.

$x^2/3-y^2/5=1$, since $a^2=3,c^2=8$.

$y^2/16-x^2/9=1$. Substitution gives $a^2=16,b^2=9$.

$(r_1-r_2)^2=36$, so $r_1^2+r_2^2=36+64=100=(2c)^2$. Apply the converse of Pythagoras.

$y=\pm\sqrt2x$. The half-chord is $4/a=2c/\sqrt3$, so $a^2(a^2+4)=12$ and $a^2=2$.

The branch nearer $B$ of a hyperbola with foci $A,B$, provided the distance difference is less than $AB$.

Set $A=(-400,0),B=(400,0)$. Since $PA-PB=680$, the equation is $x^2/115600-y^2/44400=1,x\ge340$.

$x^2-y^2=1$. For $C:x^2-y^2/b^2=1$ and $P=(u,v)$, the orthocenter is $(u,-v/b^2)$. Substitution gives $b^4=1$, hence $b=1$.