2.2(2): Properties of Ellipses (1) | Brief Solutions

Major axis $10$, minor axis $6$, $e=4/5$; foci $(\pm4,0)$; vertices $(\pm5,0),(0,\pm3)$. Use $x^2/25+y^2/9=1$.

$x^2/2+y^2=1$. Here $c=1$ and $b=c$, hence $a^2=2$.

$\frac{\sqrt2}{2}<e<1$, since $b^2-c^2<0$ implies $2c^2>a^2$.

(1) $x^2/4+y^2/3=1$ or $x^2/3+y^2/4=1$; (2) $S=\sqrt3$. Use $a=2,c=1,b^2=3$ and $S=b^2\tan(\theta/2)$.

Major axis $4$, minor axis $2\sqrt3$, $e=1/2$; foci $(\pm1,0)$; vertices $(\pm2,0),(0,\pm\sqrt3)$.

Major axis $10$, minor axis $4$, $e=\sqrt{21}/5$; foci $(0,\pm\sqrt{21})$; vertices $(\pm2,0),(0,\pm5)$.

(1) $x^2/16+y^2/12=1$; (2) $5x^2+3y^2=30$. The second ellipse in each pair has smaller eccentricity.

$x^2/3+y^2=1$, from $a^2=3b^2$ and $a^2-b^2=2$.

$5$;$6$.

$(0,\pm5)$.

$x^2/9+y^2=1$ or $x^2/9+y^2/81=1$.

$x^2/12+y^2/3=1$.

$(\pm\sqrt2,0)$.

$10$;$2$.

$x^2/9+y^2/16=1$.

$x^2/68+y^2/17=1$ or $x^2/8+y^2/32=1$.

B

(1) $3\sqrt3$; (2) $90^\circ$. Use $S=9\tan(\theta/2)$.

$x^2/16+y^2/7=1$: the distances to $(-3,0),(3,0)$ sum to $8$.

$x^2/36+y^2/16=1$, excluding $(\pm6,0)$, from $y^2/(x^2-36)=-4/9$.

$x^2/9+4y^2/5=1$, since the midpoint is $(x_P,y_P/2)$.

$e=\sqrt5/3$. The midpoint tangency implies $OP=OF=c$ and $OM=b$. Combining with the ellipse equation gives $5b^2=4c^2$, hence $e^2=5/9$.