2.2(3): Properties of Ellipses (2) | Brief Solutions

$b=\pm5\sqrt{13}$, from $\Delta=12(325-b^2)=0$.

$-5\sqrt{13}<b<5\sqrt{13}$, since $\Delta>0$.

$|b|>5\sqrt{13}$, since $\Delta<0$.

$b=\pm18$. The squared chord length is $60(325-b^2)/169$.

$y=-x/4$, $|x|<4/\sqrt5$. Subtract the endpoint equations and require the midpoint to lie inside the ellipse.

$x_0=\pm5\sqrt7/4$. Combine $x_0^2+y^2=c^2=16$ with the ellipse.

$-5\sqrt7/4<x_0<5\sqrt7/4$, from $x_0^2+y^2<16$.

Center the axes at the orbit center with the major axis along $x$. Then $a=7783,c=972,b\approx7722$ km, so $x^2/7783^2+y^2/7722^2\approx1$.

Maximum $10$, minimum $2$, using $a\pm c=6\pm4$.

$40$. Subtract $r_1^2+r_2^2=4c^2$ from $(r_1+r_2)^2=4a^2$ to get $r_1r_2=2b^2$.

$m<-1/2$ or $m>1/2$, since $4<4m^2+3$.

$k\in(-\infty,-\sqrt2)\cup(\sqrt2,+\infty)$

$b\in[-\sqrt6,\sqrt6]$

$b=\pm5\sqrt{119}/6$

$y=3x/2,\quad -\sqrt2<x<\sqrt2$

$m\ge1,\quad m\ne5$

$\sqrt2$

B

$(-9/5,1/5)$. Substitute into $x^2/9+y^2=1$ and use the root sum.

The ellipse $x^2/16+y^2/12=1$, obtained from $4[(x-2)^2+y^2]=(x-8)^2$.

In units of $10^8$ km, $a=5.875,c=1.175$, giving $x^2/34.516+y^2/33.135=1$ with denominators rounded to three decimals using the printed data.

$(2x-3)^2+8y^2=1$, since $P=(2x-3,2y)$.

$2\sqrt2-\sqrt3/2$, since $\max(x-y)=\sqrt{3/2}$.

$x^2/4+y^2=1$. The midpoint is $(1,1/2)$ and the endpoints are $(0,1),(2,0)$.

$x^2+y^2/9=1$, since $a=c/e=3,b^2=1$.

Such lines exist. For midpoint $(-1/2,y_0)$, $|y_0|<3\sqrt3/2$ and $k=9/(2y_0)$, hence $\alpha\in(\pi/3,\pi/2)\cup(\pi/2,2\pi/3)$. This assumes a unique intersection with the bisecting line; if only midpoint membership is required, include $\alpha=\pi/2$.