Weekend Paper 2 (September 12): Lines, Circles, and Ellipses | Brief Solutions

$\pm4/3$

$(2,-1)$

$13/10$

$x^2/36+y^2/20=1$

$\arctan(9/2)$

The question is incomplete; omitted.

$1$

$k\in\{-\sqrt2\}\cup(-1,1]$

$\sqrt7/4$

$\sqrt{2+\sqrt3}$

$r\in[3,7]$

$4/5$

C; $m<1/4$ or $m>1$

D

C

C

$3/2<m<4$, from $m+1>4-m>0$.

$2<m<4$. Intersect the previous range with $m^2-4>0$.

Maximum $11/5$, minimum $1/5$. The center-line distance is $6/5$; add or subtract radius $1$.

Maximum $49$, minimum $9$. The fixed point $(-2,2)$ is distance $5$ from the center, so use $(5\pm2)^2$.

$x^2/4+y^2=1$. From $b=1$ and $1-b^2/a^2=3/4$, obtain $a=2$.

The constant is $5$. Set $P=(t,0)$; substitution gives $2x^2-2tx+t^2-4=0$. Then $PA^2+PB^2=\frac54[(x_1-t)^2+(x_2-t)^2]=5$.

$y=1$ and $3x+4y-1=0$. Set $y=k(x+1)+1$ and equate the center-line distance to $1$, giving $k=0,-3/4$.

$4\sqrt2/3$, attained at $t=0$. Since $PM^2=t^2+9$, $AB=2\sqrt{1-1/(t^2+9)}$.

$\sqrt2/2$, attained at $t=0$. Tangent slopes satisfy $8k^2+6tk+t^2-1=0$, so $ST=|k_1-k_2|=\sqrt{t^2+8}/4$.

$2$. Since $CD=5$, the moving radius is $3$ and the endpoints are $(0,0),(0,6)$.

$12/5$, attained at $D=(0,0)$. Put $s=CD\ge4$; the radius is $s-2$ and the tangent is $6(s-2)/(4s-11)$, decreasing in $s$.

Yes: $Q=(\pm2\sqrt3,0)$, with constant angle $60^\circ$. For $Q=(q,0)$ the tangent is $2|q|(s-2)/(q^2+4s-20)$. Proportionality to $s-2$ requires $q^2=12$.