2.2(1): Standard Equation of an Ellipse | Brief Solutions

$\frac{x^2}{25}+\frac{y^2}{16}=1$ or $\frac{x^2}{16}+\frac{y^2}{25}=1$. Use $a=5,c=3,b^2=16$ and both axis orientations.

$\frac{x^2}{25}+\frac{y^2}{16}=1$, $y\ne0$, since $AB+AC=10$ and the triangle must be nondegenerate.

$\frac{x^2}{9}+\frac{y^2}{3}=1$, from $a^2-b^2=6$ and $3/a^2+2/b^2=1$.

$\frac{x^2}{25}+\frac{y^2}{9}=1$: the distances to $(-4,0),(4,0)$ sum to $10$.

$\frac{x^2}{25}+\frac{y^2}{16}=1$, using $a=5,c=3$.

$\frac{x^2}{32}+\frac{y^2}{36}=1$, using $a=6,c=2$.

$y=0,-4\le x\le4$: equality in the triangle inequality gives the segment.

$x^2+\frac{y^2}{16}=1$, since $a=4,b^2=16-15=1$.

$\frac{x^2}{5}+y^2=1$ or $\frac{x^2}{5}+\frac{y^2}{9}=1$, considering both focal axes.

$m=5$, from $m-4=1$.

$-6<k<4$, $k\ne-1$: the denominators must be positive and unequal.

$m\in(\sqrt2,2)\cup(2,+\infty)$.

$2\sqrt{16-m^2}$,$0<m<4$.

The question is incomplete; omitted.

$\frac{x^2}{12}+\frac{y^2}{16}=1$,$x\ne0$.

Major-axis length $10$ (one possible answer).

D

$\frac{x^2}{100}+\frac{y^2}{64}=1$, with $b=8,c=6$.

$\frac{x^2}{36}+\frac{y^2}{4}=1$; the points give the semiaxes.

$\frac{x^2}{8}+\frac{y^2}{4}=1$ or $\frac{x^2}{4}+\frac{y^2}{8}=1$. Substitute the point and use $a^2-b^2=4$.

$\frac{x^2}{6}+\frac{y^2}{10}=1$. Substitute both points into $Ax^2+By^2=1$.

$\frac{16x^2}{625}+\frac{y^2}{34}=1$, using $a=25/4,c=9/4$.

$18$, since the perimeter is $2a+2c=10+8$.

With $A(-12,0),B(12,0)$, the locus is $x^2/169+y^2/25=1$, $y\ne0$. The centroid divides medians in ratio $2:1$, hence $GA+GB=26$.

Two arcs of $x^2/4+y^2/3=1$ with $-2<x<0$. Use $a+c=2b=4$, $BC>BA$, and exclude the degenerate vertex.

$x^2/4+y^2/3=1$, excluding $(-2,0)$. The focal distances sum to $(r+1)+(3-r)=4$; require $r>0$.

$2\sqrt2$. The maximum area is $bc=1$, so $a^2=b^2+c^2\ge2$, with equality at $b=c=1$.