2.3(3): Properties of Hyperbolas (2) | Brief Solutions

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Two: $|k|<1$; one: impossible; none: $|k|\ge1$. Substitute to obtain $(1-k^2)x^2=1$.

$8\sqrt2/3$. Substitution gives $3x^2-2x-5=0$ with roots $-1,5/3$.

Put the origin at the narrowest cross-section center, with vertical y-axis. Then $a=12$, $b=660/(5+\sqrt{481})\approx24.5$ m, so $x^2/12^2-y^2/24.5^2\approx1$. The top and bottom are approximately at $y=10.2,-44.8$ m.

$k<-1/\sqrt3$ or $k>1/\sqrt3$. Substituting $y=k(x-2)$ gives two positive roots when $3k^2>1$. A vertical chord also exists.

$(x-1)^2-3y^2=1,x\ge2$. Eliminate $k$ from $x=3ky$ and $y=k(x-2)$. The point $(2,0)$ corresponds to the vertical chord.

$x^2/25-y^2/3=1$. Use $a=5$ and substitute the point to get $b^2=3$.

$x^2/16-y^2/9=1$, using $b/a=3/4$ and $a^2+b^2=25$.

$x^2/a^2-y^2/b^2=\lambda,\lambda>0$. Positive scaling preserves the asymptotes.

$|y|\ge a$; symmetric about both axes and the origin; vertices $(0,\pm a)$, foci $(0,\pm c)$ with $c^2=a^2+b^2$; axis lengths $2a,2b$; $e=c/a>1$; asymptotes $y=\pm ax/b$.

For focus $(c,0)$ and asymptote $bx-ay=0$, $d=bc/\sqrt{a^2+b^2}=b$. The vertical case is analogous.

$x-y+1=0$

$(5/4,-3/4)$

$2\sqrt{26}$

$x^2-4y^2=1$

$a\in(0,1)\cup(1,\sqrt2)$. Require a nonzero quadratic coefficient and $\Delta=4a^2(2-a^2)>0$.

$a=17/13$. Put the x-coordinates at $5t,12t$. Vieta gives $17t=60t^2=-2a^2/(1-a^2)$.

$|a|<\sqrt6,\ a\ne\pm\sqrt3$. The equation is $(3-a^2)x^2-2ax-2=0$ with discriminant $24-4a^2>0$.

$a=\pm1$, since $\overrightarrow{OA}\cdot\overrightarrow{OB}=(1-a^2)/(3-a^2)=0$.

For the acute angle, $\tan\theta=4/3$. From $-n/(m+n)=-1/3$, obtain $m=2n$, so the asymptote slopes are $\pm2$.

$-2<k<-\sqrt2$. Require a positive discriminant and positive root sum and product.

Yes, $k=-(6+\sqrt6)/5$. With $F=(\sqrt6/2,0)$, the condition $\overrightarrow{FA}\cdot\overrightarrow{FB}=0$ gives $5k^2+2\sqrt6k-6=0$; retain the root in the required interval.

$k=\sqrt3-1$. The midpoint condition is $k/(2-k^2)=1/2$; the discriminant excludes $-1-\sqrt3$.

$b\in(-\infty,-\sqrt2)\cup(\sqrt2,+\infty)$. The midpoint condition gives $b=k/2+1/k$; two intersections require $0<k^2<2$, yielding the range.

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2.3(3): Properties of Hyperbolas (2) | Brief Solutions

E1Conditions for two, one or no intersections of $y=kx$ with $x^2-y^2=1$.

E2Chord length cut by $y=x+1$ in $x^2-y^2/4=1$.

E3Cooling tower: minimum radius $12$ m, top and bottom radii $13,25$ m, height $55$ m. Find the generating hyperbola.

E4.1A line through the right focus of $x^2/3-y^2=1$ cuts the right branch twice. Find possible slopes.

E4.2Same setting: find the chord-midpoint locus.

C1.1Vertices on the x-axis, separation $10$, passing through $(10,3)$: find the equation.

C1.2Focus $(5,0)$ and asymptote $3x-4y=0$: find the equation.

C2Find all horizontal hyperbolas with asymptotes $y=\pm bx/a$.

C3State the geometric properties of $y^2/a^2-x^2/b^2=1$.

C4Prove that the distance from a focus to an asymptote equals the conjugate semiaxis.

H1Find the chord line with midpoint $(1,2)$ in $x^2-y^2/2=1$.

H2Point on the right branch of $x^2-y^2=1$ at distance $\sqrt2$ from $y=x$.

H3Asymptotes $y=\pm2x/3$, passing through $(9/2,-1)$: find focal distance.

H4$P$ lies on $x^2/4-y^2=1$. Find the midpoint locus of $OP$.

H5Number of intersections of $y=x+3$ and $y^2/9-x|x|/4=1$.

H6Focus $(-\sqrt5,0)$ and midpoint of $PF_1$ at $(0,2)$: choose the hyperbola.

H7.1$x^2/a^2-y^2=1$ meets $x+y=1$ at two distinct points. Find $a>0$.

H7.2Same setting, $P=(0,1)$ and $\overrightarrow{PA}=\frac5{12}\overrightarrow{PB}$. Find $a$.

H8.1Find $a$ for two intersections of $y=ax+1$ and $3x^2-y^2=1$.

H8.2Same setting: the circle with the chord as diameter passes through the origin. Find $a$.

H9The chord $y=x+1$ of $mx^2+ny^2=1$ has midpoint x-coordinate $-1/3$. Find the tangent of the angle between the asymptotes of $y^2/m^2-x^2/n^2=1$.

H10.1$y=kx+1$ cuts the right branch of $2x^2-y^2=1$ twice. Find $k$.

H10.2Same setting: can the circle with the chord as diameter pass through the right focus?

H11.1A nonhorizontal line through $(0,1)$ cuts $x^2-y^2/2=1$ with midpoint on $x=1/2$. Find its slope.

H11.2A nonhorizontal chord through $(1,b)$ has midpoint on $x=1/2$. Find possible $b$.