Question

In each of the following find the equation of the hyperbola satisfying the given conditions.:

$\left(i\right)vertices(\pm 2,0),foci(\pm 3,0)$

$\left(ii\right)vertices(0,\pm ,5),foci(0,\pm ,8)$

$\left(iii\right)vertices(0,\pm ,3),foci(0,\pm ,5)$

$\left(iv\right)vertices(\pm 5,0),transverseaxis=8$

$\left(v\right)foci(0,\pm 13),conjugateaxis=24$

$\left(vi\right)foci(\pm 3\sqrt{5}),thelatus-rectum=8$

$\left(vii\right)foci(\pm ,4,0),thelatus-rectum=12$

$\left(viii\right)vertices(0,\pm 6),e=\frac{5}{3}.$

$\left(ix\right)foci(0,\pm \sqrt{1}0),passingthrough(2,3)$

$\left(x\right)foci(0,\pm 12),latus-rectum=36$

Answer 1

$\text{(i) Since,vertics}(\pm 2,0)andfoci(\pm ,0)lieson\text{X-axis,as coefficient of Y-axis is zero.}$

$\text{Hence,equation of hyperbola will be of the form}$

$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1...\left(i\right)$ $\text{Where it is given that}$

$a=2andc=3$

$\because c^2=a^2+b^2$

$\Rightarrow 9=4+b^2$

$\Rightarrow b^2=5$

$Putthevaluesof{a}^2=4and{b}^2=5inEq,\left(i\right),weget$

$\frac{x^2}{4}-\frac{y^2}{5}=1$

$\text{(ii) Since,vertics}(0,\pm 5)andfoci(0,\pm 8)lieson\text{Y-axis,as coefficient of X-axis is zero.}$

$\text{Hence,equation of hyperbola will be of the form}$

$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1...\left(i\right)$ $\text{Where it is given that}(0,\pm a)=(0,\pm 5)andfoci(0,\pm c)=(0,\pm 8)$

$a=5andc=8$

$\because c^2=a^2+b^2\Rightarrow 64=25+b^2$

$\Rightarrow b^2=64-25$ $\Rightarrow b^2=39$

$Put{a}^2=25and{b}^2=39inEq,\left(i\right),weget$

$\frac{y^2}{25}-\frac{x^2}{39}=1$

$\text{(iii) Since,vertics}(0,\pm 3)andfoci(0,\pm 5)lieson\text{Y-axis,as coordinate of X-axis is zero.}$

$\text{Hence,equation of hyperbola will be of the form}$

$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1...\left(i\right)$ $\text{Where it is given that}(0,\pm 3)=(0,\pm a)andfoci(0,\pm 5)=(0,\pm c)$

$a=3andc=5$

$\because c^2=a^2+b^2$

$\Rightarrow b^2=64-25$

$\Rightarrow 25=9+b^2\Rightarrow b^2=16$

$Put{a}^2=9and{b}^2=16inEq,\left(i\right),weget$

$\frac{y^2}{9}-\frac{x^2}{16}=1$

$\left(iv\right)Since,foci(\pm 5,0)\text{lies on X-axis,as coordinate of Y is zero.}$

$\text{Hence,equation of hyperbola wil be of the form}$

$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1...\left(i\right)$

$Whereitisgiventhatfoci(\pm 5,0)=(\pm c,0)\text{and length of transverse axis is 2a=8}$

$c=5and2a=8\Rightarrow a=4$

$\because c^2=a^2+b^2$

$\Rightarrow 25=16+b^2\Rightarrow b^2=9$

$Put{a}^2=16and{b}^2=9inEq.\left(i\right),weget$

$\frac{x^2}{16}-\frac{y^2}{9}=1$

$\left(v\right)Since,foci(0,\pm 13)\text{lies on Y-axis,as coordinate of X is zero.}$

$\text{Hence,equation of hyperbola wil be of the form}$

$\frac{y^2}{a^2}-\frac{x^2}{b^2}=1...\left(i\right)$

$Whereitisgiventhatfoci(0,\pm 13)=(0,\pm c)\text{and conjugaete length 2b=24}$

$c=13andb=12$

$\because c^2=a^2+b^2$

$\Rightarrow 169=a^2+144\Rightarrow a^2=169-144$

$\Rightarrow a^2=25$

$Put{a}^2=25and{b}^2=144inEq.\left(i\right),weget$

$\frac{y^2}{25}-\frac{x^2}{144}=1$

$\left(vi\right)Since,foci(\pm 3\sqrt{5},0)\text{lies on X-axis,as y-coordinate is zero.}$

$\text{Hence,equation of hyperbola wil be of the form}$

$\frac{y^2}{a^2}-\frac{x^2}{b^2}=1...\left(i\right)$

$Whereitisgiventhatfoci(\pm 3\sqrt{5},0)=(\pm c,0)\text{and length oflatusrectum}$

$\frac{2b^2}{a}=8$

$\because b^2=4a$

$\because c^2=a^2+b^2$

$\therefore (3\sqrt{5}{)}^2=a^2+(4a)$

$\Rightarrow 9\times 5=a^2+4a$

$\Rightarrow a^2+4a-45=0$

$\Rightarrow \text{On spliting the middle term}$

$\Rightarrow a^2+9a-5a-45=0$

$\Rightarrow a(a+9)-5(a+9)=0$

$\Rightarrow (a-5)(a+9)=0$

$\Rightarrow a-5=0,a+9=0$

$\Rightarrow a=5,-9$

$\because aca{n}^{\prime }tbenegative.$

$So,a=5\Rightarrow b^2=4\times 5=20$

$Put{a}^2=25and{b}^2=20inEq.\left(i\right),weget$

$\frac{x^2}{25}-\frac{y^2}{20}=1$

$\left(vii\right)Sincefoci(\pm 4,0)\text{lies on X-axis,Hence,equation,of hyperbola will be of the form}$

$\frac{y^2}{a^2}-\frac{x^2}{b^2}=1...\left(i\right)$

$\text{Where it is given that foci}(\pm 4,0)=(\pm c,0)andlengthoflatusrectum,$

\(\frac{2b^2}{a}=12\\)

$\therefore c=4and{b}^2=6a...\left(ii\right)$

$\therefore c^2=a^2+b^2$

$\Rightarrow 16=a^2+6a$

$\Rightarrow a^2+6a-16=0$

$\text{Splitting the middle term}$

$a^2+8a-2a-16=0$

$\Rightarrow a\left(a\right(a+8)-2(a+8)=0$

$\Rightarrow (a-2)(a+8)=0$

$\Rightarrow a-2=0,a+8=0\Rightarrow a=2,-8$

$\because aca{n}^{\prime }tbenegative.$ $So,a=2$

$\Rightarrow b^2=6a=6\times 2=12$

$Put{a}^2=4and{b}^2=12inEq.\left(i\right),weget$

$\frac{x^2}{4}-\frac{y^2}{12}=1$

$\text{(viii)Let the equation of the hyperbola be}$

$-\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

$Thenvertices=(0,\pm b)=(0,\pm 6)$

$\therefore b=6ande=\frac{5}{4}$

$\because e=\sqrt{1+\frac{a^2}{b^2}}\Rightarrow \frac{25}{9}=1+\frac{a^2}{36}$

$\Rightarrow \frac{25-9}{9}=\frac{a^2}{36}\Rightarrow 16=\frac{a^2}{4}+48$

$\text{So,the equation of hyperbola is,}$

$\frac{-x^2}{48}+\frac{y^2}{36}=1\Rightarrow \frac{y^2}{36}-\frac{x^2}{48}=1$

$\because Foci=(0,\pm be)=(0,\pm \frac{5}{3}\times 6)=(0,\pm 10)$

$Since,foci(0,\pm \sqrt{10})\text{lies on Y-axis,as x-coordinate is zero}.$

$\text{Hence,equation of hyperbola will be of the form}$

$\frac{y^2}{a^2}-\frac{x^2}{b^2}=1...\left(i\right)$

$\text{Where it is given that foci}(0,\pm \sqrt{10})=(0,\pm c)\Rightarrow c=\sqrt{10}$ $\text{Also hyperbola passes through the point (1,3)i.e.,pont~will~satisfy~Eq.(i).}$

$\therefore \frac{(3{)}^2}{a^2}-\frac{(2{)}^2}{b^2}=1$

$\Rightarrow \frac{9}{a^2}-\frac{4}{b^2}=1...\left(ii\right)$

$\because c=\sqrt{10}$

$\Rightarrow c^2=\sqrt{10}$

$\Rightarrow a^2+b^2=10(\because c^2=a^2+b^2)$

$\Rightarrow b^2=10-a^2...\left(iii\right)$

$FromEp.\left(ii\right),\frac{9}{a^2}-\frac{4}{10-a^2}=1$

$\Rightarrow \frac{90-9a^2-4a}{a^2(10-a^2)}=1$

$\Rightarrow 90-13a^2=10a^2-a^4$

$\Rightarrow a^4-23a^2+90=0$

$\Rightarrow a^4-18a^2-5a^2+90=0$

$\Rightarrow a^2(a^2-18)-5(a^2-18)=0$

$\Rightarrow a^2=5or{a}^2=18$

$\Rightarrow b^2=10-5=5$ $If{a}^2=18$

$\Rightarrow b^2=10-18=-8$

$\text{Which in not possible.}$

$So,a^2=b^2=5$

$Put{a}^2=b^2=5inEq.\left(i\right)weget$

$\Rightarrow \frac{y^2}{5}-\frac{x^2}{5}=1\Rightarrow y^2-x^2=5$

$\left(x\right)Sincefociare(0,\pm 12),itfollowsthatc=12$ $\text{Length of the latus rectum}=\frac{2b^2}{a}=36$ $or{b}^2=18a$

$Therefore{c}^2=a^2+b^2,gives$

$144=a^2+18a$ $i.e.,a^2+18a-144=0,$

$So,a=-24,6$

$\text{Sine a cannot be negative,we take a=6 and so}{b}^2=108$

$\text{Therefore,the equation of the required}$

$hyperbolais\frac{y^2}{36}-\frac{x^2}{108}=1,$

$i.e.,3y^2-x^2=108$