Question

Match the equation of hyperbola given in Column I with their corresponding foci $(f)$, vertices $(v)$, eccentricity $(\ell )$ and length of latusrectum $(l)$ given in Column II and choose the correct option from the codes given below.Match the equation of hyperbola given in Column I with their corresponding foci $(f)$, vertices $(v)$, eccentricity $(\ell )$ and length of latusrectum $(l)$ given in Column II and choose the correct option from the codes given below.

Column I		Column II
A	$\frac{x^2}{16}-\frac{y^2}{9}=1$	1	$f(0,\pm \sqrt{13}),v(0,\pm 2),e=\frac{\sqrt{13}}{2}$ and $l=9$
B	$\frac{y^2}{9}-\frac{x^2}{27}=1$	2	$f\left(0,\pm \frac{2\sqrt{14}}{\sqrt{5}}\right),v\left(0,\pm \frac{6}{\sqrt{5}}\right)$ $e=\frac{\sqrt{14}}{3}$ and $l=\frac{4\sqrt{5}}{3}$
C	$9y^2-4x^2=36$	3	$f(\pm 5,0),v(\pm 4,0),e=\frac{5}{4}$ and $I=\frac{9}{2}$
D	$6x^2-9y^2=576$	4	$f(0,\pm 6),v(0,\pm 3),e=2$ and $l=18$
E	$5y^2-9x^2=36$	5	$f(0,\pm \sqrt{65}),v(0,\pm 4),e=\frac{\sqrt{65}}{4}$ and $l=\frac{49}{2}$
F	$49y^2-16x^2=784$	6	$f(\pm 10,0),v(\pm 6,0),e=\frac{5}{3}$ and $I=\frac{64}{3}$

• A. A= 4 ,B= 3 ,C= 1 ,D= 6 ,E= 2 ,F= 5
• B. A= 3 ,B=4 ,C=1 ,D= 6 ,E=2 ,F= 5
• C. A=3 ,B= 4 ,C= 6 ,D= 1 ,E=2 ,F= 5
• D. A= 3 ,B= 4 ,C= 1 ,D=6 ,E=5 ,F= 2

Answer 1

Answer: (B)

A. Comparing the given equation $\frac{x^2}{16}-\frac{y^2}{9}=1$ with $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$, we get $a^2=16$ and $b^2=9$
$\to a=4$ and $b=3$
Now, $c^2=a^2+b^2=16+9=25$
$\to c=5(\because c$ must be positive $)$
Here, in hyperbolic equation coefficient of $x^2$ is positive, so transverse axis is along $X$-axis.
$\text{ Foci }-(\pm c,0)=(\pm 5,0)$
$\text{ Vertices }=(\pm a,0)=(\pm 4,0)$
$\text{ Eccentricity, }e=\frac{c}{a}-\frac{5}{4}$
$\text{ usrectum }=\frac{2b^2}{a}=\frac{2\times 9}{4}=\frac{9}{2}$
B. Comparing the given equation
$\frac{y^2}{9}-\frac{x^2}{27}=1\text{ with }\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$
$\text{ we get, }{a}^2=9\text{ and }{b}^2=27$
$\to a=3\text{ and }b=3\sqrt{3}(\because a,b>0)$
$\because c^2=a^2+b^2=9+27=36$
$\to c=6(\because c\text{ must be positive) }$
Here, in hyperbolic equation coefficient of $y^2$ is positive, so major axis is along $Y$-axis.
$\text{ Foci }=(0,\pm c)=(0,\pm 6)$
$\text{ Vertices }=(0,\pm a)=(0,\pm 3)$
$\text{ Eccentricity, }e=\frac{c}{a}=\frac{6}{3}=2$
$\text{ Latusrectum }=\frac{2b^2}{a}=\frac{2\times 27}{3}=18$
Note In hyperbolic equation, if the coefficient of $x^2$ is positive, then its major axis is $X$-axis and if the coefficient of $y^2$ is positive, then its major axis is $y$-axis.
C. Given, equation is $9y^2-4x^2=36$, divide it by 36 , we get
$\frac{9y^2}{36}-\frac{4x^2}{36}=\frac{36}{36}$
$\to \frac{y^2}{4}-\frac{x^2}{9}=1$
$\text{ Now, comparing }\frac{y^2}{4}-\frac{x^2}{9}=1\text{ with }\frac{y^2}{a^2}-\frac{x^2}{b^2}=1\text{, we get }$
$a^2-4\text{ and }{b}^2-9$
$\to a=2\text{ and }b=3$
$\because c^2=a^2+b^2$
$\to c^2=4+9=13$
$\to c=\sqrt{13}(\because c>0)$
Here, in hyperbolic equation coefficient of $y^2$ is positive, so transverse axis is along $Y$-axis.
$\text{ Foci }=(0,\pm c)=(0,\pm \sqrt{13})$
$\text{ Vertices }=(0,\pm a)=(0,\pm 2)$
$\text{ Eccentricity, e }=\frac{c}{a}=\frac{\sqrt{13}}{2}$
$\text{ Latusrectum }=\frac{2b^2}{a}=\frac{2\times 9}{2}=9$
D. Given, equation is $16x^2-9y^2-576$, divide it by 576 , we get
$\frac{16x^2}{576}-\frac{9y^2}{576}=\frac{576}{576}\to \frac{x^2}{36}-\frac{y^2}{64}=1$
Now, comparing $\frac{x^2}{36}-\frac{y^2}{64}=1$ with $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$, we get
$a^2=36\text{ and }{b}^2=64$
$\to a=6\text{ and }b=8$
$\because c^2=a^2+b^2=36+64=100$
$\to c=10$
Here, in hyperbolic equation coefficient of $x^2$ is positive, so transverse axis is along $X$-axis.
$\text{ Foci }=(\pm c,0)=(\pm 10,0)$
$\text{ Vertices }=(\pm a,0)=(\pm 6,0)$
$\text{ Eccentricity, }e=\frac{c}{a}=\frac{10}{6}=\frac{5}{3}$
$\text{ Latusrectum }=\frac{2b^2}{a}=\frac{2\times 64}{6}=\frac{64}{3}$
E. Given, equation is $5y^2-9x^2-36$, divide it by 36 ,
we get, $\frac{5y^2}{36}-\frac{9x^2}{36}=\frac{36}{36}\to \frac{y^2}{\frac{36}{5}}-\frac{x^2}{4}=1$
On cornparing $\frac{y^2}{36}-\frac{x^2}{4}=1$ wilh $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$, we get
$a^2=\frac{36}{5}$ and $b^2=4$
$\to a=\frac{6}{\sqrt{5}}\text{ and }b=2$
$\because c^2=a^2+b^2=\frac{36}{5}+4$
$\to c^2=\frac{36+20}{5}=\frac{56}{5}$
$\to c=\frac{2\sqrt{14}}{\sqrt{5}}$
Here, in hyperbolic equation coefficient of $y^2$ is positive, so transverse axis is along $Y$-axis.
$\text{ Foci }=(0,\pm c)=\left(0,\pm \frac{2\sqrt{14}}{\sqrt{5}}\right)$
$\text{ Vertices }-(0,\pm a)=\left(0,\pm \frac{6}{\sqrt{5}}\right)$
$\text{ Eccentricity, e }=\frac{c}{a}=\frac{\frac{2\sqrt{14}}{\sqrt{5}}}{\frac{6}{\sqrt{5}}}=\frac{\sqrt{14}}{3}$
$\text{ Latusrectum }=\frac{2b^2}{a}=\frac{2\times 4}{\frac{6}{\sqrt{5}}}=\frac{8\sqrt{5}}{6}=\frac{4\sqrt{5}}{3}$
F. Given, equation is $49y^2-16x^2=784$, divide it by 784 , we get, $\frac{49y^2}{784}-\frac{16x^2}{784}=\frac{784}{784}$
$\to \frac{y^2}{16}-\frac{x^2}{49}=1$
On comparing $\frac{y^2}{16}-\frac{x^2}{49}-1$ with $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$, we get $a^2=16$ and $b^2=49$
$\to a=4\text{ and }b=7$
$\because c^2=a^2+b^2-16+49=65$
$\to c=\sqrt{65}(\because c>0)$
Here, in hyperbolic equation coefficient of $y^2$ is positive, so transverse axis is along $Y$-axis.
$\text{ Foci }=(0,\pm c)-(0,\pm \sqrt{65})$
$\text{ Vertices }=(0,\pm a)-(0,\pm 4)$
$\text{ Eccentricity,e }=\frac{c}{a}=\frac{\sqrt{65}}{4}$
$\text{ Latusrectum }=\frac{2b^2}{a}-\frac{2\times 49}{4}=\frac{49}{2}$