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Q.
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 $l=9$
$f(\pm 5,0), v(\pm 4,0), e=\frac{5}{4}$ 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
$5 y^2-9 x^2=36$
5
$f(0, \pm \sqrt{65}), v(0, \pm 4), e=\frac{\sqrt{65}}{4}$
and $l=\frac{49}{2}$
F
$49 y^2-16 x^2=784$
6
$f(\pm 10,0), v(\pm 6,0), e=\frac{5}{3}$ and
$I=\frac{64}{3}$
Conic Sections
Solution:
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 $b^2=9$
$\rightarrow a=4$ and $b=3$
Now, $ c^2=a^2+b^2=16+9=25$
$\rightarrow c=5 (\because c$ 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{2 b^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 $
$ \rightarrow a=3 \text { and } b=3 \sqrt{3} (\because a, b>0)$
$ \because c^2=a^2+b^2=9+27=36 $
$ \rightarrow 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 $
$\text { Latusrectum } =\frac{2 b^2}{a}=\frac{2 \times 27}{3}=18$
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 $9 y^2-4 x^2=36$, divide it by 36 , we get
$\frac{9 y^2}{36}-\frac{4 x^2}{36}=\frac{36}{36} $
$\rightarrow \frac{y^2}{4}-\frac{x^2}{9}=1$
$ \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 $
$ \rightarrow a=2 \text { and } b=3 $
$ \because c^2=a^2+b^2$
$ \rightarrow c^2=4+9=13 $
$ \rightarrow 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) $
$\text { Eccentricity, e } =\frac{c}{a}=\frac{\sqrt{13}}{2}$
$\text { Latusrectum } =\frac{2 b^2}{a}=\frac{2 \times 9}{2}=9$
D. Given, equation is $16 x^2-9 y^2-576$, divide it by 576 , we get
$\frac{16 x^2}{576}-\frac{9 y^2}{576}=\frac{576}{576} \rightarrow \frac{x^2}{36}-\frac{y^2}{64}=1$
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 $
$\rightarrow a=6 \text { and } b=8$
$\because c^2=a^2+b^2=36+64=100$
$\rightarrow c=10$
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{2 b^2}{a}=\frac{2 \times 64}{6}=\frac{64}{3} $
E. Given, equation is $5 y^2-9 x^2-36$, divide it by 36 ,
we get, $\frac{5 y^2}{36}-\frac{9 x^2}{36}=\frac{36}{36} \rightarrow \frac{y^2}{\frac{36}{5}}-\frac{x^2}{4}=1$
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 $b^2=4$
$\rightarrow a=\frac{6}{\sqrt{5}} \text { and } b=2$
$\because c^2=a^2+b^2=\frac{36}{5}+4 $
$\rightarrow c^2=\frac{36+20}{5}=\frac{56}{5} $
$\rightarrow c=\frac{2 \sqrt{14}}{\sqrt{5}}$
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{2 b^2}{a}=\frac{2 \times 4}{\frac{6}{\sqrt{5}}}=\frac{8 \sqrt{5}}{6}=\frac{4 \sqrt{5}}{3}$
F. Given, equation is $49 y^2-16 x^2=784$, divide it by 784 , we get, $ \frac{49 y^2}{784}-\frac{16 x^2}{784}=\frac{784}{784}$
$\rightarrow \frac{y^2}{16}-\frac{x^2}{49}=1$
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 $b^2=49$
$\rightarrow a=4 \text { and } b=7$
$\because c^2=a^2+b^2-16+49=65$
$\rightarrow c=\sqrt{65} (\because c>0)$
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} $
$\text { Latusrectum } =\frac{2 b^2}{a}-\frac{2 \times 49}{4}=\frac{49}{2}$