Exercise 10.4 (Revised) - Chapter 11 - Conic Sections - Ncert Solutions class 11 - Maths
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Chapter 10: Conic Sections Class 11 NCERT Solutions - Maths
In each of the Exercises 1 to 6, find the coordinates of the foci, and the vertices, the eccentricity and the length of the latus rectum of the following hyperbolas.
Ex 10.4 Question 1.
$\frac{x^2}{16}-\frac{y^2}{9}=1$
Answer.
Given: Equation of hyperbola $\frac{x^2}{16}-\frac{y^2}{9}=1$ in the form $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
Since, the foci and vertices of the hyperbola lies on $x$-axis.
$
\begin{aligned}
& \therefore a^2=16 \\
& \Rightarrow a=4 \text { and } b^2=9 \\
& \Rightarrow b=3 \\
& \therefore c^2=a^2+b^2 \\
& \Rightarrow c^2=16+9=25 \\
& \Rightarrow c=5
\end{aligned}
$
$\therefore$ Coordinates of foci are $( \pm c, 0)$
$
\Rightarrow( \pm 5,0)
$
Coordinates of vertices are $( \pm a, 0)$
$
\Rightarrow( \pm 4,0)
$$
Eccentricity $(e)=\frac{c}{a}=\frac{5}{4}$
Length of latus rectum $=\frac{2 b^2}{a}=\frac{2 \times 9}{4}=\frac{9}{2}$ units
Ex 10.4 Question 2.
$\frac{y^2}{9}-\frac{x^2}{27}=1$
Answer.
Given: Equation of hyperbola $\frac{y^2}{9}-\frac{x^2}{27}=1$ is in the form $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$
Since, the foci and vertices of the hyperbola lies on $y$-axis.
$
\begin{aligned}
& \therefore a^2=9 \\
& \Rightarrow a=3 \text { and } b^2=27 \\
& \Rightarrow b=3 \sqrt{3} \\
& \therefore c^2=a^2+b^2 \\
& \Rightarrow c^2=9+27=36 \\
& \Rightarrow c=6
\end{aligned}
$
$\therefore$ Coordinates of foci are $(0, \pm c)$
$
\Rightarrow(0, \pm 6)
$
Coordinates of vertices are $(0, \pm a)$
$
\Rightarrow(0, \pm 3)
$
Eccentricity $(e)=\frac{c}{a}=\frac{6}{3}=2$
Length of latus rectum $=\frac{2 b^2}{a}=\frac{2 \times 27}{3}=18$ units
Ex 10.4 Question3.
$9 y^2-4 x^2=36$
Answer.
Given: Equation of hyperbola $9 y^2-4 x^2=36$
$\Rightarrow \frac{9 y^2}{36}-\frac{4 x^2}{36}=1$ (divide both sides of equation by 36 )
$\Rightarrow \frac{y^2}{4}-\frac{x^2}{9}=1$ in the form $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$
Since, the foci and vertices of the hyperbola lies on $y$-axis.
$
\begin{aligned}
& \therefore a^2=4 \\
& \Rightarrow a=2 \text { and } b^2=9 \\
& \Rightarrow b=3 \\
& \therefore c^2=a^2+b^2 \\
& \Rightarrow c^2=4+9=13 \\
& \Rightarrow c=\sqrt{13}
\end{aligned}
$
$\therefore$ Coordinates of foci are $(0, \pm c)$
$
\Rightarrow(0, \pm \sqrt{13})
$
Coordinates of vertices are $(0, \pm a)$
$
\Rightarrow(0, \pm 2)
$
Eccentricity $(e)=\frac{c}{a}=\frac{\sqrt{13}}{2}$
Length of latus rectum $=\frac{2 b^2}{a}=\frac{2 \times 9}{2}=9$ units
Ex 10.4 Question4.
$16 x^2-9 y^2=576$
Answer.
Given: Equation of hyperbola $16 x^2-9 y^2=576$
$\Rightarrow \frac{16 x^2}{576}-\frac{9 y^2}{576}=1$ (divide both sides of equation by 576 )
$\Rightarrow \frac{x^2}{36}-\frac{y^2}{64}=1$ in the form $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
Since, the foci and vertices of the hyperbola lies on $x$-axis.
$
\begin{aligned}
& \therefore a^2=36 \\
& \Rightarrow a=6 \text { and } b^2=64 \\
& \Rightarrow b=8 \\
& \therefore c^2=a^2+b^2 \\
& \Rightarrow c^2=36+64=100 \\
& \Rightarrow c=10
\end{aligned}
$
$\therefore$ Coordinates of foci are $( \pm c, 0)$
$
\Rightarrow( \pm 10,0)
$
Coordinates of vertices are $( \pm a, 0)$
$
\Rightarrow( \pm 6,0)
$
Eccentricity $(e)=\frac{c}{a}=\frac{10}{6}=\frac{5}{3}$
Length of latus rectum $=\frac{2 b^2}{a}=\frac{2 \times 64}{6}=\frac{64}{3}$ units
Ex 10.4 Question 5.
$5 y^2-9 x^2=36$
Answer.
Given: Equation of hyperbola $5 y^2-9 x^2=36$
$\Rightarrow \frac{5 y^2}{36}-\frac{9 x^2}{36}=1$ (divide both sides of eqaution by 36 )
$\Rightarrow \frac{y^2}{36 / 5}-\frac{x^2}{4}=1$ in the form $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$
Since, the foci and vertices of the hyperbola lies on $y$-axis.
$
\begin{aligned}
& \therefore a^2=\frac{36}{5} \\
& \Rightarrow a=\frac{6}{\sqrt{5}} \text { and } b^2=4 \\
& \Rightarrow b=2 \\
& \therefore c^2=a^2+b^2
\end{aligned}
$
$
\begin{aligned}
& \Rightarrow c^2=\frac{36}{5}+4=\frac{56}{5} \\
& \Rightarrow c=\sqrt{\frac{56}{5}}
\end{aligned}
$
$\therefore$ Coordinates of foci are $(0, \pm c)$
$
\Rightarrow\left(0, \pm \sqrt{\frac{56}{5}}\right)
$
Coordinates of vertices are $(0, \pm a)$
$
\Rightarrow\left(0, \pm \frac{6}{\sqrt{5}}\right)
$
Eccentricity $(e)=\frac{c}{a}=\frac{\sqrt{\frac{56}{5}}}{\frac{6}{\sqrt{5}}}=\frac{\sqrt{56}}{6}=\frac{2 \sqrt{14}}{6}=\frac{\sqrt{14}}{3}$
Length of latus rectum $=\frac{2 b^2}{a}=\frac{2 \times 4}{6 / \sqrt{5}}=\frac{2 \times 4 \times \sqrt{5}}{6}=\frac{4 \sqrt{5}}{3}$ units
Ex 10.4 Question 6.
$49 y^2-16 x^2=784$
Answer.
Given: Equation of hyperbola $49 y^2-16 x^2=784$
$\Rightarrow \frac{49 y^2}{784}-\frac{16 x^2}{784}=1$ (divide both sides of eqaution by 784 )
$\Rightarrow \frac{y^2}{16}-\frac{x^2}{49}=1$ in the form $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$
Since, the foci and vertices of the hyperbola lies on $y$-axis.
$
\therefore a^2=16
$
$\begin{aligned}
& \Rightarrow a=4 \text { and } b^2=49 \\
& \Rightarrow b=7 \\
& \therefore c^2=a^2+b^2 \\
& \Rightarrow c^2=16+49=65 \\
& \Rightarrow c=\sqrt{65}
\end{aligned}$
$\therefore$ Coordinates of foci are $(0, \pm c)$
$
\Rightarrow(0, \pm \sqrt{65})
$
Coordinates of vertices are $(0, \pm a)$
$
\Rightarrow(0, \pm 4)
$
Eccentricity $(e)=\frac{c}{a}=\frac{\sqrt{65}}{4}$
Length of latus rectum $=\frac{2 b^2}{a}=\frac{2 \times 49}{4}=\frac{49}{2}$ units
In each of the Exercises 7 to 15, find the equation of the hyperbola satisfying the given conditions.
Ex 10.4 Question 7.
Vertices $( \pm 2,0)$, foci $( \pm 3,0)$
Answer.
Given: The vertices $( \pm 2,0)$ lie on $x$-axis.
$\therefore$ The equation of hyperbola in standard form is $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
$
\therefore( \pm a, 0)=( \pm 2,0)
$
$\begin{aligned}
& \Rightarrow a=2 \\
& \text { Foci }( \pm a e, 0)=( \pm 3,0) \\
& \Rightarrow a e=3 \\
& \Rightarrow e=\frac{3}{a}=\frac{3}{2}
\end{aligned}$
Also $b=a \sqrt{e^2-1}=2 \sqrt{\frac{9}{4}-1}=2 \times \frac{\sqrt{5}}{2}=\sqrt{5}$
Therefore, the required equation of hyperbola is $\frac{x^2}{(2)^2}-\frac{y^2}{(\sqrt{5})^2}=1$
$
\Rightarrow \frac{x^2}{4}-\frac{y^2}{5}=1
$
Ex 10.4 Question 8.
Vertices $(0, \pm 5)$, foci $(0, \pm 8)$
Answer.
Given: The vertices $(0, \pm 5)$ lie on $y$-axis.
$\therefore$ The equation of hyperbola in standard form is $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$
$
\begin{aligned}
& \therefore(0, \pm a)=(0, \pm 5) \\
& \Rightarrow a=5
\end{aligned}
$
Foci $(0, \pm a e)=(0, \pm 8)$
$\Rightarrow a e=8$ and $\mathrm{a}=5$
$
\Rightarrow \frac{8}{5}
$
Also $b=a \sqrt{e^2-1}=5 \sqrt{\frac{64}{25}-1}=5 \times \frac{\sqrt{39}}{5}=\sqrt{39}$ Therefore, the required equation of hyperbola is $\frac{y^2}{(5)^2}-\frac{x^2}{(\sqrt{39})^2}=1$ $\Rightarrow \frac{y^2}{25}-\frac{x^2}{39}=1$
Ex 10.4 Question 9.
Vertices $(0, \pm 3)$, foci $(0, \pm 5)$
Answer.
Given: The vertices $(0, \pm 3)$ lie on $y$-axis.
$\therefore$ The equation of hyperbola in standard form is $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$
$
\begin{aligned}
& \therefore(0, \pm a)=(0, \pm 3) \\
& \Rightarrow a=3
\end{aligned}
$
Foci $(0, \pm a e)=(0, \pm 5)$
$
\begin{aligned}
& \Rightarrow a e=5 \\
& \Rightarrow e=\frac{5}{a}=\frac{5}{3}
\end{aligned}
$
Also $b=a \sqrt{e^2-1}=3 \sqrt{\frac{25}{9}-1}=3 \times \frac{4}{3}=4$
Therefore, the required equation of hyperbola is $\frac{y^2}{(3)^2}-\frac{x^2}{(4)^2}=1$
$
\Rightarrow \frac{y^2}{9}-\frac{x^2}{16}=1
$
Ex 10.4 Question 10.
Foci $( \pm 5,0)$, the transverse axis is of length 8 .
Answer.
Given: The foci $( \pm 5,0)$ lie on $x$-axis.
$\therefore$ The equation of hyperbola in standard form is $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
$\therefore$ Length of transverse axis: $2 a=8$
$
\begin{aligned}
& \Rightarrow a=4 \\
& \text { Foci }( \pm a e, 0)=( \pm 5,0) \\
& \Rightarrow a e=5 \\
& \Rightarrow e=\frac{5}{a}=\frac{5}{4}
\end{aligned}
$
Also $b=a \sqrt{e^2-1}=4 \sqrt{\frac{25}{16}-1}=4 \times \frac{3}{4}=3$
Therefore, the required equation of hyperbola is $\frac{x^2}{(4)^2}-\frac{y^2}{(3)^2}=1$
$
\Rightarrow \frac{x^2}{16}-\frac{y^2}{9}=1
$
Ex 10.4 Question 11.
Foci $(0, \pm 13)$, the conjugate axis is of length 24 .
Answer.
Given: The foci $(0, \pm 13)$ lie on $y$ - axis.
$\therefore$ The equation of hyperbola in standard form is $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$
$\therefore$ Length of conjugate axis: $2 b=24$
$
\begin{aligned}
& \Rightarrow b=12 \\
& \text { Foci : }( \pm c, 0) \\
& =( \pm 5,0) c=5
\end{aligned}
$
Also $c^2=a^2+b^2$
$
\Rightarrow(13)^2=a^2+(12)^2
$
$
\Rightarrow a^2=169-144=25
$
Therefore, the required equation of hyperbola is $\frac{y^2}{25}-\frac{x^2}{(12)^2}=1$
$
\Rightarrow \frac{y^2}{25}-\frac{x^2}{144}=1
$
Ex 10.4 Question 12.
Foci $( \pm 3 \sqrt{5}, 0)$, the latus rectum is of length 8 .
Answer.
Given: The foci $( \pm 3 \sqrt{5}, 0)$ lie on $x$ - axis.
$\therefore$ The equation of hyperbola in standard form is $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
$\therefore$ Foci $( \pm c, 0)=( \pm 3 \sqrt{5}, 0)$
$
\Rightarrow c=3 \sqrt{5}
$
Length of latus rectum: $\frac{2 b^2}{a}=8$
$
\Rightarrow b^2=4 a
$
$\begin{aligned}
& \text { Also } c^2=a^2+b^2 \\
& \Rightarrow(3 \sqrt{5})^2=a^2+4 a \\
& \Rightarrow a^2+4 a-45=0 \\
& \Rightarrow(a+9)(a-5)=0 \\
& \Rightarrow a=-9, a=5[a=-9 \text { not possible] } \\
& \Rightarrow a=5
\end{aligned}$
$
\Rightarrow a^2=25
$
Length of latus rectum: $b^2=4 a$
$
\Rightarrow b^2=4 \times 5=20
$
Therefore, the required equation of hyperbola is $\frac{x^2}{25}-\frac{y^2}{20}=1$.
Ex 10.4 Question 13.
Foci $( \pm 4,0)$, the latus rectum is of length 12 .
Answer.
Given: The foci $( \pm 4,0)$ lie on $x$ - axis.
$\therefore$ The equation of hyperbola in standard form is $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
$\therefore$ Foci $( \pm c, 0)=( \pm 4,0)$
$
\Rightarrow c=4
$
Length of latus rectum: $\frac{2 b^2}{a}=12$
$
\Rightarrow b^2=6 a
$
Also $c^2=a^2+b^2$
$\begin{aligned}
& \Rightarrow(4)^2=a^2+6 a \\
& \Rightarrow a^2+6 a-16=0 \\
& \Rightarrow(a+8)(a-2)=0 \\
& \Rightarrow a=-8, a=2[a=-8 \text { not possible] } \\
& \Rightarrow a=2 \\
& \Rightarrow a^2=4
\end{aligned}$
Length of latus rectum: $b^2=6 a$
$
\Rightarrow b^2=6 \times 2=12
$
Therefore, the required equation of hyperbola is $\frac{x^2}{4}-\frac{y^2}{12}=1$.
Ex 10.4 Question 14.
Vertices $( \pm 7,0)=e=\frac{4}{3}$
Answer.
Given: The vertices $( \pm 7,0)$ lie on $x$-axis.
$\therefore$ The equation of hyperbola in standard form is $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
$\therefore$ Vertices $( \pm a, 0)=( \pm 7,0)$
$\Rightarrow a=7$
$e=\frac{4}{3}$
$\Rightarrow \frac{c}{a}=\frac{4}{3}$
$
\begin{aligned}
& \Rightarrow \frac{c}{7}=\frac{4}{3} \\
& \Rightarrow c=\frac{28}{3}
\end{aligned}
$
Also $c^2=a^2+b^2$
$
\begin{aligned}
& \Rightarrow\left(\frac{28}{3}\right)^2=(7)^2+b^2 \\
& \Rightarrow b^2=\frac{784}{9}-49=\frac{343}{9}
\end{aligned}
$
Therefore, the required equation of hyperbola is $\frac{x^2}{(7)^2}-\frac{y^2}{343 / 9}=1$
$
\Rightarrow \frac{x^2}{49}-\frac{9 y^2}{343}=1
$
Ex 10.4 Question 15.
Foci $(0, \pm \sqrt{10})$, passing through $(2,3)$.
Answer.
Given: The foci $(0, \pm \sqrt{10})$ lie on $y$-axis.
$\therefore$ The equation of hyperbola in standard form is $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$
$\therefore$ Foci $(0, \pm c)=(0, \pm \sqrt{10})$
$
\Rightarrow c=\sqrt{10}
$
Also $c^2=a^2+b^2$
$
\begin{aligned}
& \Rightarrow(\sqrt{10})^2=a^2+b^2 \\
& \Rightarrow b^2=10-a^2
\end{aligned}
$
Since given hyperbola passes through $(2,3)$
$
\begin{aligned}
& \therefore \frac{9}{a^2}-\frac{4}{b^2}=1 \\
& \Rightarrow \frac{9}{a^2}-\frac{4}{10-a^2}=1 \\
& \Rightarrow \frac{9\left(10-a^2\right)-4 a^2}{a^2\left(10-a^2\right)}=1
\end{aligned}
$
$
\begin{aligned}
& \Rightarrow 9\left(10-a^2\right)-4 a^2=a^2\left(10-a^2\right) \\
& \Rightarrow a^4-23 a^2+90=0 \\
& \Rightarrow a^4-18 a^2-5 a^2+90=0 \\
& \Rightarrow\left(a^2-18\right)\left(a^2-5\right)=0 \\
& \Rightarrow a^2=18, a^2=5 \\
& \therefore b^2=10-a^2=10-18=-8 \text { [which is not possible] } \\
& \text { And } b^2=10-a^2=10-5=5
\end{aligned}
$
Therefore, the required equation of hyperbola is $\frac{y^2}{5}-\frac{x^2}{5}=1$.
