Hyperbola (S.N.Dey) |Part-3

In the previous article , we have solved few VSA type questions of Hyperbola chapter of S.N.Dey mathematics, Class 11. In this article , we will solve few Short Answer Type questions of Hyperbola related problems of s n dey mathematics class 11.

Hyperbola Solutions -S N Dey — Hyperbola- S N Dey Solutions

1. Find (i) the length of axes (ii) length of latus rectum (iii) co-ordinates of vertices (iv) eccentricity (v) co-ordinates of foci and (vi) equations of the directrices of each of the following hyperbolas :

$(a)~ 4x^2-9y^2=36~~(b)~ 9y^2-25x^2=225.$

Solution (a)

$~4x^2-9y^2=72 \\ \text{or,}~~ \frac{x^2}{18}-\frac{y^2}{8}=1\\ \text{or,}~~\frac{x^2}{(3\sqrt{2})^2}-\frac{y^2}{(2\sqrt{2})^2}=1 \rightarrow(1)$

Comparing $(1)$ with the general equation of hyperbola $~\frac{y^2}{a^2}-\frac{x^2}{b^2}=1~$ we get, $\therefore~ a=3\sqrt{2},~~b=2\sqrt{2}.$

(i) The length of the transverse axis $~(2a)=2 \times 3\sqrt{2}=6\sqrt{2}~~\text{unit}~$ and the length of the conjugate axis $~(2b)=2 \times 2\sqrt{2}=4\sqrt{2}~~\text{unit.}$

(ii) The length of latus rectum

$~\frac{2b^2}{a}=\frac{2 \times (2\sqrt{2})^2}{3\sqrt{2}}=\frac{16}{3\sqrt{2}}=\frac{8\sqrt{2}}{3}~~\text{unit}$

(iii) The co-ordinates of vertices: $(\pm a,0)=(\pm 3\sqrt{2},0).$

(iv) Eccentricity $(e)=\sqrt{1+\frac{b^2}{a^2}}=\sqrt{1+\frac{8}{18}}=\sqrt{\frac{26}{18}}=\sqrt{\frac{13}{9}}=\frac{\sqrt{13}}{3}.$

(v) The co-ordinates of the foci :

$(\pm ae,0)=\left(\pm 3\sqrt{2} \times \frac{\sqrt{13}}{3},0\right)=(\pm \sqrt{26},0).$

(vi) Equations of directrices are

$x=\pm \frac ae=\pm \frac{3\sqrt{2}}{\frac{\sqrt{13}}{3}} \\ \therefore~\sqrt{13}x= \pm 9\sqrt{2}.$

Solution(b)

$~9y^2-25x^2=225 \\ \text{or,}~~\frac{y^2}{25}-\frac{x^2}{9}=1\rightarrow(1)$

Comparing $(1)$ with the general equation of hyperbola $~\frac{y^2}{a^2}-\frac{x^2}{b^2}=1~$ we get, $~~a^2=25 \Rightarrow a=5,~b^2=9\Rightarrow b=3.$

(i) The length of the transverse axis $~(2a)=2 \times 5=10~~\text{unit}~$ and the length of the conjugate axis $~(2b)=2 \times 3=6~~\text{unit.}$

(ii) The length of latus rectum

$~\frac{2b^2}{a}=\frac{2 \times 9}{5}=\frac{18}{5}~~\text{unit}$

(iii) The co-ordinates of vertices: $(0, \pm a)=(0, \pm 5).$

(iv) Eccentricity $(e)=\sqrt{1+\frac{b^2}{a^2}}=\sqrt{1+\frac{9}{25}}=\sqrt{\frac{34}{25}}=\frac{\sqrt{34}}{5}.$

(v) The co-ordinates of the foci :

$(0, \pm ae)=\left(0, \pm 5 \times \frac{\sqrt{34}}{5}\right)=(0, \pm \sqrt{34}).$

(vi) Equations of directrices are

$y=\pm \frac ae=\pm \frac{5}{\sqrt{34}/5}=\pm \frac{25}{\sqrt{34}} \\ \therefore~\sqrt{34}y= \pm 25.$

2. Find the length of the transverse and conjugate axes of the hyperbola $~9x^2-16y^2=144.$ Write down the equation of the hyperbola conjugate to it and find the eccentricities of both the hyperbolas.

Solution.

$9x^2-16y^2=144 \\ \text{or,}~~ \frac{x^2}{16}-\frac{y^2}{9}=1\rightarrow(1)$

Comparing $(1)$ with the general equation of hyperbola $~\frac{y^2}{a^2}-\frac{x^2}{b^2}=1~$ we get,

$a^2=16 \Rightarrow a=4,~~ b^2=9 \Rightarrow b=3.$

So, the length of the transverse axis $~(2a)=2 \times 4=8~~\text{unit}~$ and the length of the conjugate axis $~(2b)=2 \times 3=6~\text{unit}.$

2nd Part :

The equation of the conjugate hyperbola is

$~\frac{y^2}{9}-\frac{x^2}{16}=1 \Rightarrow 16y^2-9x^2=144.$

3rd Part :

The eccentricity $(e_1)$ of the hyperbola is

$~e_1=\sqrt{1+\frac{b^2}{a^2}}=\sqrt{1+\frac{9}{16}}=\sqrt{\frac{25}{16}}=\frac 54.$

The eccentricity $(e_2)$ of the conjugate hyperbola is

$~e_2=\sqrt{1+\frac{16}{9}}=\sqrt{\frac{25}{9}}=\frac 53.$

3. Find the equation of the hyperbola, whose axes are axes of co-ordinates and

(i) length of transverse and conjugate axes are $5$ and $6$ respectively.

Solution.

Since the transverse axis is along $~x$ -axis and conjugate axis is along $~y$ -axis, so the equation of the hyperbola can be written in the form of $~\frac{x^2}{a^2}-\frac{y^2}{b^2}=1.$

So, by question,

$(i)~~ 2a=5 \Rightarrow a=\frac 52,~~ 2b=6 \Rightarrow b=\frac 62=3.$

Hence, the equation of the hyperbola is

$\frac{x^2}{(5/2)^2}-\frac{y^2}{3^2}=1 \\ \text{or,}~~ \frac{4x^2}{25}-\frac{y^2}{9}=1 \\ \text{or,}~~ 36x^2-25y^2=225.$

(ii) lengths of conjugate axis and latus rectum are $2$ and $~\frac 83~$ respectively.

Solution.

By question, the length of conjugate axis $~2b=2 \Rightarrow b=1~~$ and the length of latus rectum $~~\frac{2b^2}{a}=\frac 83 \Rightarrow \frac{2\times 1^2}{a}=\frac 83 \Rightarrow a=\frac 34.$

Hence, the equation of the hyperbola is

$~\frac{x^2}{(3/4)^2}-\frac{y^2}{1^2}=1 \\ \text{or,}~~ \frac{16x^2}{9}-y^2=1 \\ \text{or,}~~ 16x^2-9y^2=9.$

(iii)which passes through the points $~(1,1)~$ and $~(2,-3).$

Solution.

The hyperbola $~\frac{x^2}{a^2}-\frac{y^2}{b^2}=1~$ passes through the points $~(1,1)~$ and $~(2,-3).$

$\therefore~\frac{1}{a^2}-\frac{1}{b^2}=1 \rightarrow(1),~~\frac{2^2}{a^2}-\frac{(-3)^2}{b^2}=1 \Rightarrow~~ \frac{4}{a^2}-\frac{9}{b^2}=1 \rightarrow(2)$

So, from $(1)$ and $(2)$ we get,

$~\left(\frac{4}{a^2}-\frac{9}{b^2}\right)-4\left(\frac{1}{a^2}-\frac{1}{b^2}\right)=1-4 \\ \text{or,}~~-\frac{9}{b^2}+\frac{4}{b^2}=-3 \\ \text{or,}~~ \frac{-9+4}{b^2}=-3 \\ \text{or,}~~ -\frac{5}{b^2}=-3 \\ \text{or,}~~ b^2=\frac 53.$

So, by $(1)$ we get,

$~\frac{1}{a^2}=1+\frac{1}{b^2}=1+\frac 35 =\frac 85 \\ \therefore~ a^2=\frac 58.$

Hence, the required equation of the hyperbola is

$~\frac{x^2}{\frac 58}-\frac{y^2}{\frac 53}=1 \\ \text{or,}~~ \frac{8x^2}{5}-\frac{3y^2}{5}=1 \\ \text{or,}~~ 8x^2-3y^2=5.$

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(iv) distance between the foci is $10$ and length of conjugate axis is $6.$

Solution.

The length of conjugate axis $~2b=6\Rightarrow ~b=\frac 62=3\rightarrow(1)$

The distance between the foci

$2ae=10 \Rightarrow ae=\frac{10}{2}=5 \\ \therefore a^2e^2=5^2=25 \\ \text{or,}~~ a^2\left(1+\frac{b^2}{a^2}\right)=25 \\ \text{or,}~~ a^2+b^2=25 \\ \text{or,}~~ a^2+3^2=25 \\ \text{or,}~~a^2=25-9=16\rightarrow(2)$

Hence, by $(1)$ and $(2)$ we get the required equation of hyperbola which is $~~\frac{x^2}{16}-\frac{y^2}{9}=1 \Rightarrow 9x^2-16y^2=144.$

(v) co-ordinates of foci are $~\left(\pm \frac 52,0\right)~$ and the length of latus rectum is $~\frac 94.$

Solution.

The co-ordinates of foci

$(\pm ae, 0)=\left(\pm \frac 52,0\right) \\ \therefore ae=\frac 52\rightarrow(1)$

The length of the latus rectum

$\frac{2b^2}{a}=\frac 94 \Rightarrow b^2=\frac 98a \rightarrow(2)$

From $(1)$ we get,

$~a^2e^2=\left(\frac 52\right)^2 \\ \text{or,}~~ a^2\left(1+\frac{b^2}{a^2}\right)=\frac{25}{4} \\ \text{or,}~~ a^2+b^2=\frac{25}{4} \\ \text{or,}~~ a^2+\frac 98a=\frac{25}{4}~~[\text{By (2)}] \\ \text{or,}~~ 8a^2+9a=50 \\ \text{or,}~~ 8a^2+9a-50=0 \\ \text{or,}~~ 8a^2+25a-16a-50=0 \\ \text{or,}~~ a(8a+25)-2(8a+25)=0 \\ \text{or,}~~(8a+25)(a-2)=0 \\ \therefore~ a=2~~[~\because a>0]$

$\therefore~b^2=\frac 98a=\frac 98 \times 2=\frac 94.$

Hence, the equation of the hyperbola is

$~\frac{x^2}{2^2}-\frac{y^2}{\frac 94}=1 \\ \text{or,}~~\frac{x^2}{4}-\frac{4y^2}{9}=1 \\ \text{or,}~~ 9x^2-16y^2=36.$

(vi)distances between the foci and directrices are $~4\sqrt{7}~$ and $~\frac{16}{\sqrt{7}}~$ respectively.

Solution.

The distance between the foci $=2ae.$

$\therefore~2ae=4\sqrt{7}\rightarrow(1)$

Again, the distance between two directrices $=\frac{2a}{e}.$

$\Rightarrow~ \frac{2a}{e}=\frac{16}{\sqrt{7}}\rightarrow(2)$

So, by $~(1) \times (2)~$ , we get

$~2ae \times \frac{2a}{e}=4\sqrt{7}\times \frac{16}{\sqrt{7}} \\ \text{or,}~~ 4a^2=64 \\ \text{or,}~~a^2=16\rightarrow(3)$

Again, by $(1)$ we get,

$a^2e^2=(2\sqrt{7})^2 \\ \text{or,}~~ a^2\left(1+\frac{b^2}{a^2}\right)=28 \\ \text{or,}~~a^2+b^2=28 \\ \text{or,}~~ 16+b^2=28 \\ \text{or,}~~ b^2=28-16=12.$

So, the equation of the hyperbola is

$~\frac{x^2}{a^2}-\frac{y^2}{b^2}=1 \\ \text{or,}~~ \frac{x^2}{16}-\frac{y^2}{12}=1 \\ \text{or,}~~ 3x^2-4y^2=48.$

(vii) eccentricity is $~\frac{\sqrt{13}}{3}$ and the sum of squares of the lengths of axes is $52.$

Solution.

By question, we have

$(2a)^2+(2b)^2=52 \\ ~~\text{or,}~~ 4a^2+4b^2=52 \\~~ \text{or,}~~a^2+b^2=\frac{52}{4}=13 \rightarrow(1)$

$e=\frac{\sqrt{13}}{3}~~\text{(Given)} \\~~ \text{or,}~~e^2=\frac{13}{9} \\ ~~\text{or,}~~1+\frac{b^2}{a^2}=\frac{13}{9} \\~~ \text{or,}~~ \frac{a^2+b^2}{a^2}=\frac{13}{9} \\~~ \text{or,}~~ \frac{13}{a^2}=\frac{13}{9}~~[\text{By (1)}] \\ ~~\text{or,}~~ a^2=9$

Putting the value of $a^2$ in $(1)$ we get,

$~9+b^2=13 \Rightarrow b^2=13-9=4.$

Hence, the equation of the hyperbola is

$~\frac{x^2}{a^2}-\frac{y^2}{b^2}=1 \\~~ \text{or,}~~ \frac{x^2}{9}-\frac{y^2}{4}=1 \\~~ \therefore~4x^2-9y^2=36.$

(viii)transverse axis is $2a$ and the vertex bisects the line segment joining the centre and focus.

Solution.

Since the vertex bisects the line segment joining the centre and the focus,

$\therefore~$ the distance of the vertex from the centre

$=\frac 12 \times$ (the distance of the focus from the centre)

$\therefore~ a=\frac{ae}{2} \Rightarrow e=2 \\ \text{or,}~~ e^2=1+\frac{b^2}{a^2}=4 \\ \text{or,}~~ \frac{b^2}{a^2}=3 \\ \text{or,}~~ b^2=3a^2$

Hence, the equation of the hyperbola is

$~\frac{x^2}{a^2}-\frac{y^2}{b^2}=1 \\ \text{or,}~~ \frac{x^2}{a^2}-\frac{y^2}{3a^2}=1 \\ \therefore~3x^2-y^2=3a^2.$

(ix) which passes through the point $(2,1)$ and whose eccentricity is $~\sqrt{\frac 32}$ .

Solution.

The hyperbola $~\frac{x^2}{a^2}-\frac{y^2}{b^2}=1~$ passes through the point $~(2,1).$

$\therefore~\frac{2^2}{a^2}-\frac{1^2}{b^2}=1 \Rightarrow \frac{4}{a^2}-\frac{1}{b^2}=1 \\ \text{or,}~~4-\frac{a^2}{b^2}=a^2 \\ \text{or,}~~ \frac{a^2}{b^2}=4-a^2 \rightarrow(1)$

$\text{Again,}~e=\sqrt{\frac 32} \Rightarrow e^2=\frac 32 \\ \text{or,}~~ 1+\frac{b^2}{a^2}=\frac 32 \Rightarrow \frac{b^2}{a^2}=\frac 32-1=\frac 12 \rightarrow(2)$

By $~(1) \times (2)~$ we get,

$~ \frac{a^2}{b^2} \times \frac{b^2}{a^2}=\frac 12(4-a^2) \\ \text{or,}~~1=\frac{4-a^2}{2} \\ \text{or,}~~ 2=4-a^2 \\ \therefore~ a^2=4-2=2.$

So, using $(2)$ we get, $~ \frac{b^2}{2}=\frac 12 \Rightarrow b^2=1.$

Hence, the equation of the hyperbola is given by

$~\frac{x^2}{a^2}-\frac{y^2}{b^2}=1 \\ \text{or,}~~ \frac{x^2}{2}-\frac{y^2}{1}=1 \\ \text{or,}~~ x^2-2y^2=2.$

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(x) eccentricity is $3$ and the co-ordinates of one focus are $\left(\frac 32,0\right).$

Solution.

Eccentricity $(e)$ is $~3~$ and the co-ordinates of one focus $(ae,0)$ are $~\left(\frac 32,0\right).$

$\therefore ae=\frac 32 \Rightarrow a \times 3=\frac 32 \Rightarrow a=\frac 12.$

$e=3 \Rightarrow e^2=3^2 \Rightarrow 1+\frac{b^2}{a^2}=9 \\ \therefore~ \frac{b^2}{a^2}=8 \Rightarrow b^2=8a^2=8 \times \frac 14=2.$

$\therefore~$ The equation of the hyperbola is

$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1 \\ \text{or,}~~ \frac{x^2}{1/4}-\frac{y^2}{2}=1\\ \text{or,}~~4x^2-\frac{y^2}{2}=1 \\ \therefore~8x^2-y^2=2.$

(xi) eccentricity is $\sqrt{\frac 32}$ and the length of semi latus rectum is $2$

Solution.

It is given that the eccentricity $(e)$ is $~3~$ and the length of semi-latus rectum $~\left(\frac{b^2}{a}\right)$ is $~2.$

$\therefore~ \frac{b^2}{a}=2 \Rightarrow b^2=2a \rightarrow(1)$

$~e=\sqrt{\frac 32} \Rightarrow e^2=\frac 32 \\ \text{or,}~~ 1+\frac{b^2}{a^2}=\frac 32 \\ \text{or,}~~ \frac{b^2}{a^2}=\frac 32-1 \\ \text{or,}~~ \frac{2a}{a^2}=\frac 12~~[\text{By (1)}] \\ \text{or,}~~ \frac 2a=\frac 12 \\ \text{or,}~~ a=4 \Rightarrow a^2=4^2=16. \\ \therefore b^2=2\times 4=8$