Hyperbola (S.N.Dey) |Part-1 | Ex-6

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In the previous articles , we have done complete solution of Ellipse chapter of S.N.Dey mathematics, Class 11. In this article , we will solve few questions of Hyperbola related problems.
Hyperbola Solutions S N Dey mathematics
Hyperbola VSA type Questions’ Solutions

1.Explain what general diagrams are represented by the equation \frac{x^2}{a^2}+\frac{y^2}{a^2(1-e^2)}=1,~ where a and e are constants. 

Solution.

We have the equation given by 

\frac{x^2}{a^2}+\frac{y^2}{a^2(1-e^2)}=1\rightarrow(1)

Now, there can be three conditions for different values of ~e.

~(a)~ 0<e<1,~~(b)~ e=0,~~(c)~e>1.

Case (a):

For ~0<e<1,~~1-e^2=c(>0)~\text{(say)}.

In this case, we get from (1),

~\frac{x^2}{a^2}+\frac{y^2}{a^2\cdot c}=1 \\ \text{or,}~~\frac{x^2}{a^2}+\frac{y^2}{b^2}=1~~(\text{where}~~ b^2=a^2\cdot c)\rightarrow(2)

Clearly, (2) represents the equation of the ellipse.

Case (b):

For ~e=0,~ we get by ~(1),

\frac{x^2}{a^2}+\frac{y^2}{a^2(1-0)}=1 \\ \text{or,}~~ x^2+y^2=a^2 \rightarrow(3)

Equation ~(3)~ represents a ‘circle.’

Case (c):

For ~e>1,~~ 1-e^2<0.

Let ~1-e^2=-c~~(c>0)

So, from (1) we get,

~\frac{x^2}{a^2}+\frac{y^2}{a^2(-c)}=1 \\ \text{or,}~~ \frac{x^2}{a^2}-\frac{y^2}{b^2}=1~~(b^2=a^2c)\rightarrow(4)

Equation (4) represents a ‘hyperbola.’

2. Show that the eccentricity of any rectangular hyperbola is \sqrt{2}.

Solution.

We know that the general form of hyperbola is ~\frac{x^2}{a^2}-\frac{y^2}{b^2}=1 \rightarrow(1)

The eccentricity (e) of  (1) is given by 

e=\sqrt{1+\frac{b^2}{a^2}}\rightarrow(2)

For rectangular hyperbola, ~a=b~ and so from (2) we get,

~e=\sqrt{1+\frac{a^2}{a^2}}=\sqrt{1+1}=\sqrt{2}.

3. Find the eccentricity, co-ordinates of the foci and the equations of the directrices of the hyperbolas :

(i)~16x^2-9y^2=144~~(ii)~4x^2-9y^2=36.

Solution.(i)

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

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

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

\therefore ~e=\sqrt{1+\frac{b^2}{a^2}}=\sqrt{1+\frac{16}{9}}=\sqrt{\frac{25}{9}}=\frac 53.

Co-ordinates of foci : ~(\pm ae,0)=\left(\pm 3 \times \frac 53,0\right)=(\pm 5,0).

The equations of directrices of the hyperbola :

x= \pm \frac ae=\pm \frac{3}{\frac 53}=\pm \frac 95 \\ \text{or,}~~ 5x=\pm 9.

Solution(ii)

4x^2-9y^2=36 \\ \text{or,}~~ \frac{x^2}{9}-\frac{y^2}{4}=1\rightarrow(1)

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

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

\therefore ~e=\sqrt{1+\frac{b^2}{a^2}}=\sqrt{1+\frac{4}{9}}=\sqrt{\frac{13}{9}}=\frac {\sqrt{13}}{3}.

Co-ordinates of foci : ~(\pm ae,0)=\left(\pm 3 \times \frac {\sqrt{13}}{3},0\right)=(\pm \sqrt{13},0).

The equations of directrices of the hyperbola :

x= \pm \frac ae=\pm \frac{3}{\frac {\sqrt{13}}{3}}=\pm \frac{9}{\sqrt{13}} \\ \text{or,}~~ \sqrt{13}x=\pm 9.

4. Show that the eccentricities of the two hyperbolas ~\frac{x^2}{16}-\frac{y^2}{9}=1~ and ~\frac{x^2}{64}-\frac{y^2}{36}=1~ are equal.

Solution.

We have two hyperbolas given by 

~\frac{x^2}{16}-\frac{y^2}{9}=1\rightarrow(1)~ and ~\frac{x^2}{64}-\frac{y^2}{36}=1~\rightarrow(2).

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

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

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

~a^2=64,~b^2=36 \\ \therefore ~e_2=\sqrt{1+\frac{b^2}{a^2}}=\sqrt{1+\frac{36}{64}}=\sqrt{\frac{64+36}{64}}=\sqrt{\frac {100}{64}} \\ \text{or,}~~ e_2=\frac{10}{8}=\frac 54.

\therefore~ e_1=e_2.

Note[*] : e_1 \rightarrow eccentricity of the first hyperbola, e_2 \rightarrow eccentricity of the second hyperbola.

5(i)~ Find the co-ordinates of the foci of the hyperbola ~x^2-y^2+1=0.

Solution.

x^2-y^2+1=0 \\ \text{or,}~~ \frac{y^2}{1}-\frac{x^2}{1}=1\rightarrow(1)

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

~a^2=1 \Rightarrow a=1, b^2=1. \\ \therefore~~e=\sqrt{1+\frac{b^2}{a^2}}=\sqrt{1+\frac 11}=\sqrt{2}

The co-ordinates of foci of the hyperbola (1):

(0,\pm ae)=(0, \pm 1 \times \sqrt{2})=(0,\pm \sqrt{2}).

(ii) Find the eccentricity and the length of latus rectum of the hyperbola ~x^2-y^2=2.

Solution.

x^2-y^2=2 \Rightarrow \frac{x^2}{2}-\frac{y^2}{2}=1\rightarrow(1)

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

a^2=b^2=2 \Rightarrow a=b=\sqrt{2}.

So, ~e=\sqrt{1+\frac{b^2}{a^2}}=\sqrt{1+\frac 22}=\sqrt{1+1}=\sqrt{2}

The length of the latus rectum is 

\frac{2b^2}{a}=\frac{2 \times 2}{\sqrt{2}}=2\sqrt{2}~~\text{unit}.

6. Find the length of latus rectum and the equations of the directrices of the hyperbola ~3y^2-4x^2=12.

Solution.

3y^2-4x^2=12 \\ \text{or,}~~ \frac{y^2}{4}-\frac{x^2}{3}=1 \rightarrow(1)

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

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

The length of the latus rectum of the hyperbola (1)~ is 

\frac{2b^2}{a}=\frac{2 \times 3}{2}=3~~\text{unit.}

Equations of the directrices of the hyperbola ~(1)

x= \pm \frac ae =\pm \frac{2}{\frac{\sqrt{7}}{2}}=\pm \frac{4}{\sqrt{7}} \\ \text{or,}~~ \sqrt{7}x= \pm 4.

7. What type of conic is represented by the equation ~x^2-y^2=4~? What is its eccentricity?

Solution.

The given equation represents the equation of rectangular hyperbola and its eccentricity is \sqrt{1+\frac{4}{4}}=\sqrt{1+1}=\sqrt{2}.

8. If the length of conjugate axis and the length of latus rectum of a hyperbola are equal, find its eccentricity.

Solution.

The length of the conjugate axis =2b~ unit and the length of the latus rectum is ~\frac{2b^2}{a}~ unit of the hyperbola ~\frac{x^2}{a^2}-\frac{y^2}{b^2}=1.

By question,

2b=\frac{2b^2}{a}\Rightarrow \frac ba=1. \\ \therefore~ e=\sqrt{1+\frac{b^2}{a^2}}=\sqrt{1+1^2}=\sqrt{1+1}=\sqrt{2}.

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