Ellipse (S.N.Dey)|Part-1|Ex-5 Examhoop

In this article, we will do complete solutions of Very Short Answer Type Questions of Ellipse Chapter of S.N.Dey mathematics, Class 11.

$1.~$ Find the length of the latus rectum and the co-ordinate of the foci of the ellipse $~25x^2+4y^2=100.$

Solution.

The given equation of ellipse can be written as

$\frac{x^2}{4}+\frac{y^2}{25}=1 \rightarrow(1)$

Comparing $~(1)~$ with $~\frac{x^2}{a^2}+\frac{y^2}{b^2}=1~$ we get

$~ a^2=25 \Rightarrow a=5, ~~b^2=4 \Rightarrow b=2.$

The length of the latus rectum is

$\frac{2b^2}{a}=2 \times \frac 45=\frac 85~~\text{unit.}$

Now, eccentricity of the ellipse

$(e)=\sqrt{1-\frac{b^2}{a^2}}=\sqrt{1-\frac{4}{25}}=\sqrt{\frac{25-4}{25}}=\frac{\sqrt{21}}{5}$

The co-ordinates of foci are

$(0, \pm ae)=(0, \pm 5 \times \frac{\sqrt{21}}{5})=(0, \pm \sqrt{21}).$

$2.~$ Find the length of the latus rectum of the ellipse $~\frac{x^2}{9}+\frac{y^2}{16}=1.$

Solution.

We have, the given equation of ellipse $~\frac{x^2}{9}+\frac{y^2}{16}=1 \rightarrow(1)$

Comparing $~(1)~$ with $~\frac{x^2}{b^2}+\frac{y^2}{a^2}=1~$ we get,

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

So, the length of the latus rectum is given by

$\frac{2b^2}{a}=2 \times \frac 94=\frac 92~~\text{unit.}$

$3.~$ Calculate the eccentricity of the ellipse $~\frac{x^2}{169}+\frac{y^2}{144}=1.$

Solution.

Comparing the given ellipse with $~\frac{x^2}{a^2}+\frac{y^2}{b^2}=1~$ we get, $~a^2=169,~b^2=144.$

So, the eccentricity of the ellipse is given by

$(e)=\sqrt{1-\frac{b^2}{a^2}}=\sqrt{1-\frac{144}{169}}=\sqrt{\frac{169-144}{169}} \\ \therefore~~ e=\sqrt{\frac{25}{169}}=\frac{5}{13}.$

$4.(i)~$ Find the equations of the directrices of the ellipse $~x^2+4y^2=4.$

Solution.

The given equation of the ellipse can be written as

$~\frac{x^2}{4}+\frac{y^2}{1}=1\rightarrow(1)$

Comparing $~(1)~$ with $~\frac{x^2}{a^2}+\frac{y^2}{b^2}=1~$ we get, $~a^2=4,~b^2=1.$

$\therefore~ e=\sqrt{1-\frac{b^2}{a^2}}=\sqrt{1-\frac 14}=\sqrt{\frac{4-1}{4}}=\frac{\sqrt{3}}{2}.$

So, the equations of the directrices of the given ellipse are

$x=\pm \frac ae=\pm \frac{2}{\sqrt{3}/2}=\pm \frac{4}{\sqrt{3}} \\ \therefore~ \sqrt{3}x= \pm 4.$

$(ii)~$ Find the distance between the foci of the ellipse $~3x^2+4y^2=12.$

Solution.

The equation of the ellipse can be written as $~\frac{x^2}{4}+\frac{y^2}{3}=1\rightarrow(1)$

Comparing $~(1)~$ with $~\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,~~(a^2>b^2)~$ we get, $~a^2=4 \Rightarrow a=2,~~b^2=3.$

Now, the eccentricity $~(e)~$ of the ellipse $(1)$ is

$~\sqrt{1-\frac{b^2}{a^2}}=\sqrt{1-\frac 34}=\sqrt{\frac{4-3}{4}}=\sqrt{\frac14}=\frac 12.$

So, the distance between the foci

$2ae=2 \times 2 \times \frac 12=2~~\text{unit.}$

$5.~$ Find the eccentricity of the ellipse if

$(i)~$ the length of the latus rectum is equal to half the minor axis of the ellipse.

Solution.

Let the equation of the ellipse be $~\frac{x^2}{a^2}+\frac{y^2}{b^2}=1.$

Now, the length of the latus rectum $=\frac{2b^2}{a}~$ and the length of the half the minor axis $=b~$ unit.

So, by condition,

$~\frac{2b^2}{a}=b \Rightarrow 2b=a.$

$\therefore~ e\\=\sqrt{1-\frac{b^2}{a^2}}\\=\sqrt{1-\frac{b^2}{4b^2}}\\=\sqrt{1-\frac 14}\\=\frac{\sqrt{3}}{2}.$

$(ii)~$ the length of the minor axis is equal to half the distance between the foci of the ellipse.

Solution.

Given :

Minor axis= $\frac 12 \times$ (Distance between the foci)

$\text{or,}~~ 2b=\frac 12 \times 2ae \\ ~~~\text{or,}~~ 2b=ae \\~~~ \therefore~ b=\frac 12 ae.$

$\text{Now,}~e=\sqrt{1-\frac{b^2}{a^2}} \\ \text{or,}~~ e^2=1-\frac{1}{a^2} \times \left(\frac 12 ae\right)^2 \\ \text{or,}~~ e^2=1-\frac{1}{a^2} \times \frac 14 a^2e^2 \\ \text{or,}~~ 4e^2=4-e^2 \\ \text{or,}~~ 5e^2=4 \\ \text{or,}~~ e=\sqrt{\frac 45}=\frac{2}{\sqrt{5}}.$

$(iii)~$ the length of minor axis is equal to the distance between the later recta.

Solution.

We know that the distance between the foci of the ellipse $=2ae,~$ where $~a=\frac 12 \times \text{(length of the major axis)},\\e=\text{eccentricity}=\sqrt{1-\frac{b^2}{a^2}}\rightarrow(1),\\2b=\text{length of the minor axis.}$

By question,

$~ 2ae=2b \\ \text{or,}~~ ae=b \\ \text{or,}~~ e=\frac ba\rightarrow(2)$

So, by $~(1)~$ and $~(2)~$ we get,

$~e^2=1-\frac{b^2}{a^2}=1-e^2 \\ \text{or,}~~ 2e^2=1 \\ \therefore~ e=\frac{1}{\sqrt{2}}.$

$6.(i)~$ If the ellipses $~\frac{x^2}{a^2}+\frac{y^2}{b^2}=1~$ and $~\frac{x^2}{p^2}+\frac{y^2}{q^2}=1~$ have same eccentricity, show that $~aq=bp.$

We have the equations of two ellipses which are

$~\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\rightarrow(1) ,~~\frac{x^2}{p^2}+\frac{y^2}{q^2}=1 \rightarrow(2)$

The eccentricity $(e_1)~$ of $~(1)$

$e_1=\sqrt{1-\frac{b^2}{a^2}}$

The eccentricity $(e_2)~$ of $~(2)$

$e_2=\sqrt{1-\frac{q^2}{p^2}}$

$\because ~ e_1=e_2 \\ \therefore~ \sqrt{1-\frac{b^2}{a^2}}=\sqrt{1-\frac{q^2}{p^2}} \\ \text{or,}~~ 1-\frac{b^2}{a^2}=1-\frac{q^2}{p^2} \\ \text{or,}~~ \frac{b^2}{a^2}=\frac{q^2}{p^2} \\ \text{or,}~~ a^2q^2=b^2p^2 \\ \text{or,}~~ aq=bp.$

$(ii)~$ The ellipse $~\frac{x^2}{169}+\frac{y^2}{25}=1~$ has the same eccentricity as the ellipse $~\frac{x^2}{a^2}+\frac{y^2}{b^2}=1~$ . Find the ratio $~\frac ab.$

Solution.

We have the given equation of ellipse $~\frac{x^2}{169}+\frac{y^2}{25}=1\rightarrow(1)$

The eccentricity of $~(1),$

$(e)=\sqrt{1-\frac{25}{169}}=\sqrt{\frac{169-25}{169}} \\ \therefore~ e=\sqrt{\frac{144}{169}}=\frac{12}{13}$

Again, the eccentricity of $~\frac{x^2}{a^2}+\frac{y^2}{b^2}=1~$ is

$~e=\sqrt{1-\frac{b^2}{a^2}} \\ \text{or,}~~ e^2=1-\frac{b^2}{a^2} \\ \text{or,}~~ \left(\frac{12}{13}\right)^2=1-\frac{b^2}{a^2} \\ \text{or,}~~ \frac{b^2}{a^2}=1-\frac{144}{169}=\frac{169-144}{169} \\ \text{or,}~~ \frac{b^2}{a^2}=\frac{25}{169} =\left(\frac{5}{13}\right)^2 \\ \text{or,}~~ \frac ba=\frac{5}{13} \\ \therefore~ \frac ab=\frac{13}{5}.$

$7.~$ Find the positions of the points $~(i)~(-3,1)~~(ii)~(-2,-3)~$ and $~(iii)~(5,-2)~$ with respect to the ellipse $~3x^2+4y^2=48.$

Solution.

The equation of the ellipse can be written as $~3x^2+4y^2-48=0$

Case-1 :

$~3(-3)^2+4 \times 1^2-48=27+4-48=-17<0$

So, the point $~(-3,1)~$ lies inside the ellipse.

Case-2:

$~3(-2)^2+4(-3)^2-48=12+36-48=0$

So, the point $~(-2,-3)~$ lies on the ellipse.

Case-3:

$~3 \times 5^2+4(-2)^2-48=75+16-48=43>0$

So, the point $~(5,-2)~$ lies outside the ellipse.

$8.~$ For what values of $~a^2~$ does the point $~(2\sqrt{3},1)~$ lie outside the ellipse $~\frac{x^2}{a^2}+\frac{y^2}{4}=1.$

$9.~$ The co-ordinates of a point on the ellipse $~9x^2+16y^2=144~$ are $~\left(2,\frac{3\sqrt{3}}{2}\right)~$ ; find the eccentric angle of the point.

Solution.

The equation of the given ellipse can be written in standard form as

$~\frac{x^2}{16}+\frac{y^2}{9}=1 \Rightarrow~(x/4)^2+(y/3)^2=1 \rightarrow(1)$

Any point on $~(1)~$ can be taken as $~(4 \cos\phi, 3\sin\phi).$

Now, by question, we can have

$~2=4 \cos\phi \Rightarrow \cos\phi=\frac 24=\frac 12,\\~\frac{3\sqrt{3}}{2}=3\sin\phi \Rightarrow \sin\phi=\frac{\sqrt{3}}{2}.$

$\therefore~\tan\phi=\frac{\sin\phi}{\cos\phi}=\sqrt{3} \\ \text{or,}~~\phi=\frac{\pi}{3}=60^{\circ}.$

Hence, the eccentric angle of the point is $~60^{\circ}.$

$10.~$ Find the co-ordinates of the point on the ellipse $~x^2+2y^2=4~$ whose eccentric angle is $~60^{\circ}.$

Solution.

The equation of the ellipse can be written as $~\frac{x^2}{4}+\frac{y^2}{2}=1\rightarrow(1)$

Comparing $~(1)~$ with the general equation of the ellipse $~\frac{x^2}{a^2}+\frac{y^2}{b^2}=1~$ we get, $~ a^2=4 \Rightarrow a=2,~~b^2=2 \Rightarrow b=\sqrt{2}.$

So, any point on the given ellipse with the eccentric angle $~60^{\circ}~$ can be written as

$~x=a \cos 60^{\circ}=2 \times \frac 12=1,~y=b\sin 60^{\circ}=\sqrt{2}\times \frac{\sqrt{3}}{2}=\frac{\sqrt{6}}{2}.$

So, the co-ordinates of the point on the ellipse $~x^2+2y^2=4~$ whose eccentric angle is $~60^{\circ}~$ is given by $~\left(1,\frac{\sqrt{6}}{2}\right).$

$11.~$ If the ellipse $~\frac{x^2}{a_1^2}+\frac{y^2}{b_1^2}=1~~(a_1^2>b_1^2)~$ prove that $~a_1b_2=a_2b_1.$

Solution.

We have the equations of two ellipses which are

$\frac{x^2}{a_1^2}+\frac{y^2}{b_1^2}=1\rightarrow(1),~~ \frac{x^2}{a_2^2}+\frac{y^2}{b_2^2}=1\rightarrow(2)$

The eccentricity $(e_1)~$ of $~(1)$

$e_1=\sqrt{1-\frac{b_1^2}{a_1^2}}$

The eccentricity $(e_2)~$ of $~(2)$

$e_2=\sqrt{1-\frac{b_2^2}{a_2^2}}$

$\because~ e_1=e_2 \\ \therefore~\sqrt{1-\frac{b_1^2}{a_1^2}}=\sqrt{1-\frac{b_2^2}{a_2^2}} \\ \text{or,}~~1-\frac{b_1^2}{a_1^2}=1-\frac{b_2^2}{a_2^2} \\ \text{or,}~~\frac{b_1^2}{a_1^2}=\frac{b_2^2}{a_2^2} \\ \text{or,}~~ a_1^2b_2^2=a_2^2b_1^2 \\ \text{or,}~~a_1b_2=a_2b_1$

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In this article, we will do complete solutions of Very Short Answer Type Questions of Ellipse Chapter of S.N.Dey mathematics, Class 11.

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