Ellipse (S.N.Dey)|Part-2|Ex-5 Examhoop

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

Ellipse S N Dey Solutions (Short Answer Type Questons )

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

$(a)~16x^2+25y^2=400~~(b)~9x^2+4y^2=36 ~~(c)~x^2+4y^2=16 \\(d)~4x^2+3y^2=1.$

Solution (a).

The given equation of the ellipse can be written as $~\frac{x^2}{25}+\frac{y^2}{16}=1\rightarrow(1)$

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

$a^2=25 \Rightarrow a=5 ~( \because a>0),\,\,\, b^2=16 \Rightarrow b=4~(\because b>0).$

Clearly, major axis of the ellipse $~(1)~$ is along $~x-$ axis , minor axis of the given ellipse is along $~y-$ axis and the centre of the ellipse is at $~(0,0).$

(i) The length of the major axis is $(2a)=2 \times 5=10~\text{unit}$ and

the length of the minor axis is $~(2b)=2 \times 4=8~\text{unit}$

(ii) The length of the latus rectum is

$\frac{2b^2}{a}=\frac{2 \times 16}{5}=\frac{32}{5}~~\text{unit.}$

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

(iv) Eccentricity of the ellipse is given by

$e=\sqrt{1-\frac{b^2}{a^2}}=\sqrt{1-\frac{16}{25}}=\sqrt{\frac{25-17}{25}}=\frac 35.$

(v) The co-ordinates of foci are given by

$~(\pm ae,0)=\left( \pm 5 \cdot \frac 35,0\right)=(\pm 3,0).$

(vi) Equations of directrices are given by

$x=\pm \frac ae=\pm \frac{5}{3/5} \\ \text{or,}~~ x= \pm \frac{25}{3} \\ \therefore ~3x= \pm 25.$

Solution(b)

The given equation of the ellipse can be written as $~\frac{x^2}{4}+\frac{y^2}{9}=1\rightarrow(1)$

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

$a^2=9 \Rightarrow a=3 ~( \because a>0),\,\,\, b^2=4 \Rightarrow b=2~(\because b>0).$

Clearly, major axis of the ellipse $~(1)~$ is along $~y-$ axis , minor axis of the given ellipse is along $~x-$ axis and the centre of the ellipse is at $~(0,0).$

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(i) The length of the major axis is $(2a)=2 \times 3=6~\text{unit}$ and

the length of the minor axis is $~(2b)=2 \times 2=4~\text{unit}$

(ii) The length of the latus rectum is

$\frac{2b^2}{a}=\frac{2 \times 4}{3}=\frac83~~\text{unit.}$

(iii) The co-ordinates of vertices $~(0,\pm a)=(0,\pm 3).$

(iv) Eccentricity of the ellipse is given by

$e=\sqrt{1-\frac{b^2}{a^2}}=\sqrt{1-\frac49}=\sqrt{\frac{9-4}{9}}=\frac {\sqrt{5}}{3}$

(v) The co-ordinates of foci are given by

$~(0,\pm ae)=\left(0,\pm 3 \cdot \frac{\sqrt{5}}{3}\right)=(0,\pm \sqrt{5}).$

(vi) Equations of directrices are given by

$y=\pm \frac ae=\pm \frac{3}{\frac{\sqrt{5}}{3}} \\ \text{or,}~~ y= \pm \frac{9}{\sqrt{5}}.$

Solution(c)

The given equation of the ellipse can be written as $~\frac{x^2}{16}+\frac{y^2}{4}=1\rightarrow(1)$

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

$a^2=16 \Rightarrow a=4 ~( \because a>0),~~b^2=4 \Rightarrow b=2~(\because b>0).$

Clearly, major axis of the ellipse $~(1)~$ is along $~x-$ axis , minor axis of the given ellipse is along $~y-$ axis and the centre of the ellipse is at $~(0,0).$

(i) The length of the major axis is $(2a)=2 \times 4=8~\text{unit}$ and

the length of the minor axis is $~(2b)=2 \times 2=4~\text{unit}$

(ii) The length of the latus rectum is

$\frac{2b^2}{a}=\frac{2 \times 2^2}{4}=2~~\text{unit.}$

(iii) The co-ordinates of vertices $~(\pm a,0)=(\pm 4,0).$

(iv) Eccentricity of the ellipse is given by

$e=\sqrt{1-\frac{b^2}{a^2}}=\sqrt{1-\frac{2^2}{4^2}}=\sqrt{1-(1/2)^2}=\sqrt{\frac 34}=\frac{\sqrt{3}}{2}.$

(v) The co-ordinates of foci are given by

$~(\pm ae,0)=\left( \pm 4 \cdot \frac{\sqrt{3}}{2},0\right)=(\pm 2\sqrt{3},0).$

(vi) Equations of directrices are given by

$x=\pm \frac ae=\pm \frac{4}{\frac{\sqrt{3}}{2}} \\ \text{or,}~~ x= \pm \frac{8}{\sqrt{3}} \\ \therefore ~\sqrt{3}x= \pm 8.$

Solution(d)

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

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

$a^2=\frac 13 \Rightarrow a=\frac{1}{\sqrt{3}} ~( \because a>0),\,\,\, b^2=\frac 14 \Rightarrow b=\frac 12~(\because b>0).$

Clearly, major axis of the ellipse $~(1)~$ is along $~y-$ axis , minor axis of the given ellipse is along $~x-$ axis and the centre of the ellipse is at $~(0,0).$

(i) The length of the major axis is $(2a)=2 \times \frac{1}{\sqrt{3}}=\frac{2}{\sqrt{3}}~\text{unit}$ and

the length of the minor axis is $~(2b)=2 \times \frac 12=1~\text{unit}$

(ii) The length of the latus rectum is

$\frac{2b^2}{a}=\frac{2 \times (1/2)^2}{\frac{1}{\sqrt{3}}}=\frac{\sqrt{3}}{2}~~\text{unit.}$

(iii) The co-ordinates of vertices $~(0,\pm a)=\left(0,\pm \frac{1}{\sqrt{3}}\right).$

(iv) Eccentricity of the ellipse is given by

$e=\sqrt{1-\frac{b^2}{a^2}}=\sqrt{1-\frac{(1/2)^2}{(1/\sqrt{3})^2}}=\sqrt{1-\frac{3}{4}}=\sqrt{\frac 14}=\frac 12.$

(v) The co-ordinates of foci are given by

$~(0,\pm ae)=\left(0,\pm \frac{1}{\sqrt{3}} \cdot \frac12\right)=(0,\pm \frac{1}{2\sqrt{3}}).$

(vi) Equations of directrices are given by

$y=\pm \frac ae=\pm \frac{\frac{1}{\sqrt{3}}}{\frac 12} \\ \text{or,}~~ y= \pm \frac{2}{\sqrt{3}} \\ \text{or,}~~ \sqrt{3}y= \pm 2.$

$2.~$ Find the eccentricity and equations of the directrices of the ellipse $~\frac{x^2}{100}+\frac{y^2}{36}=1.~$ Show that the sum of the focal distances of any point on this ellipse is a constant.

Solution.

Comparing the given equation of ellipse with the general form of the ellipse $~\frac{x^2}{b^2}+\frac{y^2}{a^2}=1~(a^2>b^2)~$ we get,

$a^2=100 \Rightarrow a=10,~~ b^2=36 \Rightarrow b=6.$

Eccentricity of the ellipse is given by

$e=\sqrt{1-\frac{b^2}{a^2}}=\sqrt{1-\frac{36}{100}}=\frac{8}{10}=\frac 45.$

Equation of the directrix is given by

$x=\pm \frac ae=\pm \frac{10}{\frac 45} \\ \text{or,}~~ 2x= \pm 25.$

Any point on the given ellipse can be written as $~P(10\cos\theta,6\sin\theta).$

The co-ordinates of the foci are given by

$~(\pm ae,0)=(\pm 10 \times \frac 45,0)=(\pm 8,0).$

Now, the sum of the focal distances of the point P is

$=\sqrt{(10\cos\theta-8)^2+36\sin^2\theta}+\sqrt{(10\cos\theta+8)^2+36\sin^2\theta}\\=\sqrt{64\cos^2\theta-160\cos\theta+100}+\sqrt{64\cos^2\theta+160\cos\theta+100}~~[*]\\=\sqrt{(8\cos\theta-10)^2}+\sqrt{(8\cos\theta+10)^2}\\=10-8\cos\theta+8\cos\theta+10~~[\because~ \cos\theta \leq 8< 10]\\=20=\text{constant}$

Note[*] :

$(10\cos\theta-8)^2+36\sin^2\theta\\=100\cos^2\theta-160\cos\theta+64+36(1-\cos^2\theta)\\=64\cos^2\theta-160\cos\theta+100.$

$3.~$ Taking major and minor axes as $~x~$ and $~y~$ -axes respectively, find the equation of the ellipse

$(i)~$ whose length of major and minor axes are $~6~$ and $~5~$ respectively.

Solution.

By question, the equation of the ellipse can be written as

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

(i) The length of the major axis $~=2a=6\Rightarrow a=\frac 62=3$

and the length of the minor axis $=~2b=5\Rightarrow b=\frac 52$

Hence, by $~(1),~$ the equation of the ellipse is

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

$(ii)~$ whose lengths of minor axis and latus rectum are $~4~$ and $~2.$

Solution.

Length of the minor axis $=2b=4 \Rightarrow b=\frac 42=2.$

Length of the latus rectum $=\frac{2b^2}{a}=2 \cdot \frac{2^2}{a}=2 \Rightarrow a=4.$

Hence, the equation of the ellipse is given by

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

$(iii)~$ whose eccentricity is $~\frac 35~$ and co-ordinates of foci are $~(\pm 3,0).$

Solution.

By question, the equation of the ellipse can be written as

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

The eccentricity $(e)=\frac 35.$

The co-ordinates of foci $~(\pm ae,0)=(\pm 3,0).$

$\therefore~ ae=3 \Rightarrow a \cdot \frac 35=3 \Rightarrow a=5.$

$~e^2=\left(\frac 35\right)^2=\frac{9}{25} \\ \text{or,}~~ 1-\frac{b^2}{a^2}=\frac{9}{25} \\ \text{or,}~~ \frac{b^2}{a^2}=1-\frac{9}{25}=\frac{16}{25} \\ \text{or,}~~ \frac{b^2}{5^2}=\frac{16}{25} \\ \text{or,}~~ \frac{b^2}{25}=\frac{16}{25} \\ \text{or,}~~ b^2=16.$

So, the equation of the ellipse is given by

$~\frac{x^2}{25}+\frac{y^2}{16}=1 \\ \therefore ~ 16x^2+25y^2=400.$

$(iv)~$ whose eccentricity is $~\frac{1}{\sqrt{2}}~$ and length of latus rectum $~3.$

Solution.

$e=\frac{1}{\sqrt{2}} \\ \text{or,}~~ e^2=\frac 12 \\ \text{or,}~~ 1-\frac{b^2}{a^2}=\frac 12 \\ \text{or,}~~ \frac {b^2}{a^2}=1-\frac 12=\frac 12 \rightarrow(1) \\ \text{or,}~~ \frac 1a \cdot \frac{2b^2}{a}=1 \\ \text{or,}~~ \frac 1a \cdot 3=1~[*]\\ \text{or,}~~ a=3.$

Note[*] : $[~ \because ~\text{latus rectum}=\frac{2b^2}{a}=3~]$

Again, $~b^2=\frac{a^2}{2}=\frac 92~[~\text{By (1)}]$

Hence, the equation of the ellipse is

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

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

Solution.

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

Since the ellipse $~(1)~$ passes through the points $~(1,3)~$ and $~(2,1),$

$\frac{1}{a^2}+\frac{9}{b^2}=1 \rightarrow(2),\,\,~ \frac{4}{a^2}+\frac{1}{b^2}=1 \rightarrow(3).$

From $~(2)~$ and $~(3),~$ we get

$4\left(\frac{1}{a^2}+\frac{9}{b^2}\right)-\left(\frac{4}{a^2}+\frac{1}{b^2}\right)=4-1 \\ \text{or,}~~ \frac{35}{b^2}=3 \\ \therefore~ b^2=\frac{35}{3}.$

Similarly, from $~(2)~$ and $~(3),~$ we get

$9\left(\frac{4}{a^2}+\frac{1}{b^2}\right)-\left(\frac{1}{a^2}+\frac{9}{b^2}\right)=9-1 \\ \text{or,}~~\frac{35}{a^2}=8 \\ \therefore~ a^2=\frac{35}{8}.$

Hence, replacing the values of $~a^2~$ and $~b^2~$ , we get from $~(1),$

$\frac{x^2}{\frac{35}{8}}+\frac{y^2}{\frac{35}{3}}=1 \\ \therefore~ 8x^2+3y^2=35\rightarrow(4)$

Finally, from $~(4)~$ we get the required equation of ellipse whose major axis is along $~x-$ axis and minor axis is along $~y$ -axis.

$(vi)~$ whose eccentricity is $~\sqrt{\frac 25}~$ and passes through the point $~(-3,1).$

Solution.

By question,

$~e=\sqrt{\frac 25} \Rightarrow e^2=\frac 25 \\~~~~~~ \therefore1-\frac{b^2}{a^2}=\frac 25 \\~~~~~~ \text{or,}~~ \frac {b^2}{a^2}=1-\frac 25=\frac 35 \\~~~~~~ \text{or,}~~ \frac{1}{a^2}=\frac{3}{5b^2}\rightarrow(1)$

Since the ellipse $~\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\rightarrow(2)~$ passes through the point $~(-3,1),~$ so

$~\frac{(-3)^2}{a^2}+\frac{1^2}{b^2}=1 \\~~~~~~ \text{or,}~~ \frac{9}{a^2}+\frac{1}{b^2}=1 \\~~~~~~ \text{or,}~~ 9 \times \frac{3}{5b^2}+\frac{1}{b^2}=1 ~~[~\text{By (1)}] \\ ~~~~~~\text{or,}~~ \frac{27+5}{5b^2}=1 \\~~~~~~ \text{or,}~~ \frac{32}{5b^2}=1 \\~~~~~~ \therefore~ \frac{1}{b^2}=\frac{5}{32}\rightarrow(3)$

Hence, using $~(1),(2)~$ and $~(3)~$ we get the required equation of ellipse which is

$\frac{3x^2}{32}+\frac{5y^2}{32}=1 \Rightarrow 3x^2+5y^2=32.$

$(vii)~$ whose co-ordinates of vertices are $~(\pm 4,0)~$ and the co-ordinates of the ends of minor axes are $~(0,\pm 2).$

Solution.

By question, the co-ordinates of the vertices $~(\pm a,0)=(\pm 4,0),~$ the co-ordinates of the end of minor axes $~(0,\pm b)=(0,\pm 2).$

$\therefore~ \pm a=\pm 4 \Rightarrow a^2=4^2=16,~ \pm b=\pm 2 \Rightarrow b^2=2^2=4.$

Hence, the equation of the ellipse is

$~\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \\ \text{or,}~~ \frac{x^2}{16}+\frac{y^2}{4}=1 \\ \therefore~ x^2+4y^2=16.$

$(viii)~$ whose distance between the foci is $~2~$ and the distance between the directrices is $~4.$

Solution.

By question, the distance between the foci is $~2ae=2\rightarrow(1)~$ and the distance between the directrices is $~\frac{2a}{e}=4 \rightarrow(2).$

From $~(1)~$ and $~(2)~$ we get,

$~2ae \times \frac{2a}{e}=2 \times 4 \\ \text{or,}~~ a^2=\frac 84=2$

Again, from $~(1)~$ we get,

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

So, the equation of the ellipse is given by

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

$(ix)~$ whose eccentricity is $~\frac{1}{\sqrt{2}}~$ and the sum of squares of major and minor axes is $~24.$

Solution.

By question,

$~(2a)^2+(2b)^2=24 \\ \text{or,}~~ 4(a^2+b^2)=24 \\ \text{or,}~~ a^2+b^2=\frac{24}{4}=6\rightarrow(1)$

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

From $~(1)~$ and $~(2)~$ we get,

$~a^2+\frac{a^2}{2}=6 \\ \text{or,}~~ \frac{3a^2}{2}=6 \\ \text{or,}~~ a^2=4 \\~~\therefore b^2=\frac 42=2.$

Hence, the equation of the ellipse is given by

$~\frac{x^2}{4}+\frac{y^2}{2}=1 \\ \text{or,}~~ x^2+2y^2=4.$

$(x)~$ whose co-ordinates of vertices are $~(\pm 5,0)~$ and the co-ordinates of one focus are $~(4,0).$

Solution.

By question, the equation of the ellipse can be taken in the form $~\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\rightarrow(1)$

The co-ordinates of the vertices $~(\pm a,0)=(\pm 5,0)~$ i.e. $~a=5.$

The co-ordinates of one focus $~(ae,0)=(4,0).$

$\therefore~ ae=4 \\ \text{or,}~~ 5e=4 \\ \text{or,}~~ e=\frac 45 \\ \text{or,}~~ e^2=\left(\frac 45\right)^2 \\ \text{or,}~~ 1-\frac{b^2}{a^2}=\frac{16}{25} \\ \text{or,}~~ \frac{b^2}{a^2}=1-\frac{16}{25} \\ \text{or,}~~ \frac{b^2}{5^2}=\frac{25-16}{25} \\ \text{or,}~~ \frac{b^2}{25}=\frac{9}{25} \\ \text{or,}~~ b^2=9.$

Hence, by $~(1)~$ we get the required equation of ellipse

$~\frac{x^2}{25}+\frac{y^2}{9}=1 \\ \text{or,}~~ 9x^2+25y^2=225.$

$(xi)~$ whose length of latus rectum is $~\frac{18}{5}~$ unit and the co-ordinates of one focus are $~(4,0).$

Solution.

The length of latus rectum

$~\frac{2b^2}{a}=\frac{18}{5} \\ \text{or,}~~ b^2=\frac 95a\rightarrow(1)$

Co-ordinates of one focus $~(ae,0)=(4,0).$

$\therefore~ ae=4 \\ \text{or,}~~ a^2e^2=4^2 \\ \text{or,}~~ a^2\left(1-\frac{b^2}{a^2}\right)=16\\ \text{or,}~~ a^2-b^2=16 \\ \text{or,}~~a^2-\frac 95a=16 \\ \text{or,}~~ 5a^2-9a-80=0 \\ \text{or,}~~ 5a^2-25a+16a-80=0 \\ \text{or,}~~ 5a(a-5)+16(a-5)=0 \\ \text{or,}~~ (5a+16)(a-5)=0 \\ \therefore a=5~~ [\because~ 5a+16 \neq 0]$

So, $~b^2=\frac 95a=\frac 95 \times 5=9.$

Hence, the equation of the ellipse is

$~\frac{x^2}{25}+\frac{b^2}{9}=1 \\ \text{or,}~~ 9x^2+25y^2=225.$

(xii) whose distance between the foci is $~4\sqrt{3}~$ unit and minor axis is of length 4 unit.

Solution.

The length of the minor axis $\,\,\, \rightarrow 2b=4 \Rightarrow b=2\rightarrow(1)$

Distance between the foci $\,\,\, \rightarrow 2ae=4\sqrt{3}$

$\therefore~ (2ae)^2=(4\sqrt{3})^2 \\ \text{or,}~~ 4a^2e^2=16 \times 3 \\ \text{or,}~~ 4a^2\left(1-\frac{b^2}{a^2}\right)=48 \\ \text{or,}~~a^2-b^2=\frac{48}{4} \\ \text{or,}~~ a^2-2^2=12 \\ \text{or,}~~ a^2=12+4 =16.$