Ellipse (S.N.Dey) |Part-7 |Ex-5 Examhoop

In the previous article , we solved few solutions of Long Answer Type Questions of Ellipse Chapter of S.N.Dey mathematics, Class 11. In this chapter, we will solve few more.

$4(i)~$ The ellipse $~\frac{x^2}{a^2}+\frac{y^2}{b^2}=1~$ passes through $~(-3,2)~$ and its eccentricity is $~\sqrt{\frac 35};~$ find the length of its latus rectum.

Solution.

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

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

Again, the eccentricity $(e)$ of the ellipse is

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

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

$\frac{9}{a^2}+\frac{4}{\frac{2a^2}{5}}=1 \\ \text{or,}~~ \frac{9}{a^2}+\frac{10}{a^2}=1 \\ \text{or,}~~ \frac{9+10}{a^2}=1 \\ \text{or,}~~ \frac{19}{a^2}=1 \\ \text{or,}~~ a^2=19.$

From $(2)$ we get, $~b^2=\frac 25 \times 19=\frac{38}{5}.$

So, the length of the latus rectum is

$\frac{2b^2}{a}=2 \times \frac{38}{5}\times \frac{1}{\sqrt{19}}=\frac{2 \times 2 \times 19}{5\sqrt{19}}=\frac{4\sqrt{19}}{5}~~\text{unit}.$

$(ii)~$ The ellipse $~\frac{x^2}{a^2}+\frac{y^2}{b^2}=1~$ passes through the point of intersection of the lines $~7x+13y-87=0~$ and $~5x-8y+7=0~$ and its length of latus rectum is $~\frac{32\sqrt{2}}{5};$ find $~a~$ and $~b.$

Solution.

Two given straight lines are $~7x+13y-87=0\rightarrow(1)~$ and $~5x-8y+7=0\rightarrow(2).$

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

$5(7x+13y-87)-7(5x-8y+7)=0 \\ ~~\text{or,}~~ 35x+65y-435-35x+56y-49=0 \\ ~~\text{or,}~~ 121y-484=0 \\~~ \text{or,}~~ y=\frac{484}{121}=4.$

From $(2)$ we get,

$5x-8 \times 4+7=0 \\ \text{or,}~~ 5x-32+7=0 \\ \text{or,}~~ 5x-25=0 \\ \text{or,}~~ x=\frac{25}{5}=5.$

So, the given ellipse $~\frac{x^2}{a^2}+\frac{y^2}{b^2}=1~$ passes through the point $~(5,4).$

$\therefore~ \frac{5^2}{a^2}+\frac{4^2}{b^2}=1 \\ \text{or,}~~ \frac{25}{a^2}+\frac{16}{b^2}=1 \rightarrow(3)$

So, the length of latus rectum is

$\frac{2b^2}{a}=\frac{32\sqrt{2}}{5}\Rightarrow b^2=\frac{16\sqrt{2}}{5}a\rightarrow(4)$

From $(3)$ and $(4)$ we get,

$\frac{25}{a^2}+\frac{16}{\frac{16\sqrt{2}}{5}a}=1 \\ \text{or,}~~ \frac{25}{a^2}+\frac{5}{\sqrt{2}a}=1 \\ \text{or,}~~ 25\sqrt{2}+5a=\sqrt{2}a^2 \\ \text{or,}~~ \sqrt{2}a^2-5a-25\sqrt{2}=0 \\ \therefore~ a=\frac{5 \pm \sqrt{(-5)^2-4\sqrt{2}(-25\sqrt{2})}}{2\sqrt{2}}=\frac{5 \pm \sqrt{25+100\times 2}}{2\sqrt{2}} \\ \text{or,}~~ a=\frac{5 \pm\sqrt{225}}{2\sqrt{2}}=\frac{5 \pm 15}{2\sqrt{2}} \\ \text{or,}~~ a=\frac{5+15}{2\sqrt{2}}=\frac{20}{2\sqrt{2}}=5\sqrt{2}~~[\because~ a >0]$

$\because~ b^2=\frac{16\sqrt{2}}{5} \times 5\sqrt{2}=32 \\ \text{or,}~~ b=\sqrt{32}=4\sqrt{2}.$

Hence $~a=5\sqrt{2},~~b=4\sqrt{2}.$

For Full Solution PDF of the Ellipse ( S N De-Chhaya Mathematics ), click here.

5. The co-ordinates of the centre and of a vertex of an ellipse are $~(-2,-2)~$ and $~(-2,4)~$ and its eccentricity is $~\frac 23;~$ find the equation of the ellipse.

Solution.

The centre $~(\alpha,\beta)$ and the vertex of the ellipse is $~(-2,-2)~$ and $~(-2,4)~$ respectively. Since the abscissa of the centre and vertex of the ellipse are equal, the major axis of the ellipse is parallel to $~y$ -axis. Hence, the equation of the ellipse can be taken in the form

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

Now, by question, vertex : $(\alpha,\beta+a)=(-2,4),~~(\alpha,\beta)=(-2,-2).$

$\therefore~ \beta+a=4 \\ \text{or,}~~ -2+a=4~~[~\because\beta=-2] \\ \text{or,}~~ -2+a=4 \\ \text{or,}~~ a=4+2=6.$

$\because~ e=\frac 23 \\ \text{or,}~~ e^2=\frac 49 \\ \text{or,}~~ 1-\frac{b^2}{a^2}=\frac 49 \\ \text{or,}~~ \frac{b^2}{a^2}=1-\frac 49=\frac 59 \\ \text{or,}~~ b^2=\frac 59 a^2=\frac 59 \cdot 6^2=\frac 59 \times 36=20.$

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

$\frac{(x+2)^2}{20}+\frac{(y+2)^2}{36}=1.$

6(i) The vertices of an ellipse are $~(-1,2)~$ and $~(9,2).~$ If the eccentricity of an ellipse be $~\frac 45,~$ find its equation.

Solution.

Since the ordinates of the two vertices of the ellipse are equal, the equation of the ellipse can be taken in the form :

$\frac{(x-\alpha)^2}{a^2}+\frac{(y-\beta)^2}{b^2}=1 (~a^2>b^2)\rightarrow(1)$

So, the co-ordinates of the vertices of the ellipse $(1)$ are $~(\alpha \pm a,\beta)=(-1,2),~(9,2).$

$\therefore~ \beta=2, \alpha+a=9 \rightarrow(2),~\alpha-a=-1\rightarrow(3)$

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

$(\alpha+a)+(\alpha-a)=9-1 \Rightarrow \alpha=\frac 82=4.$

Similarly, $~(\alpha+a)-(\alpha-a)=9+1 \Rightarrow a=\frac{10}{2}=5.$

$~e=\frac 45 \Rightarrow e^2=\frac{16}{25} \\ \text{or,}~~ 1-\frac{b^2}{a^2}=\frac{16}{25} \\ \text{or,}~~ \frac{b^2}{a^2}==1-\frac{16}{25} =\frac{9}{25} \\ \text{or,}~~ \frac ba=\frac 35 \\ \text{or,}~~ b=\frac 35\times a=\frac 35 \times 5=3.$

Hence, the equation of the ellipse is given by

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

(ii) Find the equation of the ellipse whose foci are $~(2,3)~$ and $~(-2,3)~$ and whose semi-minor axis is $~\sqrt{5}.$

Solution.

The co-ordinates of foci of the given ellipse are

$(\alpha \pm ae,\beta)=(2,3),~(-2,3).$

So, $~\beta=3,~\alpha=\frac{2+(-2)}{2}=0.$

The distance between the two foci is

$(\alpha+ae)-(\alpha-ae)=2-(-2)=4 \\ \text{or,}~~ 2ae=4 \Rightarrow ae=\frac 42=2\rightarrow(1).$

From the co-ordinates of the foci of the ellipse, we can clearly say that the major axis of the ellipse is parallel to the $~x$ -axis.

So, the equation of the ellipse can be taken in the form :

$\frac{(x-\alpha)^2}{a^2}+\frac{(y-\beta)^2}{b^2}=1~~(a^2>b^2)\rightarrow(2)$

So, the length of semi-minor axis is $~\sqrt{5}~$ unit, so $~b=\sqrt{5}.$

By $(1)$ we get,

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

Hence, using $(2)$ we get the required equation of ellipse

$\frac{(x-0)^2}{9}+\frac{(y-3)^2}{5}=1 \\ \text{or,}~~ 5x^2+9(y-3)^2=45.$

(iii) The eccentricity of an ellipse is $~\frac 23~$ and the co-ordinates of its one focus and the corresponding vertex are $~(1,2)~$ and $~(2,2)~$ respectively. Find the equation of the ellipse . Also find the co-ordinate of the point of intersection of its major axis and the directrix in the same direction.

Solution.

Since the ordinates of the given focus and vertex of the ellipse are equal, hence its major axis is parallel to $~x$ -axis. Therefore, let us assume that the equation of the required ellipse be

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

The co-ordinates of one focus and the corresponding vertex of the ellipse $(1)$ are $~(\alpha+ae,\beta)~$ and $~(\alpha+a,\beta)~$ respectively. By question, $\alpha+ae=1,~~\beta=2.$

$\therefore~ \alpha+ae=1 \\~~~ \text{or,}~~ \alpha + a \times \frac 23=1 \\ ~~~\text{or,}~~ 3\alpha+2a=3 \rightarrow(2)$

$~\alpha+a=2 \rightarrow(3)$

Solving $(2)$ and $(3)$ we get, $~ \alpha=-1,~~a=3.$

$~e=\frac 23 \Rightarrow e^2=\frac 49 \\~~~ \text{or,}~~ 1-\frac{b^2}{a^2}=\frac 49 \\ ~~~\text{or,}~~ \frac{b^2}{a^2}=1-\frac 49=\frac 59 \\ ~~~\therefore~ b^2=\frac 59 \cdot a^2=\frac 59 \cdot 3^2=5.$

So, by $(1)$ we can say that the equation of the ellipse is

$~\frac{(x+1)^2}{9}+\frac{(y-2)^2}{5}=1.$

2nd Part :

The equation of the major axis : $~y=2.$

The directrix of the ellipse :

$x=\alpha+\frac ae=-1+\frac{3}{\frac 23}=-1+\frac 92=\frac 72 \\~~~ \therefore~ x=\frac 72.$

Hence, the co-ordinates of the point of intersection of its major axis and the directrix in the same direction : $~\left(\frac 72,2\right).$

7. The distance of a point of the ellipse $~x^2+3y^2=6~$ from the centre $~2~;$ find the eccentric angle of the point.

Solution.

The equation of the given ellipse is

$x^2+3y^2=6 \\ \text{or,}~~ \frac{x^2}{6}+\frac{y^2}{2}=1 \\ \text{or,}~~\frac{x^2}{(\sqrt{6})^2}+\frac{y^2}{(\sqrt{2})^2}=1 \rightarrow(1)$

Any point on the ellipse $(1)$ can be taken as $~(\sqrt{6}\cos\phi, \sqrt{2}\sin\phi)~$ where $\phi~$ is the eccentric angle of that point.

Now, since the distance of the point of the point of the given ellipse from the centre is $~2,$

$(\sqrt{6}\cos\phi-0)^2+(\sqrt{2}\sin\phi)^2=(2)^2 \\ \text{or,}~~ 6\cos^2\phi+2\sin^2\phi=4 \\ \text{or,}~~ 6\cos^2\phi+2(1-\cos^2\phi)=4 \\ \text{or,}~~ 4\cos^2\phi+2=4 \\ \text{or,}~~ 4\cos^2\phi=4-2 \\ \text{or,}~~ \cos^2\phi=\frac 24 \\ \text{or,}~~ \cos^2\phi=\frac 12 \\ \text{or,}~~ \cos\phi=\pm \frac{1}{\sqrt{2}} \Rightarrow \phi=\frac{\pi}{4},\frac{3\pi}{4},~\frac{5\pi}{4},~\frac{7\pi}{4}.$

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In the previous article , we solved few solutions of Long Answer Type Questions of Ellipse Chapter of S.N.Dey mathematics, Class 11. In this chapter, we will solve few more.

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