Ellipse (S.N.Dey) |Part-6 |Ex-5 Examhoop

In the previous article , we solved few solutions of Short Answer Type Questions of Ellipse Chapter of S.N.Dey mathematics, Class 11. In this chapter, we will solve few long answer type questions.

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

$(a)~\frac{(x+1)^2}{9}+\frac{(y-2)^2}{5}=1~(b)~ 3x^2+4y^2+6x-8y=5~(c)~9x^2+5y^2-30y=0.$

Solution (a).

We have the given equation of ellipse $~\frac{(x+1)^2}{9}+\frac{(y-2)^2}{5}=1\rightarrow(1)$

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

$~a^2=9 \Rightarrow a =3,~~b^2=5 \Rightarrow b =\sqrt{5},~~ \alpha=-1,~ \beta=2.$

So, the major axis of the ellipse is parallel to the $~x-$ axis and minor axis is parallel to $~y$ -axis.

(i) Centre $~(-1,2)~$

(ii) Vertex $~(\alpha \pm a,\beta)=(-1 \pm 3,2)~$ i.e. $~(2,2)~$ and $~(-4,2).$

(iii) Major axis is given by $~y=2 \Rightarrow y-2=0~$ and minor axis is given by $~x=-1 \Rightarrow x+1=0.$

(iv) Length of major axis $=2a=2 \times 3=6~$ unit and length of minor axis $=2b=2\sqrt{5}~$ unit.

(v) Eccentricity $(e)=\sqrt{1-\frac{b^2}{a^2}}=\sqrt{1-\frac 59}=\frac 23.$

(vi) Length of latus rectum $~\left(\frac{2b^2}{a}\right)=\frac{2 \times 5}{3}=\frac{10}{3}~$ unit.

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

$(\alpha \pm ae, \beta)=(-1 \pm 3 \times \frac 23, 2)=(-1 \pm 2,2)$

i.e. $~(1,2)~$ and $~(-3,2).$

(viii) The equation of the directrix is

$x=\alpha \pm \frac ae=-1 \pm \frac{3}{2/3}=-1 \pm \frac 92 \\ \text{or,}~~ 2x=-2 \pm 9 \Rightarrow 2x=-2+9,~~ 2x=-2-9 \\ \text{or,}~~ 2x=7,~~2x+11=0.$

Solution (b).

We have the given equation of ellipse $~\frac{(x+1)^2}{9}+\frac{(y-2)^2}{5}=1\rightarrow(1)$

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

$~a^2=4 \Rightarrow a =2,~~b^2=3 \Rightarrow b =\sqrt{3}, \alpha=-1,~ \beta=2.$

Since $~a^2>b^2,$ the major axis of the ellipse is parallel to the $~x-$ axis and minor axis is parallel to $~y$ -axis.

(i) Centre $~(\alpha,\beta)=(-1,1)~$

(ii) Vertex $~(\alpha \pm a,\beta)=(-1 \pm 2,1)~$ i.e. $~(1,1)~$ and $~(-3,1).$

(iii) Major axis is given by $~y=\beta ~~\text{or,}~~y=1 \Rightarrow y-1=0~$ and minor axis is given by $~x=\alpha~~\text{or,}~~x=-1 \Rightarrow x+1=0.$

(iv) Length of major axis $=2a=2 \times 2=4~$ unit and length of minor axis $=2b=2\sqrt{3}~$ unit.

(v) Eccentricity $(e)=\sqrt{1-\frac{b^2}{a^2}}=\sqrt{1-\frac 34}=\frac 12.$

(vi) Length of latus rectum $~\left(\frac{2b^2}{a}\right)=\frac{2 \times 3}{2}=3~$ unit.

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

$(\alpha \pm ae, \beta)=(-1 \pm 2 \times \frac 12, 1)=(-1 \pm 1,1)$

i.e. $~(0,1)~$ and $~(-2,1).$

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

Solution (c).

We have the given equation of ellipse

$9x^2+5y^2-30y=0 \\ \text{or,}~~ 9x^2+5(y^2-6y+3^2)=45 \\ \text{or,}~~ \frac{x^2}{5}+\frac{(y-3)^2}{9}=1 \rightarrow(1)$

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

$~a^2=9 \Rightarrow a =3,~~b^2=5 \Rightarrow b =\sqrt{5}, \alpha=0,~ \beta=3.$

Clearly, the major axis of the ellipse is parallel to the $~y-$ axis and minor axis is parallel to $~x$ -axis.

(i) Centre $~(\alpha,\beta)=(0,3)~$

(ii) Vertex $~(\alpha ,\beta \pm a)=(0, 3 \pm 3)~$ i.e. $~(0,6)~$ and $~(0,0).$

(iii) Minor axis is given by $~y=\beta ~~\text{or,}~~y=3 \Rightarrow y-3=0~$ and major axis is given by $~x=\alpha~~\text{or,}~~x=0 .$

(iv) Length of major axis $=2a=2 \times 3=6~$ unit and length of minor axis $=2b=2\sqrt{5}~$ unit.

(v) Eccentricity $(e)=\sqrt{1-\frac{b^2}{a^2}}=\sqrt{1-\frac 59}=\frac 23.$

(vi) Length of latus rectum $~\left(\frac{2b^2}{a}\right)=\frac{2 \times 5}{3}=\frac{10}{3}~$ unit.

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

$(\alpha, \beta \pm ae)=(0, 3 \pm 3 \times \frac 23)=(0, 3 \pm 2)$

i.e. $~(0,5)~$ and $~(0,1).$

(viii) The equation of the directrix is

$y=\beta \pm \frac ae=3 \pm \frac{3}{2/3}=3 \pm \frac 92 \\ \text{or,}~~ 2y=6 \pm 9 \\ \text{or,}~~2y= 6+9,~~ 2y=6-9 \\ \text{or,}~~ 2y-15=0,~~ 2y+3=0.$

(viii) The equation of the directrix is

$x=\alpha \pm \frac ae=-1 \pm \frac{2}{1/2}=-1 \pm 4 \\ \text{or,}~~ x=-1+4=3,~~x=-1-4 \\ \text{or,}~~ x-3=0,~~ x+5=0.$

$2(i)~$ Find the eccentricity, the lengths of latus rectum and the centre of the ellipse $~9x^2+16y^2-54x+64y+1=0.$

Solution.

We have the given equation of ellipse

$~9x^2+16y^2-54x+64y+1=0 \\ \text{or,}~~ 9(x^2-6x+9)+16(y^2+4y+4)=81+64-1 \\ \text{or,}~~ 9(x-3)^2+16(y+2)^2=144 \\ \text{or,}~~ \frac{(x-3)^2}{16}+\frac{(y+2)^2}{9}=1 \rightarrow(1)$

From $~(1),$ we can say that the major axis of the given ellipse is parallel to $~x-$ axis and minor axis of the ellipse is parallel to $~y-$ axis.

The eccentricity of the ellipse is

$(e)=\sqrt{1-\frac{9}{16}}=\sqrt{\frac{16-9}{16}}=\frac{\sqrt{7}}{4}.$

The length of the latus rectum is

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

Finally, the centre of the ellipse is $~(3,-2).$

$(ii)~$ Find the latus rectum, the eccentricity and the co-ordinates of the foci of the ellipse $~9x^2+5y^2+30y=0.$

Solution.

The equation of the ellipse

$~9x^2+5y^2+30y=0 \\ \text{or,}~~ 9x^2+5(y^2+6y+9)=45 \\ \text{or,}~~9x^2+5(y+3)^2=45 \\ \text{or,}~~ \frac{x^2}{5}+\frac{(y+3)^2}{9}=1 \rightarrow(1)$

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

$~a^2=9 \Rightarrow a=3, b^2=5,~\alpha=0,~\beta=-3.$

The length of the latus rectum

$=\frac{2b^2}{a}=2 \times \frac 53=\frac{10}{3}~\text{unit}$ .

The eccentricity of the ellipse is

$(e)=\sqrt{1-\frac{b^2}{a^2}}=\sqrt{1-\frac 59}=\sqrt{\frac{9-5}{9}}=\frac 23.$

The co-ordinates of foci are

$(\alpha, \beta \pm ae)=(0,-3 \pm 3 \times \frac 23)$

i.e., $~(0,-3+2)~$ and $~(0,-3-2)$

i.e., $~(0,-1)~$ and $~(0,-5)$ .

$3.~$ Examine, with reasons, the validity of the following statement :

$x^2+4y^2+2x-24y+33=0~$ represents the equation of an ellipse whose eccentricity is $~\frac{\sqrt{3}}{2}.$

Solution.

$x^2+4y^2+2x-24y+33=0 \\ \text{or,}~~ (x^2+2x+1)+4(y^2-6y+8)=0\\ \text{or,}~~ (x^2+2x+1)+4(y^2-6y+9-1)=0 \\ \text{or,}~~ (x+1)^2+4(y-3)^2=4 \\ \text{or,}~~ \frac{(x+1)^2}{4}+\frac{(y-3)^2}{1}=1 \rightarrow(1)$

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

The eccentricity of $~(1)~$ is

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

So, the given equation represents the equation of the ellipse with eccentricity $~(e)=\frac{\sqrt{3}}{2}.$

Hence, the aforesaid statement is valid.

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In the previous article , we solved few solutions of Short Answer Type Questions of Ellipse Chapter of S.N.Dey mathematics, Class 11. In this chapter, we will solve few long answer type questions.

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