Question

Find the centre, the lengths of the axes, eccentricity, foci of the following ellipse:
(i) $x^2+2y^2-2x+12y+10=0$
(ii) $x^2+4y^2-4x+24y+31=0$
(iii) $4x^2+y^2-8x+2y+1=0$
(iv) $3x^2+4y^2-12x-8y+4=0$
(v) $4x^2+16y^2-24x-32y-12=0$
(vi) $x^2+4y^2-2x=0$

Answer 1

(i) We have,
$x^2+2y^2-2x+12y+10=0$
$\Rightarrow x^2-2x+2y^2+12y+10=0$
$\Rightarrow (x^2-2x+1-1)+2(y^2+6y)+10=0$
$\Rightarrow \left[\right(x-1{)}^2-1]+2[(y^2+2\times y\times 3+9)-9]+10=0$
$\Rightarrow (x-1{)}^2-1+2[(y+3{)}^2-9]+10=0$
$\Rightarrow (x-1{)}^2+2(y+3{)}^2-18-1+10=0$
$\Rightarrow (x-1{)}^2+2(y+3{)}^2-19+10=0$
$\Rightarrow (x-1{)}^2+2(y+3{)}^2-9=0$
$\Rightarrow (x-1{)}^2+2(y+3{)}^2=9$
$\Rightarrow \frac{(x-1{)}^2}{9}+2\frac{(y+3{)}^2}{9}=1$
$\Rightarrow \frac{(x-1{)}^2}{9}+\frac{(y+3{)}^2}{\frac{9}{2}}=1$
$\Rightarrow \frac{(x-1{)}^2}{(3{)}^2}+\frac{(y+3{)}^2}{{\left(\frac{3}{\sqrt{2}}\right)}^2}$ ...(i)
$\therefore$ The coordinates of centre of the ellipse are (1,-3)
Shifting the origin at (1,-3) without rotating the coordinate axes and denoting the new coordinates with respect to the new axes by X and Y, we have Shifting the origin at (1,-1) without rotating the coordinate axes and denoting the new coordinates with respect to the new axes by X and Y, we have x= X+1 and y= Y-1 ...(ii)
Using these rotation, equation (i) reduces to $\frac{X^2}{1^2}+\frac{Y^2}{2^2}=1,$ where
This is of the form
$\frac{X^2}{a^2}+\frac{Y^2}{b^2}=1$
a=1 and b=2
Clearly, b>a, so, the given equation represents on ellipse whose major and minor axes are along X and Y axes respectively.
Length of the axes:
Major-axis = 2b= $2\times 2=4$
and, Minor-axis = 2a= $2\times 1=2$
Eccentricity : The eccentricity e is given by
x=X+1 and y=Y-3 ...(ii)
Using these relations, equation (i) reduces to
$\frac{X^2}{3^2}+\frac{Y^2}{{\left(\frac{3}{\sqrt{2}}\right)}^2}$ ...(iii)
This is of the form
$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,$ where a=3,
and $b=\frac{3}{\sqrt{2}}$
Clearly, a>b, so, the given equation represents an ellipse whose major and minor axes are along X and Y axes respectively.
Length of the axes:
major-axis =$2a=2\times 3=6$
and, minor-axis= $2b=\frac{2\times 3}{\sqrt{2}}=3\sqrt{2}$
Eccentricity : The eccentricity e is given by
$e=\sqrt{1-\frac{b^2}{a^2}}$
$=1-\frac{{\left(\frac{3}{\sqrt{2}}\right)}^2}{3^2}$
$=\sqrt{1-\frac{9}{2\times 9}}$
$=\sqrt{1-\frac{1}{2}}$
$=\frac{1}{\sqrt{2}}$
Foci: The coordinates of the foci with respect to the new axes are given by $(X=\pm ae,Y=0)$
i.e., $(X=\pm \frac{3}{\sqrt{2}},Y=0)$
Putting $X=\pm \frac{3}{\sqrt{2}}$ and Y=0 in equation (ii),
We get,
$X=\pm \frac{3}{\sqrt{2}}+1andy=0-3$
$\Rightarrow X=1\pm \frac{3}{\sqrt{2}}$ and y=-3
So, the coordinates of foci with respect to the old axes are given by $(1\pm \frac{3}{\sqrt{2}},-3)$
(ii) We have,
$x^2+4y^2-4x+24y+31=0$
$\Rightarrow x^2-4x+4(y^2+6y)+31=0$
$\Rightarrow [x^2-2\times x\times 2+2^2-2^2]+4[y^2+2\times 3\times y+3^2-3^2]+31=0$
$\Rightarrow \left[\right(x-2{)}^2-2^2]+4[(y+3{)}^2-9]+31=0$
$\Rightarrow (x-2{)}^2-4+4(y+3{)}^2-36+31=0$
$\Rightarrow (x-2{)}^2+4(y+3{)}^2=9$
$\Rightarrow \frac{(x-2{)}^2}{9}+\frac{4(y+3{)}^2}{9}=1$
$\Rightarrow \frac{(x-2{)}^2}{9}+\frac{(y+3{)}^2}{\frac{9}{4}}$
$\Rightarrow \frac{(x-2{)}^2}{9}+\frac{(y+3{)}^2}{{\left(\frac{3}{2}\right)}^2}$ ...(i)
$\therefore$ The coordinates of centre of the ellipse are (2,-3).
Shifting the origin at(2,-3) without rotating the coordinates axes and denoting the new coordinates with respect to the new axes by X and Y, we have
x=X+2 and y=Y-3 ...(ii)
Using these rotation, equation (i) reduces to
$\frac{X^2}{3^2}+\frac{Y^2}{{\left(\frac{3}{2}\right)}^2}=1$ ...(iii)
This is of the form
$\frac{X^2}{a^2}+\frac{Y^2}{b^2}=1,$ where
a=3 and $b=\frac{3}{2}$
Clearly, a>b, so, the given equation represents on ellipse whose major and minor axes are along X and Y axes respectively.
Length of the axes:
Major-axis = 2a = $2\times 3=6$
and, Minor-axis = 2b $2\times \frac{3}{2}=3$
Eccentricity: The eccentricity e is given by
$e=\sqrt{1-\frac{b^2}{a^2}}$
$=\sqrt{1-\frac{9}{\frac{4}{9}}}$
$=\sqrt{1-\frac{9}{4\times 9}}$
$=\sqrt{1-\frac{1}{4}}$
$=\frac{\sqrt{3}}{2}$
Foci: The coordinates of the foci with respect to the new axes are given by $(X=\pm ae,Y=0)$
i.e., $(X=\pm \frac{3\sqrt{3}}{2},Y=0)$
Putting $X=\pm \frac{3\sqrt{3}}{2}$ and Y=0 in equation (ii),
we get
$X=\pm 2$ and y=0-3
X= +2 and y=-3
$\Rightarrow X=2\pm \frac{3\sqrt{3}}{2}$ and y=-3
So, the coordinates of foci with respect to old axes are given by $(2+\frac{3\sqrt{3}}{2},-3)$
(iii) We have,
$4x^2+y^2-8x+2y+1=0$
$\Rightarrow 4(x^2-2x)+(y^2+2y)+1=0$
$\Rightarrow 4\left[\right(x^2-2x+1)-1]+\left[\right(y^2+2y+1)-1]+1=0$
$\Rightarrow 4\left[\right(x-1{)}^2-1+\left[\right((y+1{)}^2-1)-1]+1=0$
$\Rightarrow 4(x-1{)}^2-4+(y+1{)}^2-1+1=0$
$\Rightarrow 4(x-1{)}^2+(y+1{)}^2-4=0$
$\Rightarrow 4(x-1{)}^2+(y+1{)}^2=4$
$\Rightarrow \frac{(x-1{)}^2}{1}+\frac{(y+1{)}^2}{4}=1$
$\Rightarrow \frac{(x-1{)}^2}{1^2}+\frac{(y+1{)}^2}{2^2}=1$ ...(i)
$\therefore$ The coordinates of centre of the ellipse are (1,-1).
Shifting the origin at (1,-1) without rotating the coordinate axes and denoting the new coordinates with respect to the new axes by X and Y, we have
x= X+1 and y= Y-1 ...(ii)
Using these rotation, equation (i) reduces to $\frac{X^2}{1^2}+\frac{Y^2}{(2{)}^2}=1,$ where
a=1 and b=2
Clearly, b>a, so, the given equation represents on ellipse whose major and minor axes are along X and Y respectively.
Length of the axes:
Major-axis =2b= $2\times 2=4$
and, Minor-axis = 2a = $2\times 1=2$
Eccentricity : The eccentricity e is given by
$e=\sqrt{1-\frac{a^2}{b^2}}$
$=\sqrt{1-\frac{1}{4}}$
$=\sqrt{\frac{4-1}{4}}$
$=\sqrt{\frac{3}{4}}$
$=\sqrt{\frac{\sqrt{3}}{2}}$
Foci: The coordinates of the foci with respcet to the new axes are given by $(X=0,Y=\pm be)$
i.e., $(X=0,Y=\pm \sqrt{3})$
Putting X=0 and $Y=\pm \sqrt{3}$ in equation (iii), we get
$X=0+1andy=\pm \sqrt{3}-1$
$\Rightarrow X=1$ and $y=-1\pm \sqrt{3}$
So, the coordinates of the foci with respect to the old axes are given by $[1,-1\pm \sqrt{3}]$
(iv) We have,
$3x^2+4y^2-12x-8y+4=0$
$\Rightarrow -12x+4y^2-8y+4=0$
$\Rightarrow 3(x^2-4x)+4(y^2-2y)+4=0$
$\Rightarrow 3[x^2-2\times x\times 2+2^2-2^2]+4[y^2-2\times y\times 1^2-1^2]+4=0$
$\Rightarrow 3\left[\right(x-2{)}^2-4]+[(y^2-2y)+4]=0$
$\Rightarrow 3(x-2{)}^2-12+4(y+1{)}^2-4+4=0$
$\Rightarrow 3(x-2{)}^2+4(y-1{)}^2-12=0$
$\Rightarrow 3(x-2{)}^2+4(y-1{)}^2=12$
$\Rightarrow \frac{3(x-2{)}^2}{12}+\frac{4(y-1{)}^2}{12}=1$
$\Rightarrow \frac{(x-2{)}^2}{4}+\frac{(y-1{)}^2}{3}=1$
$\Rightarrow \frac{(x-2{)}^2}{2^2}+\frac{(y-1{)}^2}{(\sqrt{3}{)}^2}=1$
$\therefore$ The coordinates of centre of the ellipse are (2,1)
Shifting the origin at (2,1) without rotating the coordinates axes and denoting the new coordinates with respect to the new axes by X and Y, we have
x=X+2 and y=Y-1 ...(ii)
Using these rotation, equation (i) reduces to
$\frac{X^2}{2^2}+\frac{Y^2}{(\sqrt{3}{)}^2}=1$, where
a=2 and $b=\sqrt{3}$
Clearly, a>b, so, the given equation represents on ellipse whose major and minor axes are along X and Y axes respectively.
Length of the axes:
Major-axis$=2a=2\times 2=4$
and, Minor-axis$=2b=2\times \sqrt{3}=2\sqrt{3}$
Eccentricity: The eccentricity e is givne by
$e=\sqrt{1-\frac{b^2}{a^2}}$
$=\sqrt{1-\frac{3}{4}}$
$=\sqrt{\frac{1}{4}}=\frac{1}{2}$
Foci: The coordinates of the foci with respect to the new axes are given by $(X=\pm ae,Y=0)$
i.e., $(X=\pm 1,Y=0)$
Putting $X=\pm 1$ and $Y=0$ in equation (ii), we get
$X=\pm 1+2$ and $y=0+1$
$\Rightarrow x=2\pm 1$ and $y=1$
So, the coordinates of foci with respect to old axes are given by $(2\pm 1,1)$ i.e., $(3,1)$ and $(1,1)$.
(v) We have,
$4x^2+16y^2-24x-32y-12=0$
$\Rightarrow 4x^2-24x+16y^2-32y-12=0$
$\Rightarrow 4(x^2-6x)+16(y^2-2y)-12=0$
$\Rightarrow 4[x^2-2\times x\times 3+3^2-3^2]+16\left[\right(y^2-2y+1^2-1^2\left)\right]-12=0$
$\Rightarrow 4\left[\right(x-3{)}^2-9]+16[(y-1{)}^2-1]-12=0$
$\Rightarrow 4(x-3{)}^2-36+16(y-1{)}^2-16-12=0$
$\Rightarrow 4(x-3{)}^2+16(y-1{)}^2-64=0$
$\Rightarrow 4(x-3{)}^2+16(y-1{)}^2=64$
$\Rightarrow \frac{4(x-3{)}^2}{64}+\frac{16(y-1{)}^2}{64}=1$
$\Rightarrow \frac{(x-3{)}^2}{16}+\frac{(y-1{)}^2}{4}=1$
$\Rightarrow \frac{(x-3{)}^2}{(4{)}^2}+\frac{(y-1{)}^2}{(2{)}^2}=1$ ...(i)
$\therefore$ The coordinates of centre of the ellipse are (3,1).
Shifting the origin at (3,1) without rotating the coordinate axes and denoting the new coordinates with respect to the new axes by X and Y, we have
$x=X+3$ and $y=Y+1$ ...(ii)
Using these rotation, equation (i) reduces to
$\frac{X^2}{4^2}+\frac{Y^2}{2^2}=1$, where
$a=4$ and $b=2$
Clearly, a>b, so, the given equation represents on ellipse whose major and minor axes are along X and Y axes respectively.
Length of the axes:
Major-axis = $2a=2\times 4=8$
and, Minor-axis = $2b=2\times 2=4$
Eccentricity : The eccentricity e is given by
$e=\sqrt{1-\frac{b^2}{a^2}}$
$=\sqrt{1-\frac{4}{16}}$
$=\sqrt{\frac{3}{4}}$
$=\frac{\sqrt{3}}{2}$
Foci: The coordinates of the foci with respect to the new axes are given by $(X=\pm ae,Y=0)$
i.e., $(X=\pm 2\sqrt{3},Y=0)$
Putting $X=\pm 2\sqrt{3}$ and y=0 in equation (ii), we get
$X=\pm 2\sqrt{3}$ and $y=0+1$
$\Rightarrow x=3\pm 2\sqrt{3}$ and $y=1$
So, the coordinates of foci with respect to the old axes are given by $(3\pm 2\sqrt{3},1)$
(vi) We have,
$x^2+4y^2-2x=0$
$\Rightarrow x^2-2x+4y^2=0$
$\Rightarrow (x^2-2x+1^2-1^2)+4y^2=0$
$\Rightarrow (x-1{)}^2-1+4y^2=0$
$\Rightarrow (x-1{)}^2+4y^2=1$
$\Rightarrow \frac{(x-1{)}^2}{1}+\frac{y^2}{\frac{1}{4}}=1$
$\Rightarrow \frac{(x-1{)}^2}{1^2}+\frac{y^2}{{\left(\frac{1}{2}\right)}^2}=1$ ...(i)
$\therefore$ The coordinates of centre of the ellipse are (1,0).
Shifting the origin axes and denoting the new coordinates with respect to the new axes by X and Y, we have
$x=X+1$ and $y=Y$ ...(ii)
Using these rotation, equation (i) reduces to
$\frac{X^2}{1^2}+\frac{Y^2}{{\left(\frac{1}{2}\right)}^2}=1$, where
$a=1$ and $b=\frac{1}{2}$
Clearly, $a>b$, so, the given equation represents on ellipse whose major and minor axes are along X and Y axes respectively.
Length of the axes:
Major-axis $=2\times a=2\times 1=2$
and, Minor-axis $=2\times b=2\times \frac{1}{2}=1$
Eccentricity : The eccentricity e is given by
$e=\sqrt{1-\frac{b^2}{a^2}}$
$=\sqrt{1-\frac{1}{4}}$
$=\sqrt{1-\frac{1}{4}}$
$=\sqrt{\frac{3}{4}}$
$=\frac{\sqrt{3}}{2}$
Foci: The coordinates of the foci with respect to the new axes are given by $(X=\pm ae,Y=0)$
i.e., $(X=\pm \frac{\sqrt{3}}{2},Y=0)$
Putting $X=\pm \frac{\sqrt{3}}{2}$ and $Y=0$ in equation (ii),
we get
$X=\pm \frac{\sqrt{3}}{2}$ and $Y=0$
$\Rightarrow x=1\pm \frac{\sqrt{3}}{2}$ and $y=0$
So, the coordinates of foci with respect to the old axes are given by $(1\pm \frac{\sqrt{3}}{2},0)$.