Question

Find the equation of the ellipse in the following cases:
(i) eccentricity $e=\frac{1}{2}$ and foci $(\pm 2,0)$
(ii) eccentricity $e=\frac{2}{3}$ snd length of latus-rectum=5
(iii) eccentricity $e=\frac{1}{2}$ and semi-major axis =4
(iv) eccentricity $e=\frac{1}{2}$ and major axis =12
(v) The ellipse passes through (1,4) and (-6,1).
(vi) Vertices $(\pm 5,0)$, foci $(\pm 4,0)$
(vii) Vertices $(0,\pm 13)$, foci $(0,\pm 5)$
(viii) Vertices $(\pm 6,0)$, foci $(\pm 4,0)$
(ix) Ends of major axis $(\pm 3,0)$, ends of minor axis $(0,\pm 2)$
(x) Ends of major axis $(0,\pm \sqrt{5})$, ends of minor axis $(\pm 1,0)$
(xi) Length of major axis 26, foci $(\pm 5,0)$
(xii) Length of minor axis 16, foci $(0,\pm 6)$
(xiii) Foci $(\pm 3,0)$, a=4

Answer 1

(i) $e=\frac{1}{2}$ and foci $=(\pm 2,0)$
Coordinate of the foci $=(\pm ae,0)$
We have $ae=2$
$\Rightarrow a\times \frac{1}{2}=2$
$\Rightarrow a=4$
Now, $e=\sqrt{1-\frac{b^2}{a^2}}$
$\Rightarrow \frac{1}{2}=\sqrt{1-\frac{b^2}{16}}$
On squaring both sides, we get:
$\frac{1}{4}=\frac{16-b^2}{16}$
$\Rightarrow 64-4b^2=16$
$\Rightarrow b^2=\frac{48}{4}$
$\Rightarrow b^2=12$
Now, $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
$\Rightarrow \frac{x^2}{16}+\frac{y^2}{12}=1$
$\Rightarrow \frac{3x^2+4y^2}{48}=1$
$\Rightarrow 3x^2+4y^2=48$
This is the required equation of the ellipse.
(ii) Let the equation of the required ellipse be $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ ...(i)
The length of latus-rectum = 5
$\therefore \frac{2b^2}{a}=5$ ...(ii)
Now, $b^2=a^2(1-e^2)$
$\Rightarrow \frac{5a}{2}=a^2[1-{\left(\frac{2}{3}\right)}^2]$ $[\because e=\frac{2}{3}]$
$\Rightarrow \frac{5a}{2}=a^2[1-\frac{4}{9}]$
$\Rightarrow \frac{5}{2}=a\left(\frac{5}{9}\right)$
$\Rightarrow \frac{5}{2}\times \frac{5}{9}=a$
$\Rightarrow a=\frac{9}{2}$
$\Rightarrow a=\frac{9}{2}$
$\Rightarrow a^2=\frac{81}{4}$
Putting $a=\frac{9}{2}$ in $b^2=\frac{5a}{2}$, we get
$b^2=\frac{5}{2}\times \frac{9}{2}$
$\Rightarrow b^2=\frac{45}{4}$
substituting $a^2=\frac{81}{4}$ and $b^2=\frac{45}{4}$ in equation (i), we get
$\frac{x^2}{\frac{81}{4}}+\frac{y^2}{\frac{45}{4}}=1$
$\Rightarrow \frac{4x^2}{81}+\frac{4y^2}{45}=1$
$\Rightarrow \frac{4x^2\times 5+4y^2\times 9}{405}=1$
$\Rightarrow 20x^2+36y^2=405$
This is the equation of the required ellipse.
(iii) Let the equation of the required ellipse be
$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ ...(i)
Then, semi-major axis = a
$\therefore a=4$ $[\because \text{semi-major axis=4}]$
$\Rightarrow a^2=16$
Now,
$b^2=a^2(1-e^2)$
$\Rightarrow b^2=16[1-{\left(\frac{1}{2}\right)}^2]$
$\Rightarrow b^2=16[1-\frac{1}{4}]$
$\Rightarrow b^2=16\times \frac{3}{4}$
$\Rightarrow b^2=12$
substituting the value of $a^2$ and $b^2$ in (i), we get
$\frac{x^2}{16}=\frac{y^2}{12}=1$
$\Rightarrow 3x^2+4y^2=48$
This is the required equation of the ellipse.
(iv) Let the equation of the required ellipse be $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ where major axis =2a ...(i)
Now,
$2a=12$ $[\because \text{Major-axis=12}]$
$\Rightarrow a=6$
$\Rightarrow a^2=36$
Now,
$b^2=a^2(1-e^2)$
$\Rightarrow b^2=36(1-\frac{1}{4})$
$\Rightarrow b^2=36\times \frac{3}{4}$
$\Rightarrow b^2=27$
Substituting the value of $a^2$ and $b^2$ in (i), we get
$\frac{x^2}{36}=\frac{y^2}{27}=1$
$\Rightarrow \frac{1}{9}[\frac{x^2}{4}+\frac{y^2}{3}]=1$
$\Rightarrow \frac{3x^2+4y^2}{12}=9$
$\Rightarrow 3x^2+4y^2=108$
This is the equation of the required ellipse.
(v) The ellipse passes through (1,4) and (-6,1)
$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
$\Rightarrow \frac{1}{a^2}+\frac{16}{b^2}=1$
$\Rightarrow \frac{1}{a^2}=\alpha$ and $\frac{1}{b^2}=\beta$
Then $\alpha +16\beta =1$ ...(1)
It also passes through (-6,1)
$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
$\Rightarrow \frac{36}{a^2}+\frac{1}{b^2}=1$
$\Rightarrow 36\alpha +\beta =1$ ...(2)
Solving eqs. (1) and (2), we get:
$\alpha =\frac{3}{115}$ and $\beta =\frac{7}{115}$
Substituting the values, we get:
$\frac{3x^2}{115}+\frac{7y^2}{115}=1$
$\Rightarrow \frac{3x^2+7y^2}{115}=1$
$\Rightarrow 3x^2+7y^2=115$
This is the required equation of ellipse.
(vi) Vertices $(\pm 5,0)$ and focus $(\pm 4,0)$
The coordinate of its vertices and foci are $(\pm a,0)$ and $(\pm ae,0)$, respectively.
i.e., $a=5$ and $ae=4$
$\therefore e=\frac{4}{5}$
Now, $b^2=a^2(1-e^2)$
$\Rightarrow b^2=25(1-\frac{16}{25})$
$\Rightarrow b^2=9$
$\therefore \frac{x^2}{25}+\frac{y^2}{9}=1$
This is the required equation of the ellipse.
(vii) Vertices $(0,\pm 13)$, foci $(0,\pm 5)$
The coordinates of its vertices and foci are $(0,\pm b)$ and $(0,\pm be)$, respectively
i.e. $b=13$ and $be=5$
$\therefore e=\frac{5}{13}$
Now, $a^2=b^2(1-e^2)$
$\Rightarrow a^2=169(1-\frac{25}{169})$
$\Rightarrow a^2=144$
$\therefore \frac{x^2}{144}+\frac{y^2}{169}=1$
This is the required equation of the ellipse.
(viii) Vertices $(\pm 6,0)$, foci $(\pm 4,0)$
The coordinates of its vertices and foci are $(\pm a,0)$ and $(\pm ae,0)$, respectively
i.e. $a=6$ and $ae=4$
$\therefore e=\frac{4}{6}$ or $\frac{2}{3}$
Now, $b^2=a^2(1-e^2)$
$\Rightarrow b^2=36(1-\frac{16}{36})$
$\Rightarrow b^2=20$
$\therefore \frac{x^2}{36}+\frac{y^2}{20}=1$
This is the required equation of the ellipse.
(ix) Ends of major axis $(\pm 3,0)$, ends of minor axis $(0,\pm 2)$
Let the equation of the ellipse be $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
Ends of major axis = $(\pm 3,0)$
Ends of minor axis = $(0,\pm 2)$
But the coordinates of the end points of the major and the minor axes are $(\pm a,0)$ and $(0,\pm b)$
$\therefore a=3$ and $b=2$
Then, $\frac{x^2}{9}+\frac{y^2}{4}=1$
This is the required equation of the ellipse.
(x) Ends of major axis $(0,\pm \sqrt{5})$, ends of minor axis $(\pm 1,0)$
Let the equation of the ellipse be $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
Ends of major axis = $(0,\pm \sqrt{5})$
Ends of minor axis = $(\pm 1,0)$
But the coordinates of the end points of the major and the minor axes are $(\pm a,0)$ and $(0,\pm b)$
$\therefore a=1\text{and}b=\sqrt{5}$
Then, $\frac{x^2}{1}+\frac{y^2}{5}=1$
This is the required equation of the ellipse.
(xi) Length of major axis 26, foci $(\pm 5,0)$
Length of major axis = 26
Foci = $(\pm 5,0)$
We have $2a=26$
Also, $ae=5$
$\Rightarrow e=\frac{5}{13}$
Now, $e=\sqrt{1-\frac{b^2}{a^2}}$
$\Rightarrow \frac{5}{13}=\sqrt{1-\frac{b^2}{169}}$
On squaring both sides, we get:
$\frac{25}{169}=\frac{169-b^2}{169}$
$\Rightarrow b^2=144$
Now, $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
$\Rightarrow \frac{x^2}{169}+\frac{y^2}{144}=1$
This is the required equation of the ellipse.
(xii) Length of minor axis 16 and foci $(0,\pm 6)$ i.e., $2b=16$
$\Rightarrow b=8$
and
$be=6$
$\Rightarrow e=\frac{6}{8}$
Now,
$e=\sqrt{1-\frac{a^2}{b^2}}$
$\Rightarrow \frac{6}{8}=\sqrt{1-\frac{a^2}{64}}$
On squaring both sides, we get:
$\frac{36}{64}=\frac{64-a^2}{64}$
$\Rightarrow a^2=28$
$\frac{x^2}{64}+\frac{y^2}{28}=1$
This is the required equation of the ellipse.
(xiii) Foci $(\pm 3,0)$ and $a=4$
i.e., $ae=3$
$\Rightarrow e=\frac{3}{4}$
Now, $e=\sqrt{1-\frac{b^2}{a^2}}$
$\Rightarrow \frac{3}{4}=\sqrt{1-\frac{b^2}{16}}$
On squaring both sides, we get:
$\frac{9}{16}=\frac{16-b^2}{16}$
$\Rightarrow b^2=7$
$\therefore \frac{x^2}{16}+\frac{y^2}{7}=1$
This is the required equation of the ellipse.