Question

Find the equation of the ellipse in the following cases:
(i) eccentricity e = $\frac{1}{2}$ and foci (± 2, 0)
(ii) eccentricity e = $\frac{2}{3}$ and 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 (± 5, 0), foci (± 4, 0)
(vii) Vertices (0, ± 13), foci (0, ± 5)
(viii) Vertices (± 6, 0), foci (± 4, 0)
(ix) Ends of major axis (± 3, 0), ends of minor axis (0, ± 2)
(x) Ends of major axis (0, ± $\sqrt{5}$), ends of minor axis (± 1, 0)
(xi) Length of major axis 26, foci (± 5, 0)
(xii) Length of minor axis 16 foci (0, ± 6)
(xiii) Foci (± 3, 0), a = 4

Answer 1

$$\begin{gathered} \left(\mathrm{i}\right)e=\frac{1}{2}\mathrm{and}\mathrm{foci}=(\pm 2,0) \\ \mathrm{Coordinates}\mathrm{of}\mathrm{the}\mathrm{foci}=\left(\pm ae,0\right) \\ \mathrm{We}\mathrm{have}ae=2 \\ \Rightarrow a\times \frac{1}{2}=2 \\ \Rightarrow a=4 \\ \mathrm{Now},e=\sqrt{1-\frac{b^2}{a^2}} \\ \Rightarrow \frac{1}{2}=\sqrt{1-\frac{b^2}{16}} \\ \text{On squaring both sides, we get:} \\ \frac{1}{4}=\frac{16-b^2}{16} \\ \Rightarrow 64-4b^2=16 \\ \Rightarrow b^2=\frac{48}{4} \\ \Rightarrow b^2=12 \\ \mathrm{Now},\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \\ \Rightarrow \frac{x^2}{16}+\frac{y^2}{12}=1 \\ \Rightarrow \frac{3x^2+4y^2}{48}=1 \\ \Rightarrow 3x^2+4y^2=48 \\ \mathrm{This}\text{is the required equation of the ellipse.} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{ii}\right)\mathit{}e=\frac{2}{3}\mathrm{and}\mathrm{l}\text{ength of the latus rectum}=5 \\ \mathrm{We}\mathrm{have}\frac{2b^2}{a}=5 \\ \Rightarrow 2b^2=5a \\ \Rightarrow b^2=\frac{5a}{2} \\ \mathrm{Now},e=\sqrt{1-\frac{b^2}{a^2}} \\ \Rightarrow \frac{2}{3}=\sqrt{1-\frac{{\displaystyle \frac{5a}{2}}}{a^2}} \\ \text{On squaring both sides, we get:} \\ \frac{4}{9}=\frac{2a-5}{2\mathrm{a}} \\ \Rightarrow 8a=18a-45 \\ \Rightarrow a=\frac{9}{2} \\ \therefore b^2=\frac{45}{4} \\ \text{Substituting the values of}{\mathit{\text{a}}}^2\text{and}{\mathit{\text{b}}}^{\mathit{2}}\mathit{\text{,}}\text{we get:} \\ \frac{{\mathrm{x}}^2}{a^2}+\frac{{\mathrm{y}}^2}{b^2}=1 \\ \Rightarrow \frac{4{\mathrm{x}}^2}{81}+\frac{4{\mathrm{y}}^2}{45}=1 \\ \Rightarrow \frac{20{\mathrm{x}}^2+36{\mathrm{y}}^2}{405}=1 \\ \Rightarrow 20{\mathrm{x}}^2+36{\mathrm{y}}^2=405 \\ \text{This is the required equation of the ellipse.} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{iii}\right)\mathit{}\mathit{}e=\frac{1}{2}\mathrm{and}\text{semi major axis}=4 \\ \mathrm{i}.\mathrm{e}.a=4 \\ \mathrm{We}\mathrm{have}e=\sqrt{1-\frac{b^2}{a^2}} \\ \Rightarrow \frac{1}{2}=\sqrt{1-\frac{{\displaystyle b^2}}{16}} \\ \text{On squaring both sides, we get:} \\ \frac{1}{4}=\frac{16-b^2}{16} \\ \Rightarrow 16=64-4b^2 \\ \Rightarrow b^2=12 \\ \text{Substituting the values of}{\mathit{\text{a}}}^2\text{and}{\mathit{\text{b}}}^2\mathit{\text{,}}\text{we get:} \\ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \\ \Rightarrow \frac{x^2}{16}+\frac{y^2}{12}=1 \\ \Rightarrow \frac{3x^2+4y^2}{48}=1 \\ \Rightarrow 3x^2+4y^2=48 \\ \text{This is the required equation of the ellipse.} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{iv}\right)e=\frac{1}{2}\mathrm{and}\text{major axis}=12 \\ \mathrm{i}.\mathrm{e}.,2a=12\mathrm{o}\mathrm{r}a=6 \\ \mathrm{We}\mathrm{have}e=\sqrt{1-\frac{b^2}{a^2}} \\ \Rightarrow \frac{1}{2}=\sqrt{1-\frac{{\displaystyle b^2}}{36}} \\ \text{On squaring both sides, we get:} \\ \frac{1}{4}=\frac{36-b^2}{36} \\ \Rightarrow 36=144-4b^2 \\ \Rightarrow b^2=27 \\ \text{Substituting the values of}{\mathit{\text{a}}}^2\text{and}{\mathit{\text{b}}}^2\mathit{\text{,}}\text{we get:} \\ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \\ \Rightarrow \frac{x^2}{36}+\frac{y^2}{27}=1 \\ \Rightarrow \frac{3x^2+4y^2}{108}=1 \\ \Rightarrow 3x^2+4y^2=108 \\ \text{This is the required equation of the ellipse.} \end{gathered}$$

$$\begin{gathered} \left(v\right)\text{The ellipse passes through}\left(\text{1,4}\right)\text{and}\left(-6,1\right)\text{.} \\ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \\ \Rightarrow \frac{1}{a^2}+\frac{16}{b^2}=1 \\ \text{Let}\frac{1}{a^2}=\alpha \mathrm{a}\mathrm{n}\mathrm{d}\frac{1}{b^2}=\beta \\ \mathrm{Then}\alpha +16\beta =1..\left(1\right) \\ \mathrm{It}\mathrm{also}\text{passes through}\left(-6,1\right)\text{.} \\ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \\ \Rightarrow \frac{36}{a^2}+\frac{1}{b^2}=1 \\ \Rightarrow 36\alpha +\beta =1...\left(2\right) \\ \text{Solving eqs. (1) and (2), we get:} \\ \alpha =\frac{3}{115}\mathrm{and}\beta =\frac{7}{115} \\ \text{Substituting the values, we get:} \\ \frac{3x^2}{115}+\frac{7y^2}{115}=1 \\ \Rightarrow \frac{3x^2+7y^2}{115}=1 \\ \Rightarrow 3x^2+7y^2=115 \\ \text{This is the required equation of ellipse.} \end{gathered}$$
$$\begin{gathered} \left(\mathrm{vi}\right)\text{Vertices}\left(\pm 5,0\right)\text{and focus}\left(\pm 4,0\right) \\ \text{The coordinates of its vertices and foci are}\left(\pm a,0\right)\text{and}\left(\pm ae,0\right)\text{, respectively.} \\ \mathrm{i}.\mathrm{e}.a=5\mathrm{a}\mathrm{n}\mathrm{d}ae=4 \\ \therefore e=\frac{4}{5} \\ \mathrm{Now},b^2=a^2\left(1-e^2\right) \\ \Rightarrow b^2=25\left(1-\frac{16}{25}\right) \\ \Rightarrow b^2=9 \\ \therefore \frac{x^2}{25}+\frac{y^2}{9}=1 \\ \text{This is the required equation of the ellipse.} \end{gathered}$$
$$\begin{gathered} \left(\mathrm{vii}\right)\text{Vertices}\left(0,\pm 13\right)\text{and focus}\left(0,\pm 5\right) \\ \text{The coordinates of its vertices and foci are}\left(0,\pm b\right)\text{and}\left(0,\pm be\right)\text{, respectively.} \\ \mathrm{i}.\mathrm{e}.b=13\mathrm{a}\mathrm{n}\mathrm{d}\mathit{}be=5 \\ \therefore e=\frac{5}{13} \\ \mathrm{Now},a^2=b^2\left(1-e^2\right) \\ \Rightarrow a^2=169\left(1-\frac{25}{169}\right) \\ \Rightarrow a^2=144 \\ \therefore \frac{x^{\mathit{2}}}{144}+\frac{y^{\mathit{2}}}{169}=1 \\ \mathrm{This}\text{is the required equation of the ellipse.} \end{gathered}$$
$$\begin{gathered} \left(\mathrm{viii}\right)\text{Vertices}\left(\pm 6,0\right)\text{and focus}\left(\pm 4,0\right) \\ \text{The coordinates of its vertices and foci are}\left(\pm a,0\right)\text{and}\left(\pm ae,0\right)\text{, respectively.} \\ \mathrm{i}.\mathrm{e}.,a=6\mathrm{a}\mathrm{n}\mathrm{d}ae=4 \\ \therefore e=\frac{4}{6}\mathrm{o}\mathrm{r}\frac{2}{3} \\ \mathrm{Now},b^2=a^2\left(1-{\mathrm{e}}^2\right) \\ \Rightarrow b^2=36\left(1-\frac{16}{36}\right) \\ \Rightarrow b^{\mathit{2}}=20 \\ \therefore \frac{x^2}{36}+\frac{y^2}{20}=1 \\ \mathrm{This}\text{is the required equation of the ellipse.} \\ \left(\mathrm{ix}\right)\mathrm{Let}\mathrm{the}\mathrm{equation}\mathrm{of}the\mathrm{ellipse}\mathrm{be}\frac{x^2}{a^2}+\frac{y^2}{b^2}=1. \\ \text{End of major axis=}\left(\pm 3,0\right) \\ \text{End of minor axis=}\left(0,\pm 2\right) \\ \text{But the coordinates of the end points of the major and the minor axes are (±}\mathit{\text{a}}\text{,0) and}\left(0,\pm b\right)\text{, respectively.} \\ \therefore a=3\mathrm{and}b=2 \\ \mathrm{Then}\frac{x^2}{9}+\frac{y^2}{4}=1 \\ \text{This is the required equation of the ellipse.} \\ \left(\mathrm{x}\right)\mathrm{Let}\mathrm{the}\mathrm{equation}\mathrm{of}the\mathrm{ellipse}\mathrm{be}\frac{x^2}{a^2}+\frac{y^2}{b^2}=1. \\ \text{End of major axis=}\left(0,\pm \sqrt{5}\right) \\ \text{End of minor axis=}\left(\pm 1,0\right) \\ \text{But the coordinates of the end points of the major and the minor axes are (±}\mathit{\text{a}}\text{,0) and}\left(0,\pm b\right)\text{, respectively.} \\ \therefore a=1\mathrm{and}b=\sqrt{5} \\ \mathrm{Then}\frac{x^2}{1}+\frac{y^2}{5}=1 \\ \text{This is the required equation of the ellipse.} \end{gathered}$$
$$\begin{gathered} (\mathrm{xi}) \\ \text{Length of major axis=26} \\ \text{Foci=}\left(\pm 5,0\right) \\ \mathrm{We}\mathrm{have}2a=26 \\ \Rightarrow a=13 \\ \mathrm{Also},ae=5 \\ \Rightarrow e=\frac{5}{13} \\ \mathrm{Now},e=\sqrt{1-\frac{b^2}{a^2}} \\ \Rightarrow \frac{5}{13}=\sqrt{1-\frac{b^2}{169}} \\ \text{On squaring both sides, we get:} \\ \frac{25}{169}=\frac{169-b^2}{169} \\ \Rightarrow b^2=144 \\ \mathrm{Now},\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \\ \Rightarrow \frac{x^2}{169}+\frac{y^2}{144}=1 \\ \text{This is the required equation of the ellipse.} \\ \left(\mathrm{xii}\right)\text{Length of minor axis=16 and foci=}\left(0,\pm 6\right) \\ \mathrm{i}.\mathrm{e}.2b=16 \\ \Rightarrow b=8 \\ \mathrm{and} \\ be=6 \\ \Rightarrow e=\frac{6}{8} \\ \mathrm{Now},e=\sqrt{1-\frac{a^{\mathit{2}}}{b^2}} \\ \Rightarrow \frac{6}{8}=\sqrt{1-\frac{a^2}{64}} \\ \text{On squaring both sides, we get:} \\ \frac{36}{64}=\frac{64-a^2}{64} \\ \Rightarrow a^2=28 \\ \frac{x^2}{64}+\frac{y^2}{28}=1 \\ \text{This is the required equation of the ellipse.} \\ \left(\mathrm{xiii}\right)\mathrm{F}\text{oci=}\left(\pm 3,0\right)\text{and}\mathit{\text{a}}\text{=4} \\ \mathrm{i}.\mathrm{e}.ae=3 \\ \Rightarrow e=\frac{3}{4} \\ \mathrm{Now},e=\sqrt{1-\frac{b^2}{a^2}} \\ \Rightarrow \frac{3}{4}=\sqrt{1-\frac{b^2}{16}} \\ \text{On squaring both sides, we get:} \\ \frac{9}{16}=\frac{16-b^2}{16} \\ \Rightarrow b^2=7 \\ \therefore \frac{x^2}{16}+\frac{y^2}{7}=1 \\ \text{This is the required equation of the ellipse.} \end{gathered}$$