Question Statement

From the given information, find the equation of the ellipse in each of the following cases:

(i) Vertices $(-4,6),(-16,6)$ and co-vertices $(-10,2),(-10,10)$ (ii) Centre at $(-8,5)$, vertex $(-8,15)$ and focus is $(-8,5+\sqrt{5})$ (iii) Eccentricity is $\frac{\sqrt{15}}{4}$; centre $(-5,5)$ and co-vertex $(-8,5)$ (iv) Eccentricity is $\frac{3}{4}$, and passes through the point $(3,1)$ with centre at $(0,0)$; major axis along $y$-axis (v) Focus at $(3,4)$, directrix is $x=5$ and eccentricity $\frac{2}{3}$ (vi) Centre at $(1,-1)$; horizontal tangents are $y=7$ and $y=-9$ and vertical tangents are $x=6$ and $x=-4$ (vii) Length of latus rectum is 4; centre at $(0,0)$ and eccentricity is $\frac{1}{3}$; major axis along $y$-axis (viii) Centre at $(1,1)$, eccentricity is $\frac{2}{5}$ and one of the directrices is $x=5$

---

Background and Explanation

An ellipse is the locus of points where the sum of distances from two fixed foci is constant. Its standard equation is $\frac{(x-h{)}^2}{a^2}+\frac{(y-k{)}^2}{b^2}=1$ for a horizontal major axis, or $\frac{(x-h{)}^2}{b^2}+\frac{(y-k{)}^2}{a^2}=1$ for a vertical major axis, where $(h,k)$ is the centre, $a$ is the semi-major axis, and $b$ is the semi-minor axis.

---

Solution

Part (i)

The vertices are $A(-4,6)$ and $A^{\prime }(-16,6)$ and co-vertices are $B(-10,2)$ and $B^{\prime }(-10,10)$. The midpoint of the vertices (or co-vertices) is the centre of the ellipse: $C=(h,k)=\left(\frac{-4-16}{2},\frac{6+6}{2}\right)=(-10,6)$

The distance between the vertices is $2a$: $2a=|A{A}^{\prime }|=\sqrt{(-16+4{)}^2+(6-6{)}^2}=\sqrt{144+0}=12⟹a=6$

The distance between the co-vertices is $2b$: $2b=|B{B}^{\prime }|=\sqrt{(-10+10{)}^2+(10-2{)}^2}=\sqrt{0+64}=8⟹b=4$

Since the $y$-coordinates of both vertices are the same, the ellipse is horizontal. The equation is: $\frac{(x-h{)}^2}{a^2}+\frac{(y-k{)}^2}{b^2}=1$ Substituting the values: $\frac{(x+10{)}^2}{6^2}+\frac{(y-6{)}^2}{4^2}=1⟹\frac{(x+10{)}^2}{36}+\frac{(y-6{)}^2}{16}=1$

Part (ii)

The $x$-coordinates of the centre, vertex, and focus are the same, indicating a vertical ellipse. Given centre $(h,k)=(-8,5)$. The distance between the centre and vertex is $a$: $a=|CV|=\sqrt{(-8+8{)}^2+(15-5{)}^2}=\sqrt{100}=10$

The distance between the centre and the focus is $c$: $c=|CF|=\sqrt{(-8+8{)}^2+(5+\sqrt{5}-5{)}^2}=\sqrt{5}$

Using the relation $b^2=a^2-c^2$: $b^2=10^2-(\sqrt{5}{)}^2=100-5=95⟹b=\sqrt{95}$

The equation for a vertical ellipse is: $\frac{(x-h{)}^2}{b^2}+\frac{(y-k{)}^2}{a^2}=1⟹\frac{(x+8{)}^2}{95}+\frac{(y-5{)}^2}{100}=1$

Part (iii)

The $y$-coordinate of the centre $(-5,5)$ and the co-vertex $(-8,5)$ is the same, meaning the minor axis is horizontal and the ellipse is vertical. The distance between the centre and the co-vertex is $b$: $b=\sqrt{(-5+8{)}^2+(5-5{)}^2}=\sqrt{9}=3$

Using the eccentricity formula $c=ae$ and $b^2=a^2-c^2$: $b^2=a^2(1-e^2)$ Given $e=\frac{\sqrt{15}}{4}$ and $b=3$: $3^2=a^2\left(1-{\left(\frac{\sqrt{15}}{4}\right)}^2\right)⟹9=a^2\left(1-\frac{15}{16}\right)$ $9=a^2\left(\frac{1}{16}\right)⟹a^2=144⟹a=12$

The equation of the vertical ellipse is: $\frac{(x+5{)}^2}{3^2}+\frac{(y-5{)}^2}{12^2}=1⟹\frac{(x+5{)}^2}{9}+\frac{(y-5{)}^2}{144}=1$

Part (iv)

Centre is $(0,0)$ and the major axis is along the $y$-axis. The equation is: $\frac{x^2}{b^2}+\frac{y^2}{a^2}=1$ The ellipse passes through $(3,1)$: $\frac{3^2}{b^2}+\frac{1^2}{a^2}=1⟹\frac{9}{b^2}+\frac{1}{a^2}=1⟹9a^2+b^2=a^2{b}^2⟹9a^2=b^2(a^2-1)$ Using $b^2=a^2(1-e^2)$ with $e=\frac{3}{4}$: $b^2=a^2\left(1-\frac{9}{16}\right)=\frac{7}{16}{a}^2$ Substitute $b^2$ into equation (2): $9a^2=\frac{7}{16}{a}^2(a^2-1)⟹144=7(a^2-1)⟹\frac{144}{7}=a^2-1⟹a^2=\frac{151}{7}$ Now find $b^2$: $b^2=\frac{7}{16}\left(\frac{151}{7}\right)=\frac{151}{16}$ Substitute $a^2$ and $b^2$ into equation (1): $\frac{x^2}{151/16}+\frac{y^2}{151/7}=1⟹\frac{16x^2}{151}+\frac{7y^2}{151}=1⟹16x^2+7y^2=151$

Part (v)

Using the focus-directrix definition of an ellipse: $\frac{\text{Distance from Focus}}{\text{Distance from Directrix}}=e$. Let $P(x,y)$ be a point on the ellipse. Focus $F(3,4)$, Directrix $x-5=0$, $e=2/3$. $\frac{\sqrt{(x-3{)}^2+(y-4{)}^2}}{|x-5|}=\frac{2}{3}$ Squaring both sides: $(x-3{)}^2+(y-4{)}^2=\frac{4}{9}(x-5{)}^2$ $9(x^2-6x+9+y^2-8y+16)=4(x^2-10x+25)$ $9x^2+9y^2-54x-72y+225=4x^2-40x+100$ $5x^2+9y^2-14x-72y+125=0$

Part (vi)

The distance between horizontal tangents $y=7$ and $y=-9$ is $7-(-9)=16$ units. The distance between vertical tangents $x=6$ and $x=-4$ is $6-(-4)=10$ units. Since the vertical distance is greater, the ellipse is vertical. $2a=16⟹a=8,2b=10⟹b=5$ With centre $(1,-1)$, the equation is: $\frac{(x-1{)}^2}{5^2}+\frac{(y+1{)}^2}{8^2}=1⟹\frac{(x-1{)}^2}{25}+\frac{(y+1{)}^2}{64}=1$

Part (vii)

Length of latus rectum $\frac{2b^2}{a}=4⟹b^2=2a$ Given $e=1/3$ and $b^2=a^2(1-e^2)$: $b^2=a^2\left(1-\frac{1}{9}\right)=\frac{8}{9}{a}^2$ Equating (i) and (ii): $\frac{8}{9}{a}^2=2a⟹4a=9⟹a=\frac{9}{4}$ Then $b^2=2\left(\frac{9}{4}\right)=\frac{9}{2}$. Major axis is along the $y$-axis, centre $(0,0)$: $\frac{x^2}{9/2}+\frac{y^2}{(9/4{)}^2}=1⟹\frac{2x^2}{9}+\frac{16y^2}{81}=1⟹18x^2+4y^2=81$

Part (viii)

Directrix $x=5$ is vertical, so the ellipse is horizontal. Centre $(h,k)=(1,1)$. The equation of the directrix is $x=h\pm \frac{a}{e}$. Given $x=5$ and $e=2/5$: $1+\frac{a}{2/5}=5⟹\frac{5a}{2}=4⟹a=\frac{8}{5}$ Find $b^2$ using $b^2=a^2(1-e^2)$: $b^2={\left(\frac{8}{5}\right)}^2\left(1-\frac{4}{25}\right)=\frac{64}{25}\cdot \frac{21}{25}=\frac{1344}{625}$ The equation is: $\frac{(x-1{)}^2}{(8/5{)}^2}+\frac{(y-1{)}^2}{1344/625}=1⟹\frac{25(x-1{)}^2}{64}+\frac{625(y-1{)}^2}{1344}=1$

---

Key Formulas or Methods Used

• Standard Equation (Horizontal): $\frac{(x-h{)}^2}{a^2}+\frac{(y-k{)}^2}{b^2}=1$
• Standard Equation (Vertical): $\frac{(x-h{)}^2}{b^2}+\frac{(y-k{)}^2}{a^2}=1$
• Eccentricity Relation: $b^2=a^2(1-e^2)$ or $c=ae$
• Latus Rectum Length: $\frac{2b^2}{a}$
• Directrix Equation: $x=h\pm \frac{a}{e}$ (horizontal) or $y=k\pm \frac{a}{e}$ (vertical)
• Focus-Directrix Definition: $PF=e\cdot PD$

---

Summary of Steps

1. Identify the centre $(h,k)$ using midpoints or given coordinates.
2. Determine the orientation (horizontal or vertical) by comparing coordinates of vertices, foci, or tangent distances.
3. Calculate $a$, $b$, or $c$ using distances between points or given lengths (latus rectum).
4. Use the eccentricity relation $b^2=a^2(1-e^2)$ to find missing parameters.
5. Substitute $h,k,a^2,$ and $b^2$ into the appropriate standard form equation.