Question Statement

Find the centre, vertices, co-vertices, foci, eccentricity, length and equation of major axis, length and equation of minor axis, directrices, and length of latus rectums for the following equations of ellipse. Also draw the ellipse in each case:

(i) $\frac{x ^{2}}{9} + \frac{y ^{2}}{16} = 1$

(ii) $2 x^{2} + 3 y^{2} = 30$

(iii) $\frac{x ^{2}}{49} + \frac{( y - 3 ) ^{2}}{64} = 1$

(iv) $x^{2} + 9 y^{2} + 6 x - 90 y + 225 = 0$

(v) $16 x^{2} + 9 y^{2} - 32 x + 36 y - 92 = 0$

Background and Explanation

An ellipse is the set of all points where the sum of distances from two fixed points (foci) is constant. The standard form is $\frac{( x - h ) ^{2}}{a ^{2}} + \frac{( y - k ) ^{2}}{b ^{2}} = 1$ , where $(h, k)$ is the center. If the larger denominator is under $y$ , the ellipse is vertical; if under $x$ , it is horizontal. The relationship between the semi-major axis ( $a$ ), semi-minor axis ( $b$ ), and the distance to the foci ( $c$ ) is given by $c^{2} = a^{2} - b^{2}$ .

Solution

Part (i)

Equation: $\frac{x ^{2}}{9} + \frac{y ^{2}}{16} = 1$

The equation can be written as: $\frac{( x - 0 ) ^{2}}{9} + \frac{( y - 0 ) ^{2}}{16} = 1$

Since $16 > 9$ , the major axis is vertical.

$a^{2} = 16 \Rightarrow a = 4$
$b^{2} = 9 \Rightarrow b = 3$
Using $b^{2} = a^{2} - c^{2}$ : $9 = 16 - c^{2} \Rightarrow c^{2} = 7 \Rightarrow c = 7$

Properties:

Centre: $(0, 0)$
Vertices: $(0, \pm a) \Rightarrow (0, 4)$ and $(0, - 4)$
Co-vertices: $(\pm b, 0) \Rightarrow (3, 0)$ and $(- 3, 0)$
Foci: $(0, \pm c) \Rightarrow (0, 7)$ and $(0, - 7)$
Eccentricity: $e = \frac{c}{a} = \frac{7}{4}$
Major Axis: Length $= 2 a = 8$ units. Equation: $x = 0$ ( $y$ -axis).
Minor Axis: Length $= 2 b = 6$ units. Equation: $y = 0$ ( $x$ -axis).
Directrices: $y = \pm \frac{a}{e} = \pm \frac{4}{7 /4} = \pm \frac{16}{7}$
Length of Latus Rectum: $\frac{2 b ^{2}}{a} = \frac{2 ( 9 )}{4} = \frac{9}{2}$ units.

Part (ii)

Equation: $2 x^{2} + 3 y^{2} = 30$

Divide both sides by 30 to get the standard form: $\frac{x ^{2}}{15} + \frac{y ^{2}}{10} = 1$

Since $15 > 10$ , the major axis is horizontal.

$a^{2} = 15 \Rightarrow a = 15$
$b^{2} = 10 \Rightarrow b = 10$
$c^{2} = a^{2} - b^{2} = 15 - 10 = 5 \Rightarrow c = 5$

Properties:

Centre: $(0, 0)$
Vertices: $(\pm a, 0) \Rightarrow (15, 0)$ and $(- 15, 0)$
Co-vertices: $(0, \pm b) \Rightarrow (0, 10)$ and $(0, - 10)$
Foci: $(\pm c, 0) \Rightarrow (5, 0)$ and $(- 5, 0)$
Eccentricity: $e = \frac{c}{a} = \frac{5}{15} = \frac{1}{3}$
Major Axis: Length $= 2 a = 215$ units. Equation: $y = 0$ .
Minor Axis: Length $= 2 b = 210$ units. Equation: $x = 0$ .
Directrices: $x = \pm \frac{a}{e} = \pm \frac{15}{1/ 3} = \pm 45 = \pm 35$
Length of Latus Rectum: $\frac{2 b ^{2}}{a} = \frac{2 ( 10 )}{15} = \frac{20}{15}$ units.

Part (iii)

Equation: $\frac{x ^{2}}{49} + \frac{( y - 3 ) ^{2}}{64} = 1$

Let $X = x$ and $Y = y - 3$ . The equation becomes $\frac{X ^{2}}{49} + \frac{Y ^{2}}{64} = 1$ . Since $64 > 49$ , the ellipse is vertical.

$a^{2} = 64 \Rightarrow a = 8$
$b^{2} = 49 \Rightarrow b = 7$
$c^{2} = a^{2} - b^{2} = 64 - 49 = 15 \Rightarrow c = 15$

Properties:

Centre: $(X, Y) = (0, 0) \Rightarrow (x, y) = (0, 3)$
Vertices: $(X, Y) = (0, \pm 8) \Rightarrow (0, 3 + 8)$ and $(0, 3 - 8) \Rightarrow (0, 11)$ and $(0, - 5)$
Co-vertices: $(X, Y) = (\pm 7, 0) \Rightarrow (7, 3)$ and $(- 7, 3)$
Foci: $(X, Y) = (0, \pm 15) \Rightarrow (0, 3 \pm 15)$
Eccentricity: $e = \frac{c}{a} = \frac{15}{8}$
Major Axis: Length $= 2 a = 16$ units. Equation: $X = 0 \Rightarrow x = 0$ .
Minor Axis: Length $= 2 b = 14$ units. Equation: $Y = 0 \Rightarrow y = 3$ .
Directrices: $Y = \pm \frac{a}{e} \Rightarrow y - 3 = \pm \frac{64}{15} \Rightarrow y = 3 \pm \frac{64}{15}$
Length of Latus Rectum: $\frac{2 b ^{2}}{a} = \frac{2 ( 49 )}{8} = \frac{49}{4}$ units.

Part (iv)

Equation: $x^{2} + 9 y^{2} + 6 x - 90 y + 225 = 0$

Complete the square for $x$ and $y$ : $(x^{2} + 6 x + 9) + 9 (y^{2} - 10 y + 25) = - 225 + 9 + 225$ $(x + 3)^{2} + 9 (y - 5)^{2} = 9$ Divide by 9: $\frac{( x + 3 ) ^{2}}{9} + \frac{( y - 5 ) ^{2}}{1} = 1$

Let $X = x + 3$ and $Y = y - 5$ .

$a^{2} = 9 \Rightarrow a = 3$
$b^{2} = 1 \Rightarrow b = 1$
$c^{2} = 9 - 1 = 8 \Rightarrow c = 8 = 22$

Properties:

Centre: $X = 0, Y = 0 \Rightarrow (- 3, 5)$
Vertices: $X = \pm 3, Y = 0 \Rightarrow (- 3 \pm 3, 5) \Rightarrow (0, 5)$ and $(- 6, 5)$
Co-vertices: $X = 0, Y = \pm 1 \Rightarrow (- 3, 5 \pm 1) \Rightarrow (- 3, 6)$ and $(- 3, 4)$
Foci: $X = \pm 22, Y = 0 \Rightarrow (- 3 \pm 22, 5)$
Eccentricity: $e = \frac{c}{a} = \frac{2 2}{3}$
Major Axis: Length $= 6$ units. Equation: $Y = 0 \Rightarrow y = 5$ .
Minor Axis: Length $= 2$ units. Equation: $X = 0 \Rightarrow x = - 3$ .
Directrices: $X = \pm \frac{a}{e} \Rightarrow x + 3 = \pm \frac{9}{2 2} \Rightarrow x = - 3 \pm \frac{9}{2 2}$
Length of Latus Rectum: $\frac{2 b ^{2}}{a} = \frac{2 ( 1 )}{3} = \frac{2}{3}$ units.

Part (v)

Equation: $16 x^{2} + 9 y^{2} - 32 x + 36 y - 92 = 0$

Complete the square: $16 (x^{2} - 2 x + 1) + 9 (y^{2} + 4 y + 4) = 92 + 16 + 36$ $16 (x - 1)^{2} + 9 (y + 2)^{2} = 144$ Divide by 144: $\frac{( x - 1 ) ^{2}}{9} + \frac{( y + 2 ) ^{2}}{16} = 1$

Let $X = x - 1$ and $Y = y + 2$ .

$a^{2} = 16 \Rightarrow a = 4$
$b^{2} = 9 \Rightarrow b = 3$
$c^{2} = 16 - 9 = 7 \Rightarrow c = 7$

Properties:

Centre: $X = 0, Y = 0 \Rightarrow (1, - 2)$
Vertices: $X = 0, Y = \pm 4 \Rightarrow (1, - 2 \pm 4) \Rightarrow (1, 2)$ and $(1, - 6)$
Co-vertices: $X = \pm 3, Y = 0 \Rightarrow (1 \pm 3, - 2) \Rightarrow (4, - 2)$ and $(- 2, - 2)$
Foci: $X = 0, Y = \pm 7 \Rightarrow (1, - 2 \pm 7)$
Eccentricity: $e = \frac{7}{4}$
Major Axis: Length $= 8$ units. Equation: $X = 0 \Rightarrow x = 1$ .
Minor Axis: Length $= 6$ units. Equation: $Y = 0 \Rightarrow y = - 2$ .
Directrices: $Y = \pm \frac{a}{e} \Rightarrow y + 2 = \pm \frac{16}{7} \Rightarrow y = - 2 \pm \frac{16}{7}$
Length of Latus Rectum: $\frac{2 b ^{2}}{a} = \frac{2 ( 9 )}{4} = \frac{9}{2}$ units.

Key Formulas or Methods Used

Standard Form: $\frac{( x - h ) ^{2}}{a ^{2}} + \frac{( y - k ) ^{2}}{b ^{2}} = 1$ (Horizontal) or $\frac{( x - h ) ^{2}}{b ^{2}} + \frac{( y - k ) ^{2}}{a ^{2}} = 1$ (Vertical).
Focal Distance: $c = a^{2} - b^{2}$ .
Eccentricity: $e = \frac{c}{a}$ .
Latus Rectum Length: $L R = \frac{2 b ^{2}}{a}$ .
Directrices: $x = h \pm \frac{a}{e}$ (Horizontal) or $y = k \pm \frac{a}{e}$ (Vertical).
Completing the Square: Used to convert general quadratic forms to standard form.

Summary of Steps

Standardize: Convert the given equation into the standard form $\frac{( x - h ) ^{2}}{\dots} + \frac{( y - k ) ^{2}}{\dots} = 1$ by dividing or completing the square.
Identify $a$ and $b$ : The larger denominator is $a^{2}$ , and the smaller is $b^{2}$ .
Determine Orientation: If $a^{2}$ is under $x$ , it's horizontal; if under $y$ , it's vertical.
Calculate $c$ and $e$ : Use $c = a^{2} - b^{2}$ and $e = c / a$ .
Find Coordinates: Apply the shifts $(h, k)$ to the basic properties (vertices, foci, etc.) of an ellipse centered at the origin.
Calculate Lengths: Use $2 a$ for major axis, $2 b$ for minor axis, and $2 b^{2} / a$ for latus rectum.

Question Statement

(i) $\frac{x ^{2}}{9} + \frac{y ^{2}}{16} = 1$

(ii) $2 x^{2} + 3 y^{2} = 30$

(iii) $\frac{x ^{2}}{49} + \frac{( y - 3 ) ^{2}}{64} = 1$

(iv) $x^{2} + 9 y^{2} + 6 x - 90 y + 225 = 0$

(v) $16 x^{2} + 9 y^{2} - 32 x + 36 y - 92 = 0$

Background and Explanation

Solution

Part (i)

Equation: $\frac{x ^{2}}{9} + \frac{y ^{2}}{16} = 1$

The equation can be written as: $\frac{( x - 0 ) ^{2}}{9} + \frac{( y - 0 ) ^{2}}{16} = 1$

Since $16 > 9$ , the major axis is vertical.

$a^{2} = 16 \Rightarrow a = 4$
$b^{2} = 9 \Rightarrow b = 3$
Using $b^{2} = a^{2} - c^{2}$ : $9 = 16 - c^{2} \Rightarrow c^{2} = 7 \Rightarrow c = 7$

Properties:

Centre: $(0, 0)$
Vertices: $(0, \pm a) \Rightarrow (0, 4)$ and $(0, - 4)$
Co-vertices: $(\pm b, 0) \Rightarrow (3, 0)$ and $(- 3, 0)$
Foci: $(0, \pm c) \Rightarrow (0, 7)$ and $(0, - 7)$
Eccentricity: $e = \frac{c}{a} = \frac{7}{4}$
Major Axis: Length $= 2 a = 8$ units. Equation: $x = 0$ ( $y$ -axis).
Minor Axis: Length $= 2 b = 6$ units. Equation: $y = 0$ ( $x$ -axis).
Directrices: $y = \pm \frac{a}{e} = \pm \frac{4}{7 /4} = \pm \frac{16}{7}$
Length of Latus Rectum: $\frac{2 b ^{2}}{a} = \frac{2 ( 9 )}{4} = \frac{9}{2}$ units.

Part (ii)

Equation: $2 x^{2} + 3 y^{2} = 30$

Divide both sides by 30 to get the standard form: $\frac{x ^{2}}{15} + \frac{y ^{2}}{10} = 1$

Since $15 > 10$ , the major axis is horizontal.

$a^{2} = 15 \Rightarrow a = 15$
$b^{2} = 10 \Rightarrow b = 10$
$c^{2} = a^{2} - b^{2} = 15 - 10 = 5 \Rightarrow c = 5$

Properties:

Centre: $(0, 0)$
Vertices: $(\pm a, 0) \Rightarrow (15, 0)$ and $(- 15, 0)$
Co-vertices: $(0, \pm b) \Rightarrow (0, 10)$ and $(0, - 10)$
Foci: $(\pm c, 0) \Rightarrow (5, 0)$ and $(- 5, 0)$
Eccentricity: $e = \frac{c}{a} = \frac{5}{15} = \frac{1}{3}$
Major Axis: Length $= 2 a = 215$ units. Equation: $y = 0$ .
Minor Axis: Length $= 2 b = 210$ units. Equation: $x = 0$ .
Directrices: $x = \pm \frac{a}{e} = \pm \frac{15}{1/ 3} = \pm 45 = \pm 35$
Length of Latus Rectum: $\frac{2 b ^{2}}{a} = \frac{2 ( 10 )}{15} = \frac{20}{15}$ units.

Part (iii)

Equation: $\frac{x ^{2}}{49} + \frac{( y - 3 ) ^{2}}{64} = 1$

Let $X = x$ and $Y = y - 3$ . The equation becomes $\frac{X ^{2}}{49} + \frac{Y ^{2}}{64} = 1$ . Since $64 > 49$ , the ellipse is vertical.

$a^{2} = 64 \Rightarrow a = 8$
$b^{2} = 49 \Rightarrow b = 7$
$c^{2} = a^{2} - b^{2} = 64 - 49 = 15 \Rightarrow c = 15$

Properties:

Centre: $(X, Y) = (0, 0) \Rightarrow (x, y) = (0, 3)$
Vertices: $(X, Y) = (0, \pm 8) \Rightarrow (0, 3 + 8)$ and $(0, 3 - 8) \Rightarrow (0, 11)$ and $(0, - 5)$
Co-vertices: $(X, Y) = (\pm 7, 0) \Rightarrow (7, 3)$ and $(- 7, 3)$
Foci: $(X, Y) = (0, \pm 15) \Rightarrow (0, 3 \pm 15)$
Eccentricity: $e = \frac{c}{a} = \frac{15}{8}$
Major Axis: Length $= 2 a = 16$ units. Equation: $X = 0 \Rightarrow x = 0$ .
Minor Axis: Length $= 2 b = 14$ units. Equation: $Y = 0 \Rightarrow y = 3$ .
Directrices: $Y = \pm \frac{a}{e} \Rightarrow y - 3 = \pm \frac{64}{15} \Rightarrow y = 3 \pm \frac{64}{15}$
Length of Latus Rectum: $\frac{2 b ^{2}}{a} = \frac{2 ( 49 )}{8} = \frac{49}{4}$ units.

Part (iv)

Equation: $x^{2} + 9 y^{2} + 6 x - 90 y + 225 = 0$

Let $X = x + 3$ and $Y = y - 5$ .

$a^{2} = 9 \Rightarrow a = 3$
$b^{2} = 1 \Rightarrow b = 1$
$c^{2} = 9 - 1 = 8 \Rightarrow c = 8 = 22$

Properties:

Centre: $X = 0, Y = 0 \Rightarrow (- 3, 5)$
Vertices: $X = \pm 3, Y = 0 \Rightarrow (- 3 \pm 3, 5) \Rightarrow (0, 5)$ and $(- 6, 5)$
Co-vertices: $X = 0, Y = \pm 1 \Rightarrow (- 3, 5 \pm 1) \Rightarrow (- 3, 6)$ and $(- 3, 4)$
Foci: $X = \pm 22, Y = 0 \Rightarrow (- 3 \pm 22, 5)$
Eccentricity: $e = \frac{c}{a} = \frac{2 2}{3}$
Major Axis: Length $= 6$ units. Equation: $Y = 0 \Rightarrow y = 5$ .
Minor Axis: Length $= 2$ units. Equation: $X = 0 \Rightarrow x = - 3$ .
Directrices: $X = \pm \frac{a}{e} \Rightarrow x + 3 = \pm \frac{9}{2 2} \Rightarrow x = - 3 \pm \frac{9}{2 2}$
Length of Latus Rectum: $\frac{2 b ^{2}}{a} = \frac{2 ( 1 )}{3} = \frac{2}{3}$ units.

Part (v)

Equation: $16 x^{2} + 9 y^{2} - 32 x + 36 y - 92 = 0$

Complete the square: $16 (x^{2} - 2 x + 1) + 9 (y^{2} + 4 y + 4) = 92 + 16 + 36$ $16 (x - 1)^{2} + 9 (y + 2)^{2} = 144$ Divide by 144: $\frac{( x - 1 ) ^{2}}{9} + \frac{( y + 2 ) ^{2}}{16} = 1$

Let $X = x - 1$ and $Y = y + 2$ .

$a^{2} = 16 \Rightarrow a = 4$
$b^{2} = 9 \Rightarrow b = 3$
$c^{2} = 16 - 9 = 7 \Rightarrow c = 7$

Properties:

Centre: $X = 0, Y = 0 \Rightarrow (1, - 2)$
Vertices: $X = 0, Y = \pm 4 \Rightarrow (1, - 2 \pm 4) \Rightarrow (1, 2)$ and $(1, - 6)$
Co-vertices: $X = \pm 3, Y = 0 \Rightarrow (1 \pm 3, - 2) \Rightarrow (4, - 2)$ and $(- 2, - 2)$
Foci: $X = 0, Y = \pm 7 \Rightarrow (1, - 2 \pm 7)$
Eccentricity: $e = \frac{7}{4}$
Major Axis: Length $= 8$ units. Equation: $X = 0 \Rightarrow x = 1$ .
Minor Axis: Length $= 6$ units. Equation: $Y = 0 \Rightarrow y = - 2$ .
Directrices: $Y = \pm \frac{a}{e} \Rightarrow y + 2 = \pm \frac{16}{7} \Rightarrow y = - 2 \pm \frac{16}{7}$
Length of Latus Rectum: $\frac{2 b ^{2}}{a} = \frac{2 ( 9 )}{4} = \frac{9}{2}$ units.

Key Formulas or Methods Used

Standard Form: $\frac{( x - h ) ^{2}}{a ^{2}} + \frac{( y - k ) ^{2}}{b ^{2}} = 1$ (Horizontal) or $\frac{( x - h ) ^{2}}{b ^{2}} + \frac{( y - k ) ^{2}}{a ^{2}} = 1$ (Vertical).
Focal Distance: $c = a^{2} - b^{2}$ .
Eccentricity: $e = \frac{c}{a}$ .
Latus Rectum Length: $L R = \frac{2 b ^{2}}{a}$ .
Directrices: $x = h \pm \frac{a}{e}$ (Horizontal) or $y = k \pm \frac{a}{e}$ (Vertical).
Completing the Square: Used to convert general quadratic forms to standard form.

Summary of Steps

Standardize: Convert the given equation into the standard form $\frac{( x - h ) ^{2}}{\dots} + \frac{( y - k ) ^{2}}{\dots} = 1$ by dividing or completing the square.
Identify $a$ and $b$ : The larger denominator is $a^{2}$ , and the smaller is $b^{2}$ .
Determine Orientation: If $a^{2}$ is under $x$ , it's horizontal; if under $y$ , it's vertical.
Calculate $c$ and $e$ : Use $c = a^{2} - b^{2}$ and $e = c / a$ .
Find Coordinates: Apply the shifts $(h, k)$ to the basic properties (vertices, foci, etc.) of an ellipse centered at the origin.
Calculate Lengths: Use $2 a$ for major axis, $2 b$ for minor axis, and $2 b^{2} / a$ for latus rectum.

7.6 Q-1

Question Statement

Background and Explanation

Solution

Part (i)

Part (ii)

Part (iii)

Part (iv)

Part (v)

Key Formulas or Methods Used

Summary of Steps

7.6 Q-1

Question Statement

Background and Explanation

Solution

Part (i)

Part (ii)

Part (iii)

Part (iv)

Part (v)

Key Formulas or Methods Used

Summary of Steps