Question Statement

Find the centre, vertices, co-vertices, foci, eccentricity, length and equation of the transverse axis, length and equation of the conjugate axis, directrices, and length of latus rectums for the given equations of hyperbolas. Also, sketch the hyperbola in each case:

(i) $\frac{( x - 5 ) ^{2}}{36} - \frac{( y - 4 ) ^{2}}{81} = 1$

(ii) $\frac{( y - 10 ) ^{2}}{10} - \frac{( x - 1 ) ^{2}}{15} = 1$

(iii) $- 16 x^{2} + 9 y^{2} + 32 x + 144 y - 16 = 0$

(iv) $x^{2} - 4 y^{2} + 2 x + 40 y - 135 = 0$

Background and Explanation

To analyze a hyperbola, we first express its equation in standard form: $\frac{( x - h ) ^{2}}{a ^{2}} - \frac{( y - k ) ^{2}}{b ^{2}} = 1$ (horizontal) or $\frac{( y - k ) ^{2}}{a ^{2}} - \frac{( x - h ) ^{2}}{b ^{2}} = 1$ (vertical). The relationship between the semi-axes $a, b$ and the distance to the foci $c$ is given by $c^{2} = a^{2} + b^{2}$ . All features like vertices and foci are then calculated by shifting the standard coordinates by the center $(h, k)$ .

Solution

Part (i)

Given Equation: $\frac{( x - 5 ) ^{2}}{36} - \frac{( y - 4 ) ^{2}}{81} = 1$

1. Identify Parameters: Let $X = x - 5$ and $Y = y - 4$ . The equation becomes $\frac{X ^{2}}{36} - \frac{Y ^{2}}{81} = 1$ .

$a^{2} = 36 \Rightarrow a = 6$
$b^{2} = 81 \Rightarrow b = 9$
Using $c^{2} = a^{2} + b^{2}$ : $c^{2} = 36 + 81 = 117 \Rightarrow c = 117$ This is a horizontal hyperbola.

2. Centre: The centre is at $(X, Y) = (0, 0)$ .

$x - 5 = 0 \Rightarrow x = 5$
$y - 4 = 0 \Rightarrow y = 4$ Centre: $(5, 4)$

3. Vertices: Vertices are $(X, Y) = (\pm a, 0) = (\pm 6, 0)$ .

$x - 5 = 6 \Rightarrow x = 11$
$x - 5 = - 6 \Rightarrow x = - 1$
$y - 4 = 0 \Rightarrow y = 4$ Vertices: $(11, 4)$ and $(- 1, 4)$

4. Co-vertices: Co-vertices are $(X, Y) = (0, \pm b) = (0, \pm 9)$ .

$x - 5 = 0 \Rightarrow x = 5$
$y - 4 = 9 \Rightarrow y = 13$
$y - 4 = - 9 \Rightarrow y = - 5$ Co-vertices: $(5, 13)$ and $(5, - 5)$

5. Foci: Foci are $(X, Y) = (\pm c, 0) = (\pm 117, 0)$ .

$x = 5 \pm 117, y = 4$ Foci: $(5 + 117, 4)$ and $(5 - 117, 4)$

6. Eccentricity: $e = \frac{c}{a} = \frac{117}{6}$

7. Transverse Axis:

Length: $2 a = 2 (6) = 12$ units
Equation: $Y = 0 \Rightarrow y - 4 = 0$

8. Conjugate Axis:

Length: $2 b = 2 (9) = 18$ units
Equation: $X = 0 \Rightarrow x - 5 = 0$

9. Directrices: $X = \pm \frac{a}{e} \Rightarrow x - 5 = \pm \frac{6}{\frac{117}{6}} = \pm \frac{36}{117}$ $x = 5 \pm \frac{36}{117}$

10. Length of Latus Rectum: $L R = \frac{2 b ^{2}}{a} = \frac{2 ( 81 )}{6} = 27$ units

Sketch:

Part (ii)

Given Equation: $\frac{( y - 10 ) ^{2}}{10} - \frac{( x - 1 ) ^{2}}{15} = 1$

1. Identify Parameters: Let $X = x - 1$ and $Y = y - 10$ . The equation is $\frac{Y ^{2}}{10} - \frac{X ^{2}}{15} = 1$ .

$a^{2} = 10 \Rightarrow a = 10$
$b^{2} = 15 \Rightarrow b = 15$
$c^{2} = a^{2} + b^{2} = 10 + 15 = 25 \Rightarrow c = 5$ This is a vertical hyperbola.

2. Centre: $X = 0, Y = 0 \Rightarrow x = 1, y = 10$ . Centre: $(1, 10)$

3. Vertices: $(X, Y) = (0, \pm a) = (0, \pm 10)$ . Vertices: $(1, 10 + 10)$ and $(1, 10 - 10)$

4. Co-vertices: $(X, Y) = (\pm b, 0) = (\pm 15, 0)$ . Co-vertices: $(1 + 15, 10)$ and $(1 - 15, 10)$

5. Foci: $(X, Y) = (0, \pm c) = (0, \pm 5)$ . Foci: $(1, 15)$ and $(1, 5)$

6. Eccentricity: $e = \frac{c}{a} = \frac{5}{10}$

7. Transverse Axis:

Length: $2 a = 210$ units
Equation: $X = 0 \Rightarrow x - 1 = 0$

8. Conjugate Axis:

Length: $2 b = 215$ units
Equation: $Y = 0 \Rightarrow y - 10 = 0$

9. Directrices: $Y = \pm \frac{a}{e} \Rightarrow y - 10 = \pm \frac{10}{\frac{5}{10}} = \pm \frac{10}{5} = \pm 2$ $y = 12$ and $y = 8$

10. Length of Latus Rectum: $L R = \frac{2 b ^{2}}{a} = \frac{2 ( 15 )}{10} = \frac{30}{10}$ units

Sketch:

Part (iii)

Given Equation: $- 16 x^{2} + 9 y^{2} + 32 x + 144 y - 16 = 0$

1. Complete the Square: $9 (y^{2} + 16 y) - 16 (x^{2} - 2 x) = 16$ $9 (y^{2} + 16 y + 64 - 64) - 16 (x^{2} - 2 x + 1 - 1) = 16$ $9 (y + 8)^{2} - 576 - 16 (x - 1)^{2} + 16 = 16$ $9 (y + 8)^{2} - 16 (x - 1)^{2} = 576$ Divide by 576: $\frac{( y + 8 ) ^{2}}{64} - \frac{( x - 1 ) ^{2}}{36} = 1$

2. Identify Parameters:

$a^{2} = 64 \Rightarrow a = 8$
$b^{2} = 36 \Rightarrow b = 6$
$c^{2} = 64 + 36 = 100 \Rightarrow c = 10$ This is a vertical hyperbola.

3. Centre: $x - 1 = 0, y + 8 = 0 \Rightarrow (1, - 8)$

4. Vertices: $(X, Y) = (0, \pm 8) \Rightarrow (1, - 8 \pm 8)$ . Vertices: $(1, 0)$ and $(1, - 16)$

5. Co-vertices: $(X, Y) = (\pm 6, 0) \Rightarrow (1 \pm 6, - 8)$ . Co-vertices: $(7, - 8)$ and $(- 5, - 8)$

6. Foci: $(X, Y) = (0, \pm 10) \Rightarrow (1, - 8 \pm 10)$ . Foci: $(1, 2)$ and $(1, - 18)$

7. Eccentricity: $e = \frac{c}{a} = \frac{10}{8} = \frac{5}{4}$

8. Transverse Axis:

Length: $2 a = 16$ units
Equation: $x - 1 = 0$

9. Conjugate Axis:

Length: $2 b = 12$ units
Equation: $y + 8 = 0$

10. Directrices: $Y = \pm \frac{a}{e} \Rightarrow y + 8 = \pm \frac{8}{5/4} = \pm \frac{32}{5}$ $y = - 8 \pm \frac{32}{5} \Rightarrow 5 y + 72 = 0$ and $5 y + 8 = 0$

11. Length of Latus Rectum: $L R = \frac{2 b ^{2}}{a} = \frac{2 ( 36 )}{8} = 9$ units

Sketch:

Part (iv)

Given Equation: $x^{2} - 4 y^{2} + 2 x + 40 y - 135 = 0$

1. Complete the Square: $(x^{2} + 2 x + 1 - 1) - 4 (y^{2} - 10 y + 25 - 25) = 135$ $(x + 1)^{2} - 1 - 4 (y - 5)^{2} + 100 = 135$ $(x + 1)^{2} - 4 (y - 5)^{2} = 36$ Divide by 36: $\frac{( x + 1 ) ^{2}}{36} - \frac{( y - 5 ) ^{2}}{9} = 1$ (Note: The raw data solution contains a sign discrepancy for $y$ in the final steps; we proceed with the standard derivation from the completed square).

2. Identify Parameters:

$a^{2} = 36 \Rightarrow a = 6$
$b^{2} = 9 \Rightarrow b = 3$
$c^{2} = 36 + 9 = 45 \Rightarrow c = 35$ This is a horizontal hyperbola.

3. Centre: $x + 1 = 0, y - 5 = 0 \Rightarrow (- 1, 5)$

4. Vertices: $(X, Y) = (\pm 6, 0) \Rightarrow (- 1 \pm 6, 5)$ . Vertices: $(5, 5)$ and $(- 7, 5)$

5. Co-vertices: $(X, Y) = (0, \pm 3) \Rightarrow (- 1, 5 \pm 3)$ . Co-vertices: $(- 1, 8)$ and $(- 1, 2)$

6. Foci: $(X, Y) = (\pm 35, 0) \Rightarrow (- 1 \pm 35, 5)$ . Foci: $(- 1 + 35, 5)$ and $(- 1 - 35, 5)$

7. Eccentricity: $e = \frac{c}{a} = \frac{3 5}{6} = \frac{5}{2}$

8. Transverse Axis:

Length: $2 a = 12$ units
Equation: $y - 5 = 0$

9. Conjugate Axis:

Length: $2 b = 6$ units
Equation: $x + 1 = 0$

10. Directrices: $X = \pm \frac{a}{e} \Rightarrow x + 1 = \pm \frac{6}{5 /2} = \pm \frac{12}{5}$ $x = - 1 \pm \frac{12}{5}$

11. Length of Latus Rectum: $L R = \frac{2 b ^{2}}{a} = \frac{2 ( 9 )}{6} = 3$ units

Sketch:

Key Formulas or Methods Used

Standard Form (Horizontal): $\frac{( x - h ) ^{2}}{a ^{2}} - \frac{( y - k ) ^{2}}{b ^{2}} = 1$
Standard Form (Vertical): $\frac{( y - k ) ^{2}}{a ^{2}} - \frac{( x - h ) ^{2}}{b ^{2}} = 1$
Focal Distance: $c = a^{2} + b^{2}$
Eccentricity: $e = \frac{c}{a}$
Latus Rectum Length: $L = \frac{2 b ^{2}}{a}$
Directrices: $X = \pm \frac{a}{e}$ or $Y = \pm \frac{a}{e}$

Summary of Steps

Rearrange: If the equation is in general form, group $x$ and $y$ terms and complete the square.
Standardize: Divide by the constant on the right to set the equation equal to 1.
Identify Orientation: Determine if the hyperbola is horizontal (positive $x^{2}$ term) or vertical (positive $y^{2}$ term).
Extract $a$ and $b$ : $a^{2}$ is always under the positive term; $b^{2}$ is under the negative term.
Calculate $c$ and $e$ : Use $c^{2} = a^{2} + b^{2}$ and $e = c / a$ .
Shift Coordinates: Use the center $(h, k)$ to find the final coordinates for vertices, foci, and axes.
Sketch: Plot the center, draw the central box using $a$ and $b$ , draw asymptotes through the corners, and sketch the curves.

Question Statement

(i) $\frac{( x - 5 ) ^{2}}{36} - \frac{( y - 4 ) ^{2}}{81} = 1$

(ii) $\frac{( y - 10 ) ^{2}}{10} - \frac{( x - 1 ) ^{2}}{15} = 1$

(iii) $- 16 x^{2} + 9 y^{2} + 32 x + 144 y - 16 = 0$

(iv) $x^{2} - 4 y^{2} + 2 x + 40 y - 135 = 0$

Background and Explanation

Solution

Part (i)

Given Equation: $\frac{( x - 5 ) ^{2}}{36} - \frac{( y - 4 ) ^{2}}{81} = 1$

1. Identify Parameters: Let $X = x - 5$ and $Y = y - 4$ . The equation becomes $\frac{X ^{2}}{36} - \frac{Y ^{2}}{81} = 1$ .

$a^{2} = 36 \Rightarrow a = 6$
$b^{2} = 81 \Rightarrow b = 9$
Using $c^{2} = a^{2} + b^{2}$ : $c^{2} = 36 + 81 = 117 \Rightarrow c = 117$ This is a horizontal hyperbola.

2. Centre: The centre is at $(X, Y) = (0, 0)$ .

$x - 5 = 0 \Rightarrow x = 5$
$y - 4 = 0 \Rightarrow y = 4$ Centre: $(5, 4)$

3. Vertices: Vertices are $(X, Y) = (\pm a, 0) = (\pm 6, 0)$ .

$x - 5 = 6 \Rightarrow x = 11$
$x - 5 = - 6 \Rightarrow x = - 1$
$y - 4 = 0 \Rightarrow y = 4$ Vertices: $(11, 4)$ and $(- 1, 4)$

4. Co-vertices: Co-vertices are $(X, Y) = (0, \pm b) = (0, \pm 9)$ .

$x - 5 = 0 \Rightarrow x = 5$
$y - 4 = 9 \Rightarrow y = 13$
$y - 4 = - 9 \Rightarrow y = - 5$ Co-vertices: $(5, 13)$ and $(5, - 5)$

5. Foci: Foci are $(X, Y) = (\pm c, 0) = (\pm 117, 0)$ .

$x = 5 \pm 117, y = 4$ Foci: $(5 + 117, 4)$ and $(5 - 117, 4)$

6. Eccentricity: $e = \frac{c}{a} = \frac{117}{6}$

7. Transverse Axis:

Length: $2 a = 2 (6) = 12$ units
Equation: $Y = 0 \Rightarrow y - 4 = 0$

8. Conjugate Axis:

Length: $2 b = 2 (9) = 18$ units
Equation: $X = 0 \Rightarrow x - 5 = 0$

9. Directrices: $X = \pm \frac{a}{e} \Rightarrow x - 5 = \pm \frac{6}{\frac{117}{6}} = \pm \frac{36}{117}$ $x = 5 \pm \frac{36}{117}$

10. Length of Latus Rectum: $L R = \frac{2 b ^{2}}{a} = \frac{2 ( 81 )}{6} = 27$ units

Sketch:

Part (ii)

Given Equation: $\frac{( y - 10 ) ^{2}}{10} - \frac{( x - 1 ) ^{2}}{15} = 1$

1. Identify Parameters: Let $X = x - 1$ and $Y = y - 10$ . The equation is $\frac{Y ^{2}}{10} - \frac{X ^{2}}{15} = 1$ .

$a^{2} = 10 \Rightarrow a = 10$
$b^{2} = 15 \Rightarrow b = 15$
$c^{2} = a^{2} + b^{2} = 10 + 15 = 25 \Rightarrow c = 5$ This is a vertical hyperbola.

2. Centre: $X = 0, Y = 0 \Rightarrow x = 1, y = 10$ . Centre: $(1, 10)$

3. Vertices: $(X, Y) = (0, \pm a) = (0, \pm 10)$ . Vertices: $(1, 10 + 10)$ and $(1, 10 - 10)$

4. Co-vertices: $(X, Y) = (\pm b, 0) = (\pm 15, 0)$ . Co-vertices: $(1 + 15, 10)$ and $(1 - 15, 10)$

5. Foci: $(X, Y) = (0, \pm c) = (0, \pm 5)$ . Foci: $(1, 15)$ and $(1, 5)$

6. Eccentricity: $e = \frac{c}{a} = \frac{5}{10}$

7. Transverse Axis:

Length: $2 a = 210$ units
Equation: $X = 0 \Rightarrow x - 1 = 0$

8. Conjugate Axis:

Length: $2 b = 215$ units
Equation: $Y = 0 \Rightarrow y - 10 = 0$

9. Directrices: $Y = \pm \frac{a}{e} \Rightarrow y - 10 = \pm \frac{10}{\frac{5}{10}} = \pm \frac{10}{5} = \pm 2$ $y = 12$ and $y = 8$

10. Length of Latus Rectum: $L R = \frac{2 b ^{2}}{a} = \frac{2 ( 15 )}{10} = \frac{30}{10}$ units

Sketch:

Part (iii)

Given Equation: $- 16 x^{2} + 9 y^{2} + 32 x + 144 y - 16 = 0$

2. Identify Parameters:

$a^{2} = 64 \Rightarrow a = 8$
$b^{2} = 36 \Rightarrow b = 6$
$c^{2} = 64 + 36 = 100 \Rightarrow c = 10$ This is a vertical hyperbola.

3. Centre: $x - 1 = 0, y + 8 = 0 \Rightarrow (1, - 8)$

4. Vertices: $(X, Y) = (0, \pm 8) \Rightarrow (1, - 8 \pm 8)$ . Vertices: $(1, 0)$ and $(1, - 16)$

5. Co-vertices: $(X, Y) = (\pm 6, 0) \Rightarrow (1 \pm 6, - 8)$ . Co-vertices: $(7, - 8)$ and $(- 5, - 8)$

6. Foci: $(X, Y) = (0, \pm 10) \Rightarrow (1, - 8 \pm 10)$ . Foci: $(1, 2)$ and $(1, - 18)$

7. Eccentricity: $e = \frac{c}{a} = \frac{10}{8} = \frac{5}{4}$

8. Transverse Axis:

Length: $2 a = 16$ units
Equation: $x - 1 = 0$

9. Conjugate Axis:

Length: $2 b = 12$ units
Equation: $y + 8 = 0$

10. Directrices: $Y = \pm \frac{a}{e} \Rightarrow y + 8 = \pm \frac{8}{5/4} = \pm \frac{32}{5}$ $y = - 8 \pm \frac{32}{5} \Rightarrow 5 y + 72 = 0$ and $5 y + 8 = 0$

11. Length of Latus Rectum: $L R = \frac{2 b ^{2}}{a} = \frac{2 ( 36 )}{8} = 9$ units

Sketch:

Part (iv)

Given Equation: $x^{2} - 4 y^{2} + 2 x + 40 y - 135 = 0$

2. Identify Parameters:

$a^{2} = 36 \Rightarrow a = 6$
$b^{2} = 9 \Rightarrow b = 3$
$c^{2} = 36 + 9 = 45 \Rightarrow c = 35$ This is a horizontal hyperbola.

3. Centre: $x + 1 = 0, y - 5 = 0 \Rightarrow (- 1, 5)$

4. Vertices: $(X, Y) = (\pm 6, 0) \Rightarrow (- 1 \pm 6, 5)$ . Vertices: $(5, 5)$ and $(- 7, 5)$

5. Co-vertices: $(X, Y) = (0, \pm 3) \Rightarrow (- 1, 5 \pm 3)$ . Co-vertices: $(- 1, 8)$ and $(- 1, 2)$

6. Foci: $(X, Y) = (\pm 35, 0) \Rightarrow (- 1 \pm 35, 5)$ . Foci: $(- 1 + 35, 5)$ and $(- 1 - 35, 5)$

7. Eccentricity: $e = \frac{c}{a} = \frac{3 5}{6} = \frac{5}{2}$

8. Transverse Axis:

Length: $2 a = 12$ units
Equation: $y - 5 = 0$

9. Conjugate Axis:

Length: $2 b = 6$ units
Equation: $x + 1 = 0$

10. Directrices: $X = \pm \frac{a}{e} \Rightarrow x + 1 = \pm \frac{6}{5 /2} = \pm \frac{12}{5}$ $x = - 1 \pm \frac{12}{5}$

11. Length of Latus Rectum: $L R = \frac{2 b ^{2}}{a} = \frac{2 ( 9 )}{6} = 3$ units

Sketch:

Key Formulas or Methods Used

Standard Form (Horizontal): $\frac{( x - h ) ^{2}}{a ^{2}} - \frac{( y - k ) ^{2}}{b ^{2}} = 1$
Standard Form (Vertical): $\frac{( y - k ) ^{2}}{a ^{2}} - \frac{( x - h ) ^{2}}{b ^{2}} = 1$
Focal Distance: $c = a^{2} + b^{2}$
Eccentricity: $e = \frac{c}{a}$
Latus Rectum Length: $L = \frac{2 b ^{2}}{a}$
Directrices: $X = \pm \frac{a}{e}$ or $Y = \pm \frac{a}{e}$

Summary of Steps

Rearrange: If the equation is in general form, group $x$ and $y$ terms and complete the square.
Standardize: Divide by the constant on the right to set the equation equal to 1.
Identify Orientation: Determine if the hyperbola is horizontal (positive $x^{2}$ term) or vertical (positive $y^{2}$ term).
Extract $a$ and $b$ : $a^{2}$ is always under the positive term; $b^{2}$ is under the negative term.
Calculate $c$ and $e$ : Use $c^{2} = a^{2} + b^{2}$ and $e = c / a$ .
Shift Coordinates: Use the center $(h, k)$ to find the final coordinates for vertices, foci, and axes.
Sketch: Plot the center, draw the central box using $a$ and $b$ , draw asymptotes through the corners, and sketch the curves.

7.8 Q-1

Question Statement

Background and Explanation

Solution

Part (i)

Part (ii)

Part (iii)

Part (iv)

Key Formulas or Methods Used

Summary of Steps

7.8 Q-1

Question Statement

Background and Explanation

Solution

Part (i)

Part (ii)

Part (iii)

Part (iv)

Key Formulas or Methods Used

Summary of Steps