Question Statement

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

(i) Vertices: $(8, 14)$ , $(8, - 10)$ ; conjugate axis of length $6$ units.

(ii) Vertices: $(- 2, \frac{5}{2})$ , $(- 16, \frac{5}{2})$ and end points of conjugate axis: $(- 9, \frac{15}{2})$ , $(- 9, - \frac{5}{2})$ .

(iii) Vertices: $(- 5, 1)$ , $(- 5, - 7)$ and foci: $(- 5, - 3 + 97)$ , $(- 5, - 3 - 97)$ .

(iv) Foci: $(8, - 5 + 53)$ , $(8, - 5 - 53)$ and end points of conjugate axis: $(15, - 5)$ , $(1, - 5)$ .

(v) Vertices: $(0, 9)$ , $(0, - 9)$ and passing through the point $(8, 15)$ .

(vi) Eccentricity is $\frac{5}{4}$ , centre at $(1, 1)$ , passes through $(- 7, 2)$ , and the focal axis is horizontal.

(vii) Focus: $(5, 3)$ ; directrix: $y = - 2$ ; eccentricity: $\frac{5}{3}$ .

(viii) Centre at $(0, 0)$ , length of latus rectum is $5$ , eccentricity is $\frac{5}{4}$ , and the conjugate axis is along the $x$ -axis.

Background and Explanation

A hyperbola is the set of all points where the absolute difference of distances to two fixed points (foci) is constant. The standard form depends on whether the transverse axis is horizontal $\frac{( x - h ) ^{2}}{a ^{2}} - \frac{( y - k ) ^{2}}{b ^{2}} = 1$ or vertical $\frac{( y - k ) ^{2}}{a ^{2}} - \frac{( x - h ) ^{2}}{b ^{2}} = 1$ . Key relationships include $c^{2} = a^{2} + b^{2}$ , eccentricity $e = \frac{c}{a}$ , and the length of the latus rectum $\frac{2 b ^{2}}{a}$ .

Solution

Part (i)

Vertices: $A (8, 14)$ and $A^{'} (8, - 10)$ ; Conjugate axis length: $6$ .

Find the Centre: The centre $C (h, k)$ is the midpoint of the vertices. $C = (\frac{8 + 8}{2}, \frac{14 - 10}{2}) = (8, 2)$
Find $a$ : The distance between vertices is $2 a$ . $2 a = (8 - 8)^{2} + (14 - (- 10))^{2} = 0 + 2 4^{2} = 24 \Rightarrow a = 12$
Find $b$ : The length of the conjugate axis is $2 b = 6$ , so $b = 3$ .
Determine Orientation: Since the $x$ -coordinates of the vertices are the same, the hyperbola is vertical.
Equation: $\frac{( y - k ) ^{2}}{a ^{2}} - \frac{( x - h ) ^{2}}{b ^{2}} = 1 \Rightarrow \frac{( y - 2 ) ^{2}}{1 2 ^{2}} - \frac{( x - 8 ) ^{2}}{3 ^{2}} = 1$ $\frac{( y - 2 ) ^{2}}{144} - \frac{( x - 8 ) ^{2}}{9} = 1$

Part (ii)

Vertices: $A (- 2, \frac{5}{2})$ , $A^{'} (- 16, \frac{5}{2})$ ; Co-vertices: $B (- 9, \frac{15}{2})$ , $B^{'} (- 9, - \frac{5}{2})$ .

Find the Centre: Midpoint of vertices. $C = (\frac{- 2 - 16}{2}, \frac{5/2 + 5/2}{2}) = (- 9, \frac{5}{2})$
Find $a$ : $2 a = ∣ A A^{'} ∣ = (- 2 + 16)^{2} + (5/2 - 5/2)^{2} = 14 \Rightarrow a = 7$ .
Find $b$ : $2 b = ∣ B B^{'} ∣ = (- 9 + 9)^{2} + (\frac{15}{2} - (- \frac{5}{2}))^{2} = 0 + 1 0^{2} = 10 \Rightarrow b = 5$ .
Determine Orientation: The $y$ -coordinates of the vertices are the same, so the hyperbola is horizontal.
Equation: $\frac{( x - h ) ^{2}}{a ^{2}} - \frac{( y - k ) ^{2}}{b ^{2}} = 1 \Rightarrow \frac{( x + 9 ) ^{2}}{7 ^{2}} - \frac{( y - 5/2 ) ^{2}}{5 ^{2}} = 1$ $\frac{( x + 9 ) ^{2}}{49} - \frac{( y - 5/2 ) ^{2}}{25} = 1$

Part (iii)

Vertices: $(- 5, 1)$ , $(- 5, - 7)$ ; Foci: $(- 5, - 3 \pm 97)$ .

Find the Centre: Midpoint of vertices. $C = (\frac{- 5 - 5}{2}, \frac{1 - 7}{2}) = (- 5, - 3)$
Find $a$ : $2 a = (- 5 + 5)^{2} + (1 + 7)^{2} = 8 \Rightarrow a = 4$ .
Find $c$ : Distance between foci is $2 c$ . $2 c = 0^{2} + ((- 3 + 97) - (- 3 - 97))^{2} = 297 \Rightarrow c = 97$
Find $b$ : $b^{2} = c^{2} - a^{2} = 97 - 16 = 81 \Rightarrow b = 9$ .
Determine Orientation: $x$ -coordinates are same; hyperbola is vertical.
Equation: $\frac{( y + 3 ) ^{2}}{16} - \frac{( x + 5 ) ^{2}}{81} = 1$

Part (iv)

Foci: $(8, - 5 \pm 53)$ ; Co-vertices: $(15, - 5)$ , $(1, - 5)$ .

Find the Centre: Midpoint of co-vertices. $C = (\frac{15 + 1}{2}, \frac{- 5 - 5}{2}) = (8, - 5)$
Find $c$ : $2 c = ∣ F F^{'} ∣ = 253 \Rightarrow c = 53$ .
Find $b$ : $2 b = ∣ B B^{'} ∣ = (15 - 1)^{2} + 0 = 14 \Rightarrow b = 7$ .
Find $a$ : $a^{2} = c^{2} - b^{2} = 53 - 49 = 4 \Rightarrow a = 2$ .
Determine Orientation: Foci share $x$ -coordinate; hyperbola is vertical.
Equation: $\frac{( y + 5 ) ^{2}}{4} - \frac{( x - 8 ) ^{2}}{49} = 1$

Part (v)

Vertices: $(0, \pm 9)$ ; Point: $(8, 15)$ .

Find the Centre: Midpoint of $(0, 9)$ and $(0, - 9)$ is $(0, 0)$ .
Find $a$ : $2 a = 18 \Rightarrow a = 9$ .
Determine Orientation: $x$ -coordinates are same; hyperbola is vertical. $\frac{y ^{2}}{a ^{2}} - \frac{x ^{2}}{b ^{2}} = 1 \Rightarrow \frac{y ^{2}}{81} - \frac{x ^{2}}{b ^{2}} = 1$
Find $b$ : Substitute $(8, 15)$ . $\frac{1 5 ^{2}}{81} - \frac{8 ^{2}}{b ^{2}} = 1 \Rightarrow \frac{225}{81} - 1 = \frac{64}{b ^{2}} \Rightarrow \frac{144}{81} = \frac{64}{b ^{2}}$ $b^{2} = \frac{64 \times 81}{144} = 36 \Rightarrow b = 6$
Equation: $\frac{y ^{2}}{81} - \frac{x ^{2}}{36} = 1$

Part (vi)

$e = 5/4$ ; Centre: $(1, 1)$ ; Point: $(- 7, 2)$ ; Horizontal focal axis.

Relationship: $b^{2} = a^{2} (e^{2} - 1) = a^{2} (\frac{25}{16} - 1) = \frac{9}{16} a^{2}$ .
Equation Form: $\frac{( x - 1 ) ^{2}}{a ^{2}} - \frac{( y - 1 ) ^{2}}{b ^{2}} = 1$ .
Substitute $b^{2}$ and Point $(- 7, 2)$ : $\frac{( - 7 - 1 ) ^{2}}{a ^{2}} - \frac{( 2 - 1 ) ^{2}}{\frac{9}{16} a ^{2}} = 1 \Rightarrow \frac{64}{a ^{2}} - \frac{16}{9 a ^{2}} = 1$ $\frac{576 - 16}{9 a ^{2}} = 1 \Rightarrow 9 a^{2} = 560 \Rightarrow a^{2} = \frac{560}{9}$
Find $b^{2}$ : $b^{2} = \frac{9}{16} \times \frac{560}{9} = 35$ .
Equation: $\frac{9 ( x - 1 ) ^{2}}{560} - \frac{( y - 1 ) ^{2}}{35} = 1 or 9 (x - 1)^{2} - 16 (y - 1)^{2} = 560$

Part (vii)

Focus: $(5, 3)$ ; Directrix: $y + 2 = 0$ ; $e = 5/3$ .

Definition: $\frac{Dist ( P , Focus )}{Dist ( P , Directrix )} = e$ .
Setup: $\frac{( x - 5 ) ^{2} + ( y - 3 ) ^{2}}{∣ y + 2∣} = \frac{5}{3}$ .
Square and Simplify: $9 [(x - 5)^{2} + (y - 3)^{2}] = 25 (y + 2)^{2}$ $9 (x^{2} - 10 x + 25 + y^{2} - 6 y + 9) = 25 (y^{2} + 4 y + 4)$ $9 x^{2} - 90 x + 225 + 9 y^{2} - 54 y + 81 = 25 y^{2} + 100 y + 100$ $9 x^{2} - 16 y^{2} - 90 x - 154 y + 206 = 0$

Part (viii)

Centre: $(0, 0)$ ; Latus Rectum: $5$ ; $e = 5/4$ ; Conjugate axis along $x$ -axis.

Orientation: Conjugate axis on $x$ -axis means transverse axis is vertical.
Latus Rectum: $\frac{2 b ^{2}}{a} = 5 \Rightarrow a = \frac{2 b ^{2}}{5}$ .
Eccentricity Relationship: $b^{2} = a^{2} (e^{2} - 1) = a^{2} (\frac{9}{16})$ .
Solve for $b$ : $b^{2} = (\frac{2 b ^{2}}{5})^{2} (\frac{9}{16}) = \frac{4 b ^{4}}{25} \cdot \frac{9}{16} = \frac{9 b ^{4}}{100}$ $1 = \frac{9 b ^{2}}{100} \Rightarrow b^{2} = \frac{100}{9}$
Solve for $a$ : $a = \frac{2}{5} (\frac{100}{9}) = \frac{40}{9} \Rightarrow a^{2} = \frac{1600}{81}$ .
Equation: $\frac{y ^{2}}{1600/81} - \frac{x ^{2}}{100/9} = 1$

Key Formulas or Methods Used

Standard Forms: $\frac{( x - h ) ^{2}}{a ^{2}} - \frac{( y - k ) ^{2}}{b ^{2}} = 1$ (Horizontal) or $\frac{( y - k ) ^{2}}{a ^{2}} - \frac{( x - h ) ^{2}}{b ^{2}} = 1$ (Vertical).
Pythagorean Relation: $c^{2} = a^{2} + b^{2}$ .
Eccentricity: $e = \frac{c}{a}$ or $b^{2} = a^{2} (e^{2} - 1)$ .
Latus Rectum: $L = \frac{2 b ^{2}}{a}$ .
Conic Definition: $\frac{P F}{P M} = e$ .

Summary of Steps

Identify the Centre: Use the midpoint of vertices, foci, or co-vertices.
Determine Orientation: Compare coordinates of vertices or foci to see if the transverse axis is horizontal or vertical.
Calculate Parameters: Find $a$ (semi-transverse), $b$ (semi-conjugate), or $c$ (focal distance) using distance formulas or given lengths.
Use Relationships: Apply $c^{2} = a^{2} + b^{2}$ or eccentricity formulas to find any missing values.
Write Equation: Substitute $(h, k)$ , $a^{2}$ , and $b^{2}$ into the appropriate standard form.