GSEB Class 11 Maths Solutions Chapter 11 શાંકવો 11.4

For Questions 1 to 6, find the foci, vertices coordinates, eccentricity, and length of the latus rectum for the given hyperbolas:

Question 1. Find the foci, vertices, eccentricity, and length of the latus rectum for the hyperbola: \( \frac { x^2 }{ 16 } - \frac { y^2 }{ 9 } = 1 \)
Answer: The given equation is \( \frac { x^2 }{ 16 } - \frac { y^2 }{ 9 } = 1 \).
Comparing this equation with the standard form \( \frac { x^2 }{ a^2 } - \frac { y^2 }{ b^2 } = 1 \), we get:
\( a^2 = 16 \implies a = 4 \)
\( b^2 = 9 \implies b = 3 \)
Now, calculate \( c \):
\( c = \sqrt{a^2+b^2} = \sqrt{16+9} = \sqrt{25} = 5 \)
Using these values, we can find the required parameters:
(1) Foci: \( (\pm c, 0) = (\pm 5, 0) \)
(2) Vertices: \( (\pm a, 0) = (\pm 4, 0) \)
(3) Eccentricity: \( e = \frac { c }{ a } = \frac { 5 }{ 4 } \)
(4) Length of latus rectum: \( = \frac { 2b^2 }{ a } = \frac { 2 \times 9 }{ 4 } = \frac { 18 }{ 4 } = \frac { 9 }{ 2 } \)
In simple words: First, we match the given equation to the standard form to find 'a' and 'b'. Then, we use these to figure out 'c'. With 'a', 'b', and 'c', we can easily find the foci, vertices, eccentricity, and the length of the latus rectum for the hyperbola.

Exam Tip: Remember the standard forms for hyperbolas (\( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) or \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \)) and the formulas for calculating 'c', foci, vertices, eccentricity, and latus rectum based on 'a' and 'b'.

Question 2. Find the foci, vertices, eccentricity, and length of the latus rectum for the hyperbola: \( \frac { y^2 }{ 9 } - \frac { x^2 }{ 27 } = 1 \)
Answer: The given equation is \( \frac { y^2 }{ 9 } - \frac { x^2 }{ 27 } = 1 \).
Comparing this equation with the standard form \( \frac { y^2 }{ a^2 } - \frac { x^2 }{ b^2 } = 1 \), we obtain:
\( a^2 = 9 \implies a = 3 \)
\( b^2 = 27 \implies b = \sqrt{27} = 3\sqrt{3} \)
Now, calculate \( c \):
\( c = \sqrt{a^2+b^2} = \sqrt{9+27} = \sqrt{36} = 6 \)
Using these values, we can find the required parameters:
(1) Foci: \( (0, \pm c) = (0, \pm 6) \)
(2) Vertices: \( (0, \pm a) = (0, \pm 3) \)
(3) Eccentricity: \( e = \frac { c }{ a } = \frac { 6 }{ 3 } = 2 \)
(4) Length of latus rectum: \( = \frac { 2b^2 }{ a } = \frac { 2(27) }{ 3 } = \frac { 54 }{ 3 } = 18 \)
In simple words: Match the equation with the standard form that has \( y^2 \) first. Find 'a', 'b', and then 'c'. Use these values to quickly get the foci, vertices, eccentricity, and length of the latus rectum.

Exam Tip: Pay close attention to whether the \( x^2 \) or \( y^2 \) term is positive, as this determines the orientation of the hyperbola and the placement of its foci and vertices on either the x-axis or y-axis.

Question 3. Find the foci, vertices, eccentricity, and length of the latus rectum for the hyperbola: \( 9y^2 - 4x^2 = 36 \)
Answer: The given equation is \( 9y^2 - 4x^2 = 36 \).
To convert it to the standard form, divide the entire equation by 36:
\( \frac { 9y^2 }{ 36 } - \frac { 4x^2 }{ 36 } = \frac { 36 }{ 36 } \)
\( \implies \frac { y^2 }{ 4 } - \frac { x^2 }{ 9 } = 1 \)
Comparing this equation with the standard form \( \frac { y^2 }{ a^2 } - \frac { x^2 }{ b^2 } = 1 \), we find:
\( a^2 = 4 \implies a = 2 \)
\( b^2 = 9 \implies b = 3 \)
Now, calculate \( c \):
\( c = \sqrt{a^2+b^2} = \sqrt{4+9} = \sqrt{13} \)
Using these values, we can find the required parameters:
(1) Foci: \( (0, \pm c) = (0, \pm \sqrt{13}) \)
(2) Vertices: \( (0, \pm a) = (0, \pm 2) \)
(3) Eccentricity: \( e = \frac { c }{ a } = \frac { \sqrt{13} }{ 2 } \)
(4) Length of latus rectum: \( = \frac { 2b^2 }{ a } = \frac { 2(9) }{ 2 } = 9 \)
In simple words: First, change the given equation into its standard form by dividing by the constant. Then, match it to find 'a' and 'b'. After that, use these to compute 'c' and the other important values like foci, vertices, eccentricity, and latus rectum length.

Exam Tip: If the equation is not in standard form (\( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) or \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \)), always divide by the constant term on the right-hand side to get 1, before identifying \( a^2 \) and \( b^2 \).

Question 4. Find the foci, vertices, eccentricity, and length of the latus rectum for the hyperbola: \( 16x^2 - 9y^2 = 576 \)
Answer: The given equation is \( 16x^2 - 9y^2 = 576 \).
To convert it to the standard form, divide the entire equation by 576:
\( \frac { 16x^2 }{ 576 } - \frac { 9y^2 }{ 576 } = \frac { 576 }{ 576 } \)
\( \implies \frac { x^2 }{ 36 } - \frac { y^2 }{ 64 } = 1 \)
Comparing this equation with the standard form \( \frac { x^2 }{ a^2 } - \frac { y^2 }{ b^2 } = 1 \), we get:
\( a^2 = 36 \implies a = 6 \)
\( b^2 = 64 \implies b = 8 \)
Now, calculate \( c \):
\( c = \sqrt{a^2+b^2} = \sqrt{36+64} = \sqrt{100} = 10 \)
Using these values, we can find the required parameters:
(1) Foci: \( (\pm c, 0) = (\pm 10, 0) \)
(2) Vertices: \( (\pm a, 0) = (\pm 6, 0) \)
(3) Eccentricity: \( e = \frac { c }{ a } = \frac { 10 }{ 6 } = \frac { 5 }{ 3 } \)
(4) Length of latus rectum: \( = \frac { 2b^2 }{ a } = \frac { 2(64) }{ 6 } = \frac { 128 }{ 6 } = \frac { 64 }{ 3 } \)
In simple words: First, transform the equation into its standard form by dividing by the constant. Then, match it to identify 'a' and 'b'. After that, use these to calculate 'c' and the other important values such as foci, vertices, eccentricity, and latus rectum length.

Exam Tip: Carefully simplify the fractions after dividing by the constant to ensure \( a^2 \) and \( b^2 \) are correctly identified. This helps avoid errors in subsequent calculations.

Question 5. Find the foci, vertices, eccentricity, and length of the latus rectum for the hyperbola: \( 5y^2 - 9x^2 = 36 \)
Answer: The given equation is \( 5y^2 - 9x^2 = 36 \).
To convert it to the standard form, divide the entire equation by 36:
\( \frac { 5y^2 }{ 36 } - \frac { 9x^2 }{ 36 } = \frac { 36 }{ 36 } \)
\( \implies \frac { y^2 }{ 36/5 } - \frac { x^2 }{ 4 } = 1 \)
Comparing this equation with the standard form \( \frac { y^2 }{ a^2 } - \frac { x^2 }{ b^2 } = 1 \), we get:
\( a^2 = \frac{36}{5} \implies a = \frac{6}{\sqrt{5}} \)
\( b^2 = 4 \implies b = 2 \)
Now, calculate \( c \):
\( c = \sqrt{a^2+b^2} = \sqrt{\frac{36}{5}+4} = \sqrt{\frac{36+20}{5}} = \sqrt{\frac{56}{5}} = \frac{\sqrt{56}}{\sqrt{5}} = \frac{2\sqrt{14}}{\sqrt{5}} \)
Using these values, we can find the required parameters:
(1) Foci: \( (0, \pm c) = (0, \pm \frac{2\sqrt{14}}{\sqrt{5}}) \)
(2) Vertices: \( (0, \pm a) = (0, \pm \frac{6}{\sqrt{5}}) \)
(3) Eccentricity: \( e = \frac { c }{ a } = \frac { 2\sqrt{14}/\sqrt{5} }{ 6/\sqrt{5} } = \frac { 2\sqrt{14} }{ 6 } = \frac { \sqrt{14} }{ 3 } \)
(4) Length of latus rectum: \( = \frac { 2b^2 }{ a } = \frac { 2(4) }{ 6/\sqrt{5} } = \frac { 8\sqrt{5} }{ 6 } = \frac { 4\sqrt{5} }{ 3 } \)
In simple words: First, divide the equation by the constant to put it in standard form. Then, find the values for 'a' and 'b' and use them to calculate 'c'. With these key values, you can then determine the foci, vertices, eccentricity, and the length of the latus rectum.

Exam Tip: When \( a^2 \) or \( b^2 \) are fractions, ensure you handle the square roots correctly and simplify any expressions involving radicals for 'c' and 'e'.

Question 6. Find the foci, vertices, eccentricity, and length of the latus rectum for the hyperbola: \( 49y^2 - 16x^2 = 784 \)
Answer: The given equation is \( 49y^2 - 16x^2 = 784 \).
To convert it to the standard form, divide the entire equation by 784:
\( \frac { 49y^2 }{ 784 } - \frac { 16x^2 }{ 784 } = \frac { 784 }{ 784 } \)
\( \implies \frac { y^2 }{ 16 } - \frac { x^2 }{ 49 } = 1 \)
Comparing this equation with the standard form \( \frac { y^2 }{ a^2 } - \frac { x^2 }{ b^2 } = 1 \), we get:
\( a^2 = 16 \implies a = 4 \)
\( b^2 = 49 \implies b = 7 \)
Now, calculate \( c \):
\( c = \sqrt{a^2+b^2} = \sqrt{16+49} = \sqrt{65} \)
Using these values, we can find the required parameters:
(1) Foci: \( (0, \pm c) = (0, \pm \sqrt{65}) \)
(2) Vertices: \( (0, \pm a) = (0, \pm 4) \)
(3) Eccentricity: \( e = \frac { c }{ a } = \frac { \sqrt{65} }{ 4 } \)
(4) Length of latus rectum: \( = \frac { 2b^2 }{ a } = \frac { 2(49) }{ 4 } = \frac { 98 }{ 4 } = \frac { 49 }{ 2 } \)
In simple words: Convert the equation to its standard form by dividing by the constant. Then, identify 'a' and 'b', and use them to find 'c'. Once you have 'a', 'b', and 'c', you can determine the foci, vertices, eccentricity, and length of the latus rectum.

Exam Tip: Remember to simplify fractions to their lowest terms whenever possible, especially for the length of the latus rectum and eccentricity, to ensure full marks.

For Questions 7 to 15, find the equation of the hyperbola satisfying the given conditions:

Question 7. Foci \( ( \pm 3, 0 ) \)
Answer: Given foci \( (\pm 3, 0) \).
This tells us that \( c = 3 \).
Since the foci are on the X-axis, the equation of the hyperbola is of the form \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \).
From the context of the solution provided by the source, an additional value for 'a' is implicitly used. We will proceed with \( a=2 \), as suggested by the steps.
We know the relationship \( c^2 = a^2 + b^2 \).
Substitute \( c=3 \) and \( a=2 \):
\( 3^2 = 2^2 + b^2 \)
\( 9 = 4 + b^2 \)
\( b^2 = 9 - 4 \)
\( b^2 = 5 \)
Therefore, the equation of the hyperbola is:
\( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \)
\( \implies \frac{x^2}{4} - \frac{y^2}{5} = 1 \)
In simple words: The foci tell us 'c' and which axis the hyperbola is on. The problem implies a value for 'a'. Using 'c' and 'a', we find 'b²' and then write the equation.

Exam Tip: When only foci are given, a unique hyperbola cannot be determined without another piece of information (like 'a', 'b', or eccentricity). In exam settings, if such a problem appears, ensure all necessary information is explicitly provided or derive it if possible.

Question 8. Vertices \( ( 0, \pm 5 ) \), Foci \( ( 0, \pm 8 ) \)
Answer: Given vertices \( (0, \pm 5) \) and foci \( (0, \pm 8) \).
From the vertices, we get \( a = 5 \).
From the foci, we get \( c = 8 \).
Since both vertices and foci are on the Y-axis, the equation of the hyperbola is of the form \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \).
We use the relationship \( c^2 = a^2 + b^2 \).
Substitute \( a=5 \) and \( c=8 \):
\( 8^2 = 5^2 + b^2 \)
\( 64 = 25 + b^2 \)
\( b^2 = 64 - 25 \)
\( b^2 = 39 \)
Therefore, the equation of the hyperbola is:
\( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \)
\( \implies \frac{y^2}{25} - \frac{x^2}{39} = 1 \)
In simple words: The given vertices give us 'a', and the foci give us 'c'. Since they are on the y-axis, we use the y-first equation. Then, we use the formula \( c^2 = a^2 + b^2 \) to find 'b²', and write the final equation.

Exam Tip: If vertices and foci share the same coordinate (x or y), it indicates the axis on which the hyperbola lies. This helps determine the correct standard form of the equation.

Question 9. Vertices \( ( 0, \pm 3 ) \), Foci \( ( 0, \pm 5 ) \)
Answer: Given vertices \( (0, \pm 3) \) and foci \( (0, \pm 5) \).
From the vertices, we get \( a = 3 \).
From the foci, we get \( c = 5 \).
Since both vertices and foci are on the Y-axis, the equation of the hyperbola is of the form \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \).
We use the relationship \( c^2 = a^2 + b^2 \).
Substitute \( a=3 \) and \( c=5 \):
\( 5^2 = 3^2 + b^2 \)
\( 25 = 9 + b^2 \)
\( b^2 = 25 - 9 \)
\( b^2 = 16 \)
Therefore, the equation of the hyperbola is:
\( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \)
\( \implies \frac{y^2}{9} - \frac{x^2}{16} = 1 \)
In simple words: The vertices give 'a', and the foci give 'c'. Since they are on the y-axis, we use the standard form with \( y^2 \) first. Then, use \( c^2 = a^2 + b^2 \) to find 'b²' and write the equation.

Exam Tip: Quickly identify 'a' from vertices and 'c' from foci. These values are crucial for finding 'b' and then constructing the hyperbola's equation.

Question 10. Foci \( ( \pm 5, 0 ) \). Length of the major axis is 8.
Answer: Given foci \( (\pm 5, 0) \).
This tells us that \( c = 5 \).
Since the foci are on the X-axis, the equation of the hyperbola is of the form \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \).
Given that the length of the major axis (transverse axis for hyperbolas) is 8.
Length of transverse axis \( 2a = 8 \implies a = 4 \).
Now, we use the relationship \( c^2 = a^2 + b^2 \).
Substitute \( a=4 \) and \( c=5 \):
\( 5^2 = 4^2 + b^2 \)
\( 25 = 16 + b^2 \)
\( b^2 = 25 - 16 \)
\( b^2 = 9 \)
Therefore, the equation of the hyperbola is:
\( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \)
\( \implies \frac{x^2}{16} - \frac{y^2}{9} = 1 \)
In simple words: The foci give 'c', and the length of the major axis gives 'a'. Since the foci are on the x-axis, we use the x-first equation. Then, use \( c^2 = a^2 + b^2 \) to find 'b²' and write the equation.

Exam Tip: For hyperbolas, the "major axis" refers to the transverse axis, which has length \( 2a \). Always distinguish this from the conjugate axis, which has length \( 2b \).

Question 11. Foci \( ( 0, \pm 13 ) \), Length of the conjugate axis is 24.
Answer: Given foci \( (0, \pm 13) \).
This tells us that \( c = 13 \).
Since the foci are on the Y-axis, the equation of the hyperbola is of the form \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \).
Given that the length of the conjugate axis is 24.
Length of conjugate axis \( 2b = 24 \implies b = 12 \).
Now, we use the relationship \( c^2 = a^2 + b^2 \).
Substitute \( b=12 \) and \( c=13 \):
\( 13^2 = a^2 + 12^2 \)
\( 169 = a^2 + 144 \)
\( a^2 = 169 - 144 \)
\( a^2 = 25 \)
Therefore, the equation of the hyperbola is:
\( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \)
\( \implies \frac{y^2}{25} - \frac{x^2}{144} = 1 \)
In simple words: The foci give 'c' and the orientation. The conjugate axis length gives 'b'. Using these, we find 'a²' with \( c^2 = a^2 + b^2 \) and then write the hyperbola's equation.

Exam Tip: Differentiate between the transverse axis (related to 'a') and the conjugate axis (related to 'b'). A common mistake is to confuse their lengths.

Question 12. Foci \( ( \pm 3\sqrt{5}, 0 ) \), Length of the latus rectum is 8.
Answer: Given foci \( (\pm 3\sqrt{5}, 0) \).
This tells us that \( c = 3\sqrt{5} \).
Since the foci are on the X-axis, the equation of the hyperbola is of the form \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \).
Given that the length of the latus rectum is 8.
Length of latus rectum \( \frac{2b^2}{a} = 8 \).
\( 2b^2 = 8a \)
\( b^2 = 4a \)
Now, we use the relationship \( c^2 = a^2 + b^2 \).
Substitute \( c=3\sqrt{5} \) and \( b^2 = 4a \):
\( (3\sqrt{5})^2 = a^2 + 4a \)
\( 45 = a^2 + 4a \)
Rearrange the terms to form a quadratic equation:
\( a^2 + 4a - 45 = 0 \)
Factorize the quadratic equation:
\( (a + 9)(a - 5) = 0 \)
This gives two possible values for \( a \): \( a = -9 \) or \( a = 5 \).
Since \( a \) represents a length, it must be positive. Therefore, \( a = 5 \).
Now find \( b^2 \):
\( b^2 = 4a = 4(5) = 20 \)
Therefore, the equation of the hyperbola is:
\( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \)
\( \implies \frac{x^2}{25} - \frac{y^2}{20} = 1 \)
In simple words: The foci provide 'c' and axis direction. The latus rectum formula gives a relationship between 'a' and 'b²'. We substitute this into \( c^2 = a^2 + b^2 \) to form a quadratic equation for 'a', solve it, pick the positive 'a', and then find 'b²' to write the equation.

Exam Tip: When solving for 'a' using a quadratic equation, remember that 'a' (like 'b' and 'c') must be a positive length. Reject any negative solutions for 'a'.

Question 13. Foci \( ( 0, \pm 4 ) \), Length of the latus rectum is 12.
Answer: Given foci \( (0, \pm 4) \).
This tells us that \( c = 4 \).
Since the foci are on the Y-axis, the equation of the hyperbola is of the form \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \).
Given that the length of the latus rectum is 12.
Length of latus rectum \( \frac{2b^2}{a} = 12 \).
\( 2b^2 = 12a \)
\( b^2 = 6a \)
Now, we use the relationship \( c^2 = a^2 + b^2 \).
Substitute \( c=4 \) and \( b^2 = 6a \):
\( 4^2 = a^2 + 6a \)
\( 16 = a^2 + 6a \)
Rearrange the terms to form a quadratic equation:
\( a^2 + 6a - 16 = 0 \)
Factorize the quadratic equation:
\( (a + 8)(a - 2) = 0 \)
This gives two possible values for \( a \): \( a = -8 \) or \( a = 2 \).
Since \( a \) represents a length, it must be positive. Therefore, \( a = 2 \).
Now find \( b^2 \):
\( b^2 = 6a = 6(2) = 12 \)
Therefore, the equation of the hyperbola is:
\( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \)
\( \implies \frac{y^2}{4} - \frac{x^2}{12} = 1 \)
In simple words: The foci give 'c' and the direction of the axis. The length of the latus rectum gives a connection between 'a' and 'b²'. We put this into the \( c^2 = a^2 + b^2 \) formula to solve for 'a', then find 'b²' to write the final equation.

Exam Tip: Always double-check the axis of the foci. If foci are \( (0, \pm c) \), the hyperbola opens along the y-axis, and the equation starts with \( \frac{y^2}{a^2} \). If \( (\pm c, 0) \), it opens along the x-axis, and the equation starts with \( \frac{x^2}{a^2} \).

Question 14. Vertices \( ( \pm 7, 0 ) \), Eccentricity \( e = \frac { 4 }{ 3 } \)
Answer: Given vertices \( (\pm 7, 0) \).
This tells us that \( a = 7 \).
Since the vertices are on the X-axis, the equation of the hyperbola is of the form \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \).
Given eccentricity \( e = \frac{4}{3} \).
We know that for a hyperbola, \( e = \frac{c}{a} \).
Substitute the known values:
\( \frac{4}{3} = \frac{c}{7} \)
\( c = \frac{4 \times 7}{3} = \frac{28}{3} \)
Now, we use the relationship \( c^2 = a^2 + b^2 \).
Substitute \( a=7 \) and \( c=\frac{28}{3} \):
\( (\frac{28}{3})^2 = 7^2 + b^2 \)
\( \frac{784}{9} = 49 + b^2 \)
\( b^2 = \frac{784}{9} - 49 \)
To subtract, find a common denominator:
\( b^2 = \frac{784}{9} - \frac{49 \times 9}{9} \)
\( b^2 = \frac{784 - 441}{9} \)
\( b^2 = \frac{343}{9} \)
Therefore, the equation of the hyperbola is:
\( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \)
\( \implies \frac{x^2}{49} - \frac{y^2}{343/9} = 1 \)
\( \implies \frac{x^2}{49} - \frac{9y^2}{343} = 1 \)
In simple words: The vertices give 'a' and the axis. The eccentricity formula helps us find 'c'. Then, using 'a' and 'c' in the \( c^2 = a^2 + b^2 \) formula, we calculate 'b²' and write the hyperbola's equation.

Exam Tip: Be careful with fractional calculations, especially when squaring and subtracting. Ensure all steps are clear and denominators are correctly managed.

Question 15. Foci \( ( 0, \pm \sqrt{10} ) \), Passing through \( ( 2, 3 ) \)
Answer: Given foci \( (0, \pm \sqrt{10}) \).
This tells us that \( c = \sqrt{10} \).
Since the foci are on the Y-axis, the equation of the hyperbola is of the form \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \).
We know the relationship \( c^2 = a^2 + b^2 \).
Substitute \( c=\sqrt{10} \):
\( (\sqrt{10})^2 = a^2 + b^2 \)
\( 10 = a^2 + b^2 \)
\( b^2 = 10 - a^2 \) ... (Equation 1)
The hyperbola passes through the point \( (2, 3) \). Substitute \( x=2 \) and \( y=3 \) into the standard equation:
\( \frac{3^2}{a^2} - \frac{2^2}{b^2} = 1 \)
\( \frac{9}{a^2} - \frac{4}{b^2} = 1 \) ... (Equation 2)
Now, substitute Equation 1 into Equation 2:
\( \frac{9}{a^2} - \frac{4}{10 - a^2} = 1 \)
To eliminate the denominators, multiply the entire equation by \( a^2(10 - a^2) \):
\( 9(10 - a^2) - 4a^2 = a^2(10 - a^2) \)
\( 90 - 9a^2 - 4a^2 = 10a^2 - a^4 \)
\( 90 - 13a^2 = 10a^2 - a^4 \)
Rearrange the terms to form a quadratic equation in \( a^2 \):
\( a^4 - 23a^2 + 90 = 0 \)
Let \( A = a^2 \). The equation becomes \( A^2 - 23A + 90 = 0 \).
Factorize the quadratic equation:
\( (A - 18)(A - 5) = 0 \)
This gives two possible values for \( A \): \( A = 18 \) or \( A = 5 \).
So, \( a^2 = 18 \) or \( a^2 = 5 \).
Case 1: If \( a^2 = 18 \).
From Equation 1, \( b^2 = 10 - a^2 = 10 - 18 = -8 \). Since \( b^2 \) cannot be negative, this case is not possible.
Case 2: If \( a^2 = 5 \).
From Equation 1, \( b^2 = 10 - a^2 = 10 - 5 = 5 \). This case is possible.
Therefore, \( a^2 = 5 \) and \( b^2 = 5 \).
The equation of the hyperbola is:
\( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \)
\( \implies \frac{y^2}{5} - \frac{x^2}{5} = 1 \)
In simple words: The foci give 'c' and the axis. We use \( c^2 = a^2 + b^2 \) to link 'a²' and 'b²'. Then, we plug the point the hyperbola passes through into the general equation. This creates a quadratic in 'a²'. We solve it, pick the valid positive 'a²', find 'b²', and write the final equation.

Exam Tip: When solving for \( a^2 \) or \( b^2 \), always check the validity of the solutions. \( a^2 \) and \( b^2 \) must always be positive numbers in the context of conic sections.