GSEB Class 11 Maths Solutions Chapter 11 Conic Sections Exercise 11.4

In each of the following questions 1 to 6, find the co-ordinates of the foci and the vertices, the eccentricity and length of latus rectum of the hyperbola:

Question 1. Find the co-ordinates of the foci and the vertices, the eccentricity and length of latus rectum of the hyperbola: \( \frac{x^{2}}{16} - \frac{y^{2}}{9} = 1 \)
Answer: The standard form of the hyperbola is \( \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 \).
Comparing the given equation \( \frac{x^{2}}{16} - \frac{y^{2}}{9} = 1 \) with the standard form, we get:
\( a^2 = 16 \implies a = 4 \)
\( b^2 = 9 \implies b = 3 \)
We know that \( c^2 = a^2 + b^2 \).
\( \implies c^2 = 16 + 9 = 25 \)
\( \implies c = 5 \)
The transverse axis lies along the x-axis.
Co-ordinates of the foci are \( (\pm c, 0) = (\pm 5, 0) \).
Co-ordinates of the vertices are \( (\pm a, 0) = (\pm 4, 0) \).
The eccentricity is \( e = \frac{c}{a} = \frac{5}{4} \).
The length of the latus rectum is \( \frac{2b^2}{a} = \frac{2 \times 9}{4} = \frac{9}{2} \).
In simple words: First, identify 'a' and 'b' from the equation. Then use \( c^2 = a^2 + b^2 \) to find 'c'. Once you have 'a', 'b', and 'c', you can easily determine the foci, vertices, eccentricity, and latus rectum using their respective formulas. The 'x' term being first means the hyperbola opens left and right.

Exam Tip: Remember to identify whether the hyperbola's transverse axis is along the x-axis or y-axis first, as this determines the format of foci and vertices coordinates.

Question 2. Find the co-ordinates of the foci and the vertices, the eccentricity and length of latus rectum of the hyperbola: \( \frac{y^{2}}{9} - \frac{x^{2}}{27} = 1 \)
Answer: The standard form of the hyperbola is \( \frac{y^{2}}{a^{2}} - \frac{x^{2}}{b^{2}} = 1 \).
Comparing the given equation \( \frac{y^{2}}{9} - \frac{x^{2}}{27} = 1 \) with the standard form, we get:
\( a^2 = 9 \implies a = 3 \)
\( b^2 = 27 \implies b = 3\sqrt{3} \)
We know that \( c^2 = a^2 + b^2 \).
\( \implies c^2 = 9 + 27 = 36 \)
\( \implies c = 6 \)
The transverse axis lies along the y-axis.
Co-ordinates of the foci are \( (0, \pm c) = (0, \pm 6) \).
Co-ordinates of the vertices are \( (0, \pm a) = (0, \pm 3) \).
The eccentricity is \( e = \frac{c}{a} = \frac{6}{3} = 2 \).
The length of the latus rectum is \( \frac{2b^2}{a} = \frac{2 \times 27}{3} = 18 \).
In simple words: For this hyperbola, 'a' and 'b' values are found from the equation. Then 'c' is calculated using \( c^2 = a^2 + b^2 \). Since the 'y' term comes first, the hyperbola opens up and down, so the foci and vertices will be on the y-axis. Finally, use the formulas for eccentricity and latus rectum.

Exam Tip: Notice that when the 'y' term is positive and comes first, the transverse axis is vertical, and the foci and vertices will have x-coordinate 0.

Question 3. Find the co-ordinates of the foci and the vertices, the eccentricity and length of latus rectum of the hyperbola: \( 9y^2 - 4x^2 = 36 \)
Answer: The given equation of the hyperbola is \( 9y^2 - 4x^2 = 36 \).
To convert it into 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 with the standard form \( \frac{y^2}{a^{2}} - \frac{x^{2}}{b^{2}} = 1 \), we get:
\( a^2 = 4 \implies a = 2 \)
\( b^2 = 9 \implies b = 3 \)
We know that \( c^2 = a^2 + b^2 \).
\( \implies c^2 = 4 + 9 = 13 \)
\( \implies c = \sqrt{13} \)
The transverse axis lies along the y-axis.
Co-ordinates of the foci are \( (0, \pm c) = (0, \pm \sqrt{13}) \).
Co-ordinates of the vertices are \( (0, \pm a) = (0, \pm 2) \).
The eccentricity is \( e = \frac{c}{a} = \frac{\sqrt{13}}{2} \).
The length of the latus rectum is \( \frac{2b^2}{a} = \frac{2 \times 9}{2} = 9 \).
In simple words: First, rewrite the equation into the standard form by dividing by the constant on the right side. Then, determine 'a' and 'b', and use them to calculate 'c'. Since 'y' squared is the positive term, the hyperbola's main axis is vertical. Finally, state the foci, vertices, eccentricity, and latus rectum.

Exam Tip: Always make sure the right-hand side of the hyperbola's equation is 1 before determining \( a^2 \) and \( b^2 \).

Question 4. Find the co-ordinates of the foci and the vertices, the eccentricity and length of latus rectum of the hyperbola: \( 16x^2 - 9y^2 = 576 \)
Answer: The given equation of the hyperbola is \( 16x^2 - 9y^2 = 576 \).
To convert it into 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 with the standard form \( \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 \), we get:
\( a^2 = 36 \implies a = 6 \)
\( b^2 = 64 \implies b = 8 \)
We know that \( c^2 = a^2 + b^2 \).
\( \implies c^2 = 36 + 64 = 100 \)
\( \implies c = 10 \)
The transverse axis lies along the x-axis.
Co-ordinates of the foci are \( (\pm c, 0) = (\pm 10, 0) \).
Co-ordinates of the vertices are \( (\pm a, 0) = (\pm 6, 0) \).
The eccentricity is \( e = \frac{c}{a} = \frac{10}{6} = \frac{5}{3} \).
The length of the latus rectum is \( \frac{2b^2}{a} = \frac{2 \times 64}{6} = \frac{64}{3} \).
In simple words: Convert the hyperbola's equation to its standard form by dividing by 576. Then, identify 'a' and 'b' from the denominators. Calculate 'c' using the formula \( c^2 = a^2 + b^2 \). Since the 'x' term is positive, the transverse axis is horizontal. Finally, list the foci, vertices, eccentricity, and the latus rectum.

Exam Tip: Always check that the denominators \( a^2 \) and \( b^2 \) are correctly identified based on the leading positive term; for hyperbola, \( a^2 \) is always under the positive term.

Question 5. Find the co-ordinates of the foci and the vertices, the eccentricity and length of latus rectum of the hyperbola: \( 5y^2 - 9x^2 = 36 \)
Answer: The given equation of the hyperbola is \( 5y^2 - 9x^2 = 36 \).
To convert it into 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 with the standard form \( \frac{y^{2}}{a^{2}} - \frac{x^{2}}{b^{2}} = 1 \), we get:
\( a^2 = \frac{36}{5} \implies a = \frac{6}{\sqrt{5}} \)
\( b^2 = 4 \implies b = 2 \)
We know that \( c^2 = a^2 + b^2 \).
\( \implies c^2 = \frac{36}{5} + 4 = \frac{36+20}{5} = \frac{56}{5} \)
\( \implies c = \sqrt{\frac{56}{5}} = \frac{\sqrt{56}}{\sqrt{5}} = \frac{2\sqrt{14}}{\sqrt{5}} \)
The transverse axis lies along the y-axis.
Co-ordinates of the foci are \( (0, \pm c) = \left(0, \pm \frac{2\sqrt{14}}{\sqrt{5}}\right) \).
Co-ordinates of the vertices are \( (0, \pm a) = \left(0, \pm \frac{6}{\sqrt{5}}\right) \).
The eccentricity is \( e = \frac{c}{a} = \frac{2\sqrt{14}/\sqrt{5}}{6/\sqrt{5}} = \frac{2\sqrt{14}}{6} = \frac{\sqrt{14}}{3} \).
The length of the latus rectum is \( \frac{2b^2}{a} = \frac{2 \times 4}{6/\sqrt{5}} = \frac{8\sqrt{5}}{6} = \frac{4\sqrt{5}}{3} \).
In simple words: Transform the given hyperbola equation into its standard form. From this, you can find 'a' and 'b'. Then, calculate 'c' using the relationship \( c^2 = a^2 + b^2 \). Since the 'y' term is positive, the hyperbola is vertical. Finally, use 'a', 'b', and 'c' to find the foci, vertices, eccentricity, and latus rectum.

Exam Tip: Be careful when \( a^2 \) or \( b^2 \) are fractions; ensure you simplify the square roots correctly for 'a', 'b', and 'c'.

Question 6. Find the co-ordinates of the foci and the vertices, the eccentricity and length of latus rectum of the hyperbola: \( 49y^2 - 16x^2 = 784 \)
Answer: The given equation of the hyperbola is \( 49y^2 - 16x^2 = 784 \).
To convert it into 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 with the standard form \( \frac{y^{2}}{a^{2}} - \frac{x^{2}}{b^{2}} = 1 \), we get:
\( a^2 = 16 \implies a = 4 \)
\( b^2 = 49 \implies b = 7 \)
We know that \( c^2 = a^2 + b^2 \).
\( \implies c^2 = 16 + 49 = 65 \)
\( \implies c = \sqrt{65} \)
The transverse axis lies along the y-axis.
Co-ordinates of the foci are \( (0, \pm c) = (0, \pm \sqrt{65}) \).
Co-ordinates of the vertices are \( (0, \pm a) = (0, \pm 4) \).
The eccentricity is \( e = \frac{c}{a} = \frac{\sqrt{65}}{4} \).
The length of the latus rectum is \( \frac{2b^2}{a} = \frac{2 \times 49}{4} = \frac{49}{2} \).
In simple words: First, change the given hyperbola equation into its standard format by dividing by 784. Identify the values for \( a^2 \) and \( b^2 \). Use these to calculate \( c^2 = a^2 + b^2 \). Since the 'y' term is positive, the transverse axis is vertical. Finally, determine the foci, vertices, eccentricity, and latus rectum using the calculated 'a', 'b', and 'c' values.

Exam Tip: Always double-check your calculations for \( a^2 \), \( b^2 \), and \( c^2 \), especially when dealing with larger numbers or square roots, to avoid arithmetic errors.

In each of the following questions 7 to 15, find the equations of hyperbola satisfying the given conditions:

Question 7. Find the equation of hyperbola satisfying the given conditions: Vertices \( (\pm 2,0) \), foci \( (\pm 3, 0) \)
Answer: The given vertices are \( (\pm 2, 0) \). This indicates that \( a = 2 \).
The given foci are \( (\pm 3, 0) \). This indicates that \( c = 3 \).
Since the foci and vertices are on the x-axis, the transverse axis is along the x-axis.
The equation of the hyperbola will be of the form \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \).
We know the relationship \( c^2 = a^2 + b^2 \).
\( \implies 3^2 = 2^2 + b^2 \)
\( \implies 9 = 4 + b^2 \)
\( \implies b^2 = 9 - 4 = 5 \)
Therefore, the equation of the hyperbola is \( \frac{x^2}{2^2} - \frac{y^2}{5} = 1 \).
\( \implies \frac{x^2}{4} - \frac{y^2}{5} = 1 \).
In simple words: From the vertices, we get 'a'. From the foci, we get 'c'. Since both are on the x-axis, the hyperbola is horizontal. Use the formula \( c^2 = a^2 + b^2 \) to find \( b^2 \). Then, substitute 'a' and \( b^2 \) into the standard hyperbola equation for the x-axis.

Exam Tip: The location of vertices and foci (e.g., on x-axis or y-axis) directly tells you the orientation of the hyperbola and the form of its equation.

Question 8. Find the equation of hyperbola satisfying the given conditions: Vertices \( (0, \pm 5) \), foci \( (0, \pm 8) \)
Answer: The given vertices are \( (0, \pm 5) \). This indicates that \( a = 5 \).
The given foci are \( (0, \pm 8) \). This indicates that \( c = 8 \).
Since the foci and vertices are on the y-axis, the transverse axis is along the y-axis.
The equation of the hyperbola will be of the form \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \).
We know the relationship \( c^2 = a^2 + b^2 \).
\( \implies 8^2 = 5^2 + b^2 \)
\( \implies 64 = 25 + b^2 \)
\( \implies b^2 = 64 - 25 = 39 \)
Therefore, the equation of the hyperbola is \( \frac{y^2}{5^2} - \frac{x^2}{39} = 1 \).
\( \implies \frac{y^2}{25} - \frac{x^2}{39} = 1 \).
In simple words: The vertices give 'a', and the foci give 'c'. As they are on the y-axis, the hyperbola is vertical. Use \( c^2 = a^2 + b^2 \) to solve for \( b^2 \). Then, put 'a' and \( b^2 \) into the standard hyperbola equation for the y-axis.

Exam Tip: When the vertices and foci have an x-coordinate of 0, it means the hyperbola opens vertically, and the \( y^2 \) term comes first in the equation.

Question 9. Find the equation of hyperbola satisfying the given conditions: Vertices \( (0, \pm 3) \), foci \( (0, \pm 5) \)
Answer: The given vertices are \( (0, \pm 3) \). This indicates that \( a = 3 \).
The given foci are \( (0, \pm 5) \). This indicates that \( c = 5 \).
Since the foci and vertices are on the y-axis, the transverse axis is along the y-axis.
The equation of the hyperbola will be of the form \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \).
We know the relationship \( c^2 = a^2 + b^2 \).
\( \implies 5^2 = 3^2 + b^2 \)
\( \implies 25 = 9 + b^2 \)
\( \implies b^2 = 25 - 9 = 16 \)
Therefore, the equation of the hyperbola is \( \frac{y^2}{3^2} - \frac{x^2}{16} = 1 \).
\( \implies \frac{y^2}{9} - \frac{x^2}{16} = 1 \).
In simple words: The 'a' value comes from the vertices, and the 'c' value comes from the foci. Since these points are on the y-axis, the hyperbola is vertical. Calculate \( b^2 \) using the equation \( c^2 = a^2 + b^2 \). Finally, write the hyperbola's equation using the standard form for a vertical hyperbola.

Exam Tip: Be sure to correctly substitute the values of \( a^2 \) and \( b^2 \) into the standard equation, especially noting which term (\( x^2 \) or \( y^2 \)) is positive.

Question 10. Find the equation of hyperbola satisfying the given conditions: Foci \( (\pm 5, 0) \), the transverse axis is of length 8.
Answer: The given foci are \( (\pm 5, 0) \). This indicates that \( c = 5 \).
Since the foci are on the x-axis, the transverse axis is along the x-axis.
The length of the transverse axis is given as 8.
We know that the length of the transverse axis is \( 2a \).
\( \implies 2a = 8 \implies a = 4 \)
The equation of the hyperbola will be of the form \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \).
We know the relationship \( c^2 = a^2 + b^2 \).
\( \implies 5^2 = 4^2 + b^2 \)
\( \implies 25 = 16 + b^2 \)
\( \implies b^2 = 25 - 16 = 9 \)
Therefore, the equation of the hyperbola is \( \frac{x^2}{4^2} - \frac{y^2}{9} = 1 \).
\( \implies \frac{x^2}{16} - \frac{y^2}{9} = 1 \).
In simple words: The foci directly provide 'c'. The transverse axis length helps us find 'a'. Since the foci are on the x-axis, the hyperbola is horizontal. With 'a' and 'c', calculate \( b^2 \) using \( c^2 = a^2 + b^2 \). Finally, substitute these values into the standard equation for a horizontal hyperbola.

Exam Tip: Distinguish between the transverse axis (length \( 2a \)) and the conjugate axis (length \( 2b \)) to correctly identify 'a' or 'b' from the given length.

Question 11. Find the equation of hyperbola satisfying the given conditions: Foci \( (0, \pm 13) \), the conjugate axis is of length 24.
Answer: The given foci are \( (0, \pm 13) \). This indicates that \( c = 13 \).
Since the foci are on the y-axis, the transverse axis is along the y-axis.
The length of the conjugate axis is given as 24.
We know that the length of the conjugate axis is \( 2b \).
\( \implies 2b = 24 \implies b = 12 \)
The equation of the hyperbola will be of the form \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \).
We know the relationship \( c^2 = a^2 + b^2 \).
\( \implies 13^2 = a^2 + 12^2 \)
\( \implies 169 = a^2 + 144 \)
\( \implies a^2 = 169 - 144 = 25 \)
Therefore, the equation of the hyperbola is \( \frac{y^2}{25} - \frac{x^2}{12^2} = 1 \).
\( \implies \frac{y^2}{25} - \frac{x^2}{144} = 1 \).
In simple words: From the foci, we determine 'c'. The length of the conjugate axis gives 'b'. As the foci are on the y-axis, the hyperbola is vertical. Use \( c^2 = a^2 + b^2 \) to solve for \( a^2 \). Finally, substitute 'a' and 'b' into the standard equation for a vertical hyperbola.

Exam Tip: Clearly differentiate between the transverse axis and conjugate axis in hyperbola problems; mixing them up is a common source of error.

Question 12. Find the equation of hyperbola satisfying the given conditions: Foci \( (\pm 3\sqrt{5}, 0) \), the latus rectum is length 8.
Answer: The given foci are \( (\pm 3\sqrt{5}, 0) \). This indicates that \( c = 3\sqrt{5} \).
Since the foci are on the x-axis, the transverse axis is along the x-axis.
The length of the latus rectum is given as 8.
We know that the length of the latus rectum is \( \frac{2b^2}{a} \).
\( \implies \frac{2b^2}{a} = 8 \)
\( \implies b^2 = 4a \) (Equation 1)
We also know the relationship \( c^2 = a^2 + b^2 \).
\( \implies (3\sqrt{5})^2 = a^2 + b^2 \)
\( \implies 45 = a^2 + b^2 \) (Equation 2)
Substitute \( b^2 = 4a \) from Equation 1 into Equation 2:
\( 45 = a^2 + 4a \)
\( \implies a^2 + 4a - 45 = 0 \)
Factor the quadratic equation:
\( (a + 9)(a - 5) = 0 \)
This gives \( a = -9 \) or \( a = 5 \).
Since 'a' must be a positive value, we take \( a = 5 \).
Now, substitute \( a = 5 \) into Equation 1 to find \( b^2 \):
\( b^2 = 4 \times 5 = 20 \)
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 give 'c', and the latus rectum length gives a relation between \( a \) and \( b^2 \). Because the foci are on the x-axis, the hyperbola is horizontal. Set up two equations: one from the latus rectum and one from \( c^2 = a^2 + b^2 \). Solve these two equations to find 'a' and \( b^2 \). Remember 'a' must be positive. Finally, write the hyperbola's equation.

Exam Tip: When solving for 'a' using a quadratic equation, always ensure that you select the positive root, as 'a' represents a length and must be positive.

Question 13. Find the equation of hyperbola satisfying the given conditions: Foci \( (\pm 4, 0) \), the latus rectum is of length 12.
Answer: The given foci are \( (\pm 4, 0) \). This indicates that \( c = 4 \).
Since the foci are on the x-axis, the transverse axis is along the x-axis.
The length of the latus rectum is given as 12.
We know that the length of the latus rectum is \( \frac{2b^2}{a} \).
\( \implies \frac{2b^2}{a} = 12 \)
\( \implies b^2 = 6a \) (Equation 1)
We also know the relationship \( c^2 = a^2 + b^2 \).
\( \implies 4^2 = a^2 + b^2 \)
\( \implies 16 = a^2 + b^2 \) (Equation 2)
Substitute \( b^2 = 6a \) from Equation 1 into Equation 2:
\( 16 = a^2 + 6a \)
\( \implies a^2 + 6a - 16 = 0 \)
Factor the quadratic equation:
\( (a + 8)(a - 2) = 0 \)
This gives \( a = -8 \) or \( a = 2 \).
Since 'a' must be a positive value, we take \( a = 2 \).
Now, substitute \( a = 2 \) into Equation 1 to find \( b^2 \):
\( b^2 = 6 \times 2 = 12 \)
Therefore, the equation of the hyperbola is \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \).
\( \implies \frac{x^2}{4} - \frac{y^2}{12} = 1 \).
In simple words: Use the foci to get 'c' and the latus rectum formula to relate 'a' and 'b'. Since the foci are on the x-axis, it's a horizontal hyperbola. Formulate two equations using \( c^2 = a^2 + b^2 \) and the latus rectum. Solve them simultaneously to find 'a' and \( b^2 \) (remembering 'a' must be positive). Then, write the final equation of the hyperbola.

Exam Tip: Be mindful of potential negative solutions when solving for 'a' or 'b' from quadratic equations; always choose the positive value since they represent lengths.

Question 14. Find the equation of hyperbola satisfying the given conditions: Vertices \( (\pm 7, 0) \), \( e = \frac{4}{3} \).
Answer: The given vertices are \( (\pm 7, 0) \). This indicates that \( a = 7 \).
Since the vertices are on the x-axis, the transverse axis is along the x-axis.
The eccentricity is given as \( e = \frac{4}{3} \).
We know that \( e = \frac{c}{a} \).
\( \implies \frac{4}{3} = \frac{c}{7} \)
\( \implies c = \frac{4 \times 7}{3} = \frac{28}{3} \)
The equation of the hyperbola will be of the form \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \).
We know the relationship \( c^2 = a^2 + b^2 \).
\( \implies \left(\frac{28}{3}\right)^2 = 7^2 + b^2 \)
\( \implies \frac{784}{9} = 49 + b^2 \)
\( \implies b^2 = \frac{784}{9} - 49 \)
\( \implies b^2 = \frac{784 - (49 \times 9)}{9} = \frac{784 - 441}{9} = \frac{343}{9} \)
Therefore, the equation of the hyperbola is \( \frac{x^2}{7^2} - \frac{y^2}{343/9} = 1 \).
\( \implies \frac{x^2}{49} - \frac{9y^2}{343} = 1 \)
To remove the fraction in the denominator of \( y^2 \), multiply the whole equation by 343:
\( \implies \frac{343x^2}{49} - 9y^2 = 343 \)
\( \implies 7x^2 - 9y^2 = 343 \).
In simple words: The vertices give 'a', and the eccentricity 'e' lets you calculate 'c'. Since the vertices are on the x-axis, the hyperbola is horizontal. Use \( c^2 = a^2 + b^2 \) to find \( b^2 \). Then, substitute these values into the standard equation for a horizontal hyperbola and simplify to a cleaner form.

Exam Tip: When 'e' is provided, use \( e = \frac{c}{a} \) as a key equation alongside \( c^2 = a^2 + b^2 \) to solve for the unknown parameters.

Question 15. Find the equation of hyperbola satisfying the given conditions: Foci \( (0, \pm \sqrt{10}) \), passing through \( (2, 3) \).
Answer: The given foci are \( (0, \pm \sqrt{10}) \). This indicates that \( c = \sqrt{10} \).
Since the foci are on the y-axis, the transverse axis is along the y-axis.
The general equation of such a hyperbola is \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \).
We know that \( c^2 = a^2 + b^2 \).
\( \implies (\sqrt{10})^2 = a^2 + b^2 \)
\( \implies 10 = a^2 + b^2 \)
\( \implies b^2 = 10 - a^2 \)
Substitute \( b^2 \) into the hyperbola equation:
\( \frac{y^2}{a^2} - \frac{x^2}{10-a^2} = 1 \)
The hyperbola passes through the point \( (2, 3) \). Substitute \( x = 2 \) and \( y = 3 \) into the equation:
\( \frac{3^2}{a^2} - \frac{2^2}{10-a^2} = 1 \)
\( \implies \frac{9}{a^2} - \frac{4}{10-a^2} = 1 \)
Multiply the entire equation by \( a^2(10-a^2) \) to clear the denominators:
\( 9(10-a^2) - 4a^2 = a^2(10-a^2) \)
\( \implies 90 - 9a^2 - 4a^2 = 10a^2 - a^4 \)
\( \implies 90 - 13a^2 = 10a^2 - a^4 \)
Rearrange the terms to form a quadratic equation in \( a^2 \):
\( a^4 - 23a^2 + 90 = 0 \)
Let \( A = a^2 \). Then the equation becomes \( A^2 - 23A + 90 = 0 \).
Factor the quadratic equation:
\( (A - 5)(A - 18) = 0 \)
This gives \( A = 5 \) or \( A = 18 \).
So, \( a^2 = 5 \) or \( a^2 = 18 \).
Case 1: If \( a^2 = 5 \).
Then \( b^2 = 10 - a^2 = 10 - 5 = 5 \). This is a valid value for \( b^2 \).
The equation of the hyperbola is \( \frac{y^2}{5} - \frac{x^2}{5} = 1 \).
\( \implies y^2 - x^2 = 5 \).
Case 2: If \( a^2 = 18 \).
Then \( b^2 = 10 - a^2 = 10 - 18 = -8 \). This is not possible because \( b^2 \) must be a positive value.
Therefore, the only valid equation for the hyperbola is \( y^2 - x^2 = 5 \).
In simple words: The foci provide 'c', and since they are on the y-axis, the hyperbola is vertical. Use \( c^2 = a^2 + b^2 \) to express \( b^2 \) in terms of \( a^2 \). Substitute this into the general hyperbola equation. Then, plug in the coordinates of the point \( (2, 3) \) that the hyperbola passes through. Solve the resulting equation to find \( a^2 \), making sure to discard any solutions that lead to a negative \( b^2 \). Finally, write the equation of the hyperbola.

Exam Tip: When a hyperbola passes through a given point, substitute its coordinates into the general equation to establish an additional relationship, which often helps in solving for the unknown parameters.