Question 62 of 70intermediate🔧 ApplyShort Answer2 marks

In each of the Exercises 7 to 15, find the equations of the hyperbola satisfying the given conditions: Foci (0, ± 10\sqrt{10}), passing through (2, 3)

Correct Answer

Since the foci are (0, ±10\sqrt{10}), the hyperbola has its transverse axis along the y-axis. Its equation is of the form y²/a² – x²/b² = 1. For a hyperbola, c² = a² + b², where c = 10\sqrt{10}, so a² + b² = 10. The hyperbola passes through (2,3). Substituting gives 9/a² – 4/b² = 1. Let a² = A, b² = B. Then A + B = 10 and 9B – 4A = AB. Solving yields A = 5, B = 5 (since A = 18 gives negative B). Thus the equation is y²/5 – x²/5 = 1, i.e., y² – x² = 5.

Exercise: EXERCISE 10.4 | Q: 15 | (Page 27)
For More Understanding

Explanation

The hyperbola has foci on the y-axis, so its standard equation is y2y^{2}/a2a^{2} - x2x^{2}/b2b^{2} = 1. Using the relation c2c^{2} = a2a^{2} + b2b^{2} with c = 10\sqrt{10} gives a2a^{2} + b2b^{2} = 10. Substituting the point (2,3) yields 9/a2a^{2} - 4/b2b^{2} = 1. Solving the system gives a2a^{2} = b2b^{2} = 5, leading to the final equation y2y^{2} - x2x^{2} = 5.

Solution Steps

  1. Step 1: Identify orientation: foci on y-axis → equation y²/a² – x²/b² = 1.

  2. Step 2: Use relation c² = a² + b² with c = 10\sqrt{10} → a² + b² = 10.

  3. Step 3: Substitute point (2,3): 9/a² – 4/b² = 1.

  4. Step 4: Let A = a², B = b². Solve system A + B = 10 and 9B – 4A = AB.

  5. Step 5: Substitute B = 10 – A into 9(10 – A) – 4A = A(10 – A) → A² – 23A + 90 = 0.

  6. Step 6: Solve: A = 5 or 18. A = 5 gives B = 5 (valid); A = 18 gives B = –8 (invalid).

  7. Step 7: Write equation: y²/5 – x²/5 = 1 → y² – x² = 5.

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