Question 61 of 70intermediate🔧 ApplyShort Answer2 marks

In each of the Exercises 7 to 15, find the equations of the hyperbola satisfying the given conditions: vertices (± 7, 0), e = 4/3.

Correct Answer

The equation of the hyperbola is x249−y2343/9=1\frac{x^2}{49} - \frac{y^2}{343/9} = 1 or x249−9y2343=1\frac{x^2}{49} - \frac{9y^2}{343} = 1.

Exercise: EXERCISE 10.4 | Q: 14 | (Page 27)
For More Understanding

Explanation

This question tests understanding of hyperbola properties. Since vertices are at (±7, 0), the hyperbola has its transverse axis along the x-axis with center at origin. The value of a is determined from vertices, then eccentricity gives c, and finally b² is calculated using the hyperbola relationship c² = a² + b².

Solution Steps

  1. Step 1: Since vertices are (±7, 0), the hyperbola has transverse axis on x-axis with a = 7 and a² = 49.

  2. Step 2: Using eccentricity formula e = c/a, we get c = e × a = (4/3) × 7 = 28/3.

  3. Step 3: For hyperbola, c² = a² + b², so b² = c² - a² = (28/3)² - 49 = 784/9 - 441/9 = 343/9.

  4. Step 4: The standard equation of hyperbola with transverse axis on x-axis is x²/a² - y²/b² = 1.

  5. Step 5: Substituting values: x²/49 - y²/(343/9) = 1 or equivalently x²/49 - 9y²/343 = 1.

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