Math, asked by chsmqrimaptopalaik, 1 year ago

The length of the major axis of the ellipse (5x?10)^2+(5y+15)^2 = (3x-4y+7)^2/4 is

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Answered by kvnmurty
10
Ellipse : 

4 (5 x + 10)² + 4 (5 y + 15)² = (3 x - 4 y + 7)²
91 x² + 24 x y + 84 y² + 1251 + 358 x + 656 y = 0

For an Ellipse: A x² + B xy + C y² + D x + E y + F = 0, the slope of major axis = m = [C - A - √{ (A-C)² + B² } ] / B      for  B ≠ 0

Substituting the values we get:  m = [84-91-√(7²+24²)]/24 = -4/3
Slope of the minor axis = -1/m = 3/4

So Ellipse equation is : (y +4x/3+c)² / a² + (y - 3x/4+d)² / b² = 1

So comparing coefficients:
       a² + b² = 84
   9 a² /16 + 16 b² / 9 = 91  =>  81 a² + 256 b² = 144 * 91
   - 3 a²/2  + 8 b² /3 = 24     =>  - 9 a² + 16 b² = 144

Solving these equations we get: b² = 36  and  a² = 48

The length of major axis = 2 a = 8 √3  units.
                     Minor axis  = 2 b = 12 units

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Answered by Neeleshcs
2

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