Find the equation of hyperbola x^2/36- y^2/9=1 at the point whose abscissa is 10
Answers
5x - 8y = 18 & 5x + 8y = 18 are tangent of hyperbola x²/36 - y²/9 = 1 at the point whose abscissa is 10
Step-by-step explanation:
complete question
Find the equation of tangent of hyperbola x²/36 - y²/9 = 1 at the point whose abscissa is 10
x²/36 - y²/9 = 1
abscissa = 10
=> x = 10
=> 10²/36 - y²/9 = 1
=> y²/9 = 10²/36 - 1
=> y²/9 = 100/36 - 1
=> y²/9 = (100 - 36)/36
=> y² = 64/4
=> y² = 16
=> y = ± 4
Points are (10 , 4) & (10 , -4)
x²/36 - y²/9 = 1
=> 2x/36 - 2y/9 * dy/dx = 0
=> x - 4ydy/dx = 0
=> dy/dx =x/4y
At (10 , 4)
dy/dx = 10/16 = 5/8
Equation of tangent at (10 , 4)
y - 4 = (5/8)(x - 10)
=> 8y - 32 = 5x - 50
=> 5x - 8y = 18
dy/dx =x/4y
At (10 , -4)
dy/dx = 10/-16 = -5/8
Equation of tangent at (10 , -4)
y + 4 = (-5/8)(x - 10)
=> 8y + 32 = -5x + 50
=> 5x + 8y = 18
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