Find the equation of the normals to the circle x²+y²-8x - 2y +12 =0 at the point whose ordinate is -1
Answers
Given : circle x²+y²-8x-2y+12=0
To Find : equation of the tangent at the points whose ordinates are "1" or - 1
Solution:
x²+y²-8x-2y+12=0
=> 2x + 2y.dy/dx - 8 - 2dy/dx = 0
=> x + y .dy/dx - 4 - dy/dx = 0
=> (dy/dx)(y - 1) = 4 - x
=> (dy/dx) =( 4 - x)/(y - 1)
y = 1
=> dy/dx = (x - 4)/0
Parallel to y axis
x²+y²-8x-2y+12=0
ordinates are "1"
=> y = 1
=> x²+1-8x-2+12=0
=> x² - 8x + 11 = 0
x = (8 ± √64 - 44)/2
=> x = ( 4 ± √5)
Parallel to y axis
Equation of tangents x = 4 + √5 , x = 4 - √5
x²+y²-8x-2y+12=0
=> 2x + 2y.dy/dx - 8 - 2dy/dx = 0
=> x + y .dy/dx - 4 - dy/dx = 0
=> (dy/dx)(y - 1) = 4 - x
=> (dy/dx) =( 4 - x)/(y - 1)
ordinates are "-1"
y = -1
=> dy/dx = (4 - x)/-2
x²+y²-8x-2y+12=0
ordinates are "-1"
=> y = -1
=> x²+1-8x+2+12=0
=> x² - 8x + 15 = 0
(x-5)(x-3)=0
=> x = 5 , 3
dy/dx = (4 - x)/-2
x = 3 , y = -1
=> dy/dx = -1/2
y + 1 = (-1/2)(x - 3)
=> 2y + 2 = -x + 3
=> x + 2y = 1
dy/dx = (4 - x)/-2
x = 5 , y = -1
=> dy/dx = -1/-2 = 1/2
y + 1 = (1/2)(x - 5)
=> 2y + 2 = x -5
=> x - 2y = 7
Equation of Tangents x + 2y = 1 , x - 2y = 7
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