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astarmathsandphysics:
1 hour

astarmathsandphysics:
find the coordinates of the point where the tangent to the curve y=x^2 + 1  at the point (2,5)
meets the normal to the same curve at the point (1,2)
tangent to curve y=x^2 + 1  dy/dx=2x=2*2=4
y-5=4(x-2) so y=4x-3 (1)
normal at (1,2) has gradient -1/2*1=-1/2
y-2=(-1/2)(x-1) so y=-0.5x+2.5 (2)
(1) -(2) gives 0=4.5x-5.5 so x=9/11 and from (1) y=4*9/11 -3=3/11
(9/11,3/11)

halosh92:

--- Quote from: astarmathsandphysics on March 21, 2010, 07:13:39 am ---find the coordinates of the point where the tangent to the curve y=x^2 + 1  at the point (2,5)
meets the normal to the same curve at the point (1,2)
tangent to curve y=x^2 + 1  dy/dx=2x=2*2=4
y-5=4(x-2) so y=4x-3 (1)
normal at (1,2) has gradient -1/2*1=-1/2
y-2=(-1/2)(x-1) so y=-0.5x+2.5 (2)
(1) -(2) gives 0=4.5x-5.5 so x=9/11 and from (1) y=4*9/11 -3=3/11
(9/11,3/11)

--- End quote ---

how did you find the gradient of the normal..on wat basis did u come up with these equation? (normal at (1,2) has gradient -1/2*1=-1/2
y-2=(-1/2)(x-1) so y=-0.5x+2.5 )

astarmathsandphysics:
Dy/dx=2x and x=1 so dy/dx=2 and gradient of normal=-1/2

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