y=2x2−5x+6 Point (2,1) does not lie on the curve. Point (1,4) does not lie on the curve.
The tangent to the curve y=2x2−5x+6 at the point (2,4) intersects the normal to the same curve at the point (1,3) at point Q. Find the coordinates of Q.
y′=4x−5Point (2,4)
slope1=m1=4(2)−5=3The equation of the tangent line at (2,4)
y=3(x−2)+4
y=3x−2 Point (1,3)
slope2=m2=−3(1)−51=−21The equation of the normal line at (1,4)
y=−21(x−1)+3
y=−21x+27
Then
3x−2=−21x+27
x=711
y=3(711)−2=719 Q (711,719)
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