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 = 2x² - 5x + 6
So slope of tangent at (2,4) is 3 and that of normal at (1,3) is 1.
So equation of tangent at (2,4) is
y - 4 = 3(x-2) => y = 3x -2 ........(1)
And equation of normal at (1,4) is
y - 3 = 1(x-1) => y = x + 2 ........(2)
Solving equation (1) and equation (2) we get
3x - 2 = x + 2
=> 3x - x = 2+2
=> 2x = 4
=> x = 2
So y = 2+2 = 4
Therefore the point of intersection of the tangent and the normal is (2,4)
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