Answer to Question #309220 in Calculus for Nurul

Question

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

Answer 1

y = 2x² - 5x + 6

$\frac{dy}{dx} = 4x - 5$

$[\frac{dy}{dx} ]_{(2,4)}=4*2-5 = 3$

$[-\frac{dx}{dy} ]_{(1,3)}=[\frac{1}{5-4x}]_{(1,3)}=\frac{1}{5-4}=1$

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)