Answer to Question #305323 in Calculus for Nurul

Question

The tangent to the curve 𝑦 = 2𝑥2 − 5𝑥 + 6 at the point (2,1) intersects the normal to the same curve at the point (1,4) at point 𝑄. Find the coordinates of 𝑄

Answer 1

$$y=2x^2-5x+6$$

Point $(2,1)$ does not lie on the curve. Point $(1,4)$ does not lie on the curve.

The tangent to the curve $y=2x^2-5x+6$ at the point $(2,4)$ intersects the normal to the same curve at the point $(1,3)$ at point $𝑄.$ Find the coordinates of $Q.$

$$y'=4x-5$$

Point $(2,4)$

$$slope_1=m_1=4(2)-5=3$$

The equation of the tangent line at $(2, 4)$

$$y=3(x-2)+4$$

$$y=3x-2$$

Point $(1,3)$

$$slope_2=m_2=-\dfrac{1}{3(1)-5}=-\dfrac{1}{2}$$

The equation of the normal line at $(1, 4)$

$$y=-\dfrac{1}{2}(x-1)+3$$

$$y=-\dfrac{1}{2}x+\dfrac{7}{2}$$

Then

$$3x-2=-\dfrac{1}{2}x+\dfrac{7}{2}$$

$$x=\dfrac{11}{7}$$

$$y=3(\dfrac{11}{7})-2=\dfrac{19}{7}$$

$Q\ (\dfrac{11}{7}, \dfrac{19}{7})$