Question #305323

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


1
Expert's answer
2022-03-03T18:06:55-0500
y=2x25x+6y=2x^2-5x+6

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


The tangent to the curve y=2x25x+6y=2x^2-5x+6 at the point (2,4)(2,4) intersects the normal to the same curve at the point (1,3)(1,3) at point 𝑄.𝑄. Find the coordinates of Q.Q.


y=4x5y'=4x-5

Point (2,4)(2,4)


slope1=m1=4(2)5=3slope_1=m_1=4(2)-5=3

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


y=3(x2)+4y=3(x-2)+4

y=3x2y=3x-2

Point (1,3)(1,3)

slope2=m2=13(1)5=12slope_2=m_2=-\dfrac{1}{3(1)-5}=-\dfrac{1}{2}

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

y=12(x1)+3y=-\dfrac{1}{2}(x-1)+3

y=12x+72y=-\dfrac{1}{2}x+\dfrac{7}{2}


Then


3x2=12x+723x-2=-\dfrac{1}{2}x+\dfrac{7}{2}

x=117x=\dfrac{11}{7}

y=3(117)2=197y=3(\dfrac{11}{7})-2=\dfrac{19}{7}

Q (117,197)Q\ (\dfrac{11}{7}, \dfrac{19}{7})



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