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AP EAMCET202018 Sep 2020Morning ShiftMathematicsApplication of DerivativesActual

Find the equation of the normal to the curve (y= (x-7) (x-2)(x-3) ) at the point where it cuts the (X )-axis.

Options

  1. A(20 x+y+140=0 )
  2. B(x-20 y-140=0 )
  3. C(x+20 y+140=0 )
  4. D(20 x+y-140=0 )

Correct answer

D. (20 x+y-140=0 )

Step-by-step solution

Given curve, (y= (x-7) (x-2)(x-3) ),cuts the (x )-axis at point (P(7,0) ). ( aligned & Now, d y d x = (x-2)(x-3)-(x-7)[(x-2)+(x-3)] (x-2)^2(x-3)^2 & . d y d x |_ x=7 = 5 4 5^2 4^2 = 1 20 aligned ) So, the slope of normal to given curve at point (P ) is ( - 2 0 ). Therefore equation of required normal is (y-0=-20(x-7) 20 x+y-140=0 )

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