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Example 18 - Chapter 6 Class 12 Application of Derivatives (Term 1)
Last updated at April 19, 2021 by Teachoo
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Chapter 6 Class 12 Application of Derivatives
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Transcript
Example 18 Find the equation of the tangent to the curve y = (π₯ β 7)/((π₯ β 2)(π₯ β 3)) at the point where it cuts the x-axis.Slope of the tangent to the curve is ππ¦/ππ₯ = ((π₯ β 7)^β² [(π₯ β 2) (π₯ β 3)]β (π₯ β 7) [(π₯ β 3) (π₯ β 2)]^β²)/((π₯ β 2)^2 (π₯ β 3)^2 ) ππ¦/ππ₯ = (1 Γ (π₯ β 2) (π₯ β 3) β (π₯ β 7)[(π₯ β 3)^β² (π₯ β 2) + (π₯ β 3) (π₯ β 2)^β² ])/((π₯ β 2)^2 (π₯ β 3)^2 ) ππ¦/ππ₯ = ((1) (π₯ β 2) (π₯ β 3) β (π₯ β 7)[1 Γ (π₯ β 2) + (π₯ β 3) Γ 1])/((π₯ β 2)^2 (π₯ β 3)^2 ) ππ¦/ππ₯ = ((π₯ β 2) (π₯ β 3) β (π₯ β 7)(2π₯ β 5))/((π₯ β 2)^2 (π₯ β 3)^2 ) ππ¦/ππ₯ = ((π₯ β 2) (π₯ β 3) )/((π₯ β 2)^2 (π₯ β 3)^2 )β(π₯ β 7)(2π₯ β 5)/((π₯ β 2)^2 (π₯ β 3)^2 ) ππ¦/ππ₯ = (1 )/((π₯ β 2) (π₯ β 3) )β((π₯ β 7))/((π₯ β 2) (π₯ β 3) ) Γ ((2π₯ β 5))/((π₯ β 2) (π₯ β 3) ) ππ¦/ππ₯ = (1 )/((π₯ β 2) (π₯ β 3) )βπ¦ Γ ((2π₯ β 5))/((π₯ β 2) (π₯ β 3) ) π π/π π = (π β π(ππ β π))/((π β π) (π β π) ) We need to find Equation of tangent at the point where the curve cuts the x axis, Thus, y = 0 We need to find value of x Putting y = 0 in equation of curve 0 = (π₯ β 7)/(π₯ β 2)(π₯ β 3) β΄ x = 7 Thus, curve cuts the x-axis at point (7, 0) Finding Slope at point (7, 0) ππ¦/ππ₯ = (1 β π¦(2π₯ β 5))/((π₯ β 2) (π₯ β 3) ) Putting x = 7, y = 0 = (1 β 0[2(7)β5])/((7 β 2) (7 β 3) ) = 1/(5 Γ 4) = π/ππ Now, Equation of the tangent at point (7, 0) with slope 1/20 is π¦ βπ¦_1= π (π₯ β π₯1) π¦β0= 1/20 (π₯β7) 20π¦=π₯β7 πππβπ+π=π
