Get live Maths 1-on-1 Classs - Class 6 to 12

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Chapter 6 Class 12 Application of Derivatives

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Last updated at March 23, 2023 by Teachoo

Example 16 Find the equation of all lines having slope 2 and being tangent Equation of curve is y + 2/(π₯ β 3) = 0 Differentiating both sides w.r.t x (ππ¦ )/ππ₯ + (π )/ππ₯ (2/(π₯ β 3))=0 (ππ¦ )/ππ₯ =β(π )/ππ₯ (2/(π₯ β 3)) (ππ¦ )/ππ₯ =β2 (π )/ππ₯ (π₯β3)^(β1) (ππ¦ )/ππ₯ =β2γ Γ β(π₯β3)γ^(β1β1) (ππ¦ )/ππ₯ =2(π₯β3)^(β2) (π π )/π π =π/(π β π)^π Given that slope = 2 ππ¦/ππ₯ = 2 2/(π₯ β 3)^2 = 2 1/(π₯ β 3)^2 = 1 (π₯β3)^2 = 1 π₯β3 = Β±1 x β 3 = 1 x = 4 x β 3 = β 1 x = 2 x β 3 = β 1 x = 2 So, x = 4 & x = 2 Finding value of y If x = 2 y = (β2)/(π₯ β 3) y = (β2)/(2 β 3) π¦=2 Thus, point is (2, 2) If x = 4 y = (β2)/(π₯ β 3) y = (β2)/(4 β 3) π¦=β2 Thus, point is (4, β2) Thus, there are 2 tangents to the curve with slope 2 and passing through points (2, 2) and (4, β 2) We know that Equation of line at (π₯1 , π¦1)& having Slope m is π¦βπ¦1=π(π₯βπ₯1) Equation of tangent through (2, 2) is π¦ β 2 = 2 (π₯ β2) π¦ β 2 = 2π₯ β 4 πβππ+π = π Equation of tangent through (4, β2) is π¦ β(β2) = 2 (π₯ β4) π¦ + 2 = 2π₯ β 8 π β ππ + ππ = π to the curve y + 2/(π₯ β 3) = 0