Finding point when tangent is parallel/ perpendicular

Chapter 6 Class 12 Application of Derivatives
Concept wise

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Ex 6.3, 25 Find the equation of the tangent to the curve β(3π₯β2) which is parallel to the line 4x β 2y + 5 = 0 . Let (β , π) be the point on Curve from tangent to be taken We know that Equation of tangent is ππ¦/ππ₯ π¦=β(3π₯ β2) Differentiating w.r.t.π₯ ππ¦/ππ₯=(π(3π₯ β2)^(1/2))/ππ₯ ππ¦/ππ₯=3/(2β(3π₯ β2)) Slope of tangent at (β , π) is γππ¦/ππ₯βγ_((β , π) )=3/(2β(3β β 2)) Given tangent is parallel to the line 4π₯β2π¦+5 So , Slope of tangent = Slope of 4π₯β2π₯+5 Now, Given line is 4π₯β2π¦+5=0 β2π¦=β4π₯β5 2π¦=4π₯+5 π¦=(4π₯ + 5)/2 π¦=2π₯+5/2 The above Equation is of the form π¦=ππ₯+π where m is Slope of line β΄ Slope of line is 2 Now, Slope of tangent at (β , π)= Slope of line 4π₯β3π¦+5=0 3/(2β(3β β2))=2 3=2 Γ2β(3β β2) 3=4β(3β β2) Squaring Both Sides (3)^2=(4β(3β β2))^2 9=(4)^2 (β(3β β2))^2 9=16(3β β2) 9/16=3β β2 3β β2=9/16 3β=9/16+2 3β=(9 + 32)/16 3β=41/16 β=41/(16 Γ 3) β=41/48 Now, π¦=β(3π₯ β2) Since Point (β , π) is on the Curve Point (β , π) Satisfies the Equation of Curve Putting π₯=β , π¦=π π=β(3β β2) Finding k when β=41/48 π=β(3 Γ41/48β2) =β(41/16β2)=β((41 β32)/16)=β(9/16)=3/4 Hence the point is (h, k) = (41/48 , 3/4) We know that Equation of line at (π₯1 , π¦1)& having Slope m is π¦βπ¦1=π(π₯βπ₯1) Equation of tangent at (41/48 , 3/4) & having Slope 2 is (π¦β3/4)=2(π₯β41/48) (4π¦ β 3)/4=2((48π₯ β 41)/48) (4π¦ β 3)/4=(48π₯ β 41)/24 24(4π¦ β 3)/4=48π₯β41 6(4π¦β3)=48π₯β41 24π¦β18=48π₯β41 48π₯β41β24π¦+18=0 48π₯β24π¦β23=0 48π₯β24π¦=23 Hence Required Equation of tangent is πππβπππ=ππ