Finding point when tangent is parallel/ perpendicular

Chapter 6 Class 12 Application of Derivatives
Concept wise

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Ex 6.3, 18 For the curve π¦=4π₯3 β2π₯5, find all the points at which the tangent passes through the origin.Let (β , π) be the Required Point on the Curve at which tangent is to be taken Given Curve is π¦=4π₯^3β2π₯^5 Since Point (β , π) is on the Curve β (β , π) will satisfy the Equation of Curve Putting π₯=β , π¦=π π=4β^3β2β^5 We know that Slope of tangent to the Curve is ππ¦/ππ₯ π¦=4π₯^3β2π₯^5 Differentiating w.r.t. π₯ ππ¦/ππ₯=π(4π₯^(3 )β 2π₯^5 )/ππ₯ ππ¦/ππ₯=12π₯^2β10π₯^4 Since tangent is taken from (β , π) Slope of tangent at (β, π) is γππ¦/ππ₯βγ_((β, π) )=12β^2β10β^4 Now, Equation of tangent at (β , π) & having Slope 12β^2β10β^4 is (π¦βπ)=12β^2β10β^4 (π₯ββ) Also, Given tangent passes through Origin β΄ (0 , 0) will satisfies the Equation of tangent Putting x = 0, y = 0 in equation (0 βπ)=12β^2β10β^4 (0ββ) βπ=12β^2β10β^4 (ββ) βπ=β12β^3+10β^5 We know that Equation of line at (π₯1 , π¦1)& having Slope m is π¦βπ¦1=π(π₯βπ₯1) Now our Equations are π=4β^3β2β^5 βπ=β12β^3+10β^5 Adding (1) and (2) πβπ=4β^3β2β^5+(β12β^3+10β^5 ) 0=4β^3β12β^3β2β^5+10β^5 0=β 8β^3+8β^5 β 8β^3+8β^5=0 β 8β^3 (1ββ^2 )=0 β 8β^3=0 β=0 1ββ^2=0 β^2=1 β=Β±1 When π=π π=4β^3β2β^5 π=4(0)^3β2(0)^5 π=0β0 π=0 Point is (π , π) When π=π π=4β^3β2β^5 π=4(1)^3β2(1)^5 π=4β2 π=2 Point is (π , π) When π=βπ π=4β^3β2β^5 π=4(β1)^3β2(β1)^5 π=β4β(β2) π=β2 Point is (βπ , βπ) Hence the Required point on the Curve are (0 , 0), (1 , 2) & (β1 , β2)