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प्रश्न
Find the equations of the tangent and normal to the curve x = a sin3θ and y = a cos3θ at θ=π/4.
उत्तर
Given: ` x=a sin^3θ and y=a cos^3θ`
`⇒dx/dθ=3a sin^2θ cosθ and dy/dθ=−3a cos^2θ sinθ`
`⇒dy/dx=(dy/dθ)/(dx/dθ)`
`=(−3a cos^2θ sinθ)/(3a sin^2θ cosθ)`
`=-costheta/sintheta`
So, the equation of the tangent is given by
`y−a cos^3θ=−cosθ/sinθ(x−a sin^3θ) [Using y−y1=(dy/dx)_(x1,y1)(x−x1)]`
Putting θ=π/4, we get:
`y-a(1/sqrt2)^3=-(1/sqrt2)/(1/sqrt2)(x-a(1/sqrt2)^3) (∵ sin(pi/4)=1/sqrt2, cos(pi/4)=1/sqrt2`
`y-a1/(2sqrt2)=-1xx(x-a/(2sqrt2))`
`x+y=a/(2sqrt2)+a/(2sqrt2)=a/sqrt2`
`x+y=a/sqrt2 or sqrt2x+sqrt2y=a`
Thus, the equation of the tangent at `θ=pi/4 is sqrt2x+sqrt2y=a.`
Now, the equation of the normal is given by
`y-acos^3theta=-1/(-costheta/sintheta)(x-asin^3theta) [`
Putting θ=π/4, we get:
`y-a(1/sqrt2)^3=-1/((-1/sqrt2)/(1/sqrt2))(x-a(1/sqrt2)^3) (∵ sin(pi/4)=1/sqrt2, cos(pi/4)=1/sqrt2)`
`y-a(1/2sqrt2)=1xx(x-a/(2sqrt2))`
`=x-y=a/(2sqrt2)-a/(2sqrt2)=0`
x-y=0
∴ The equation of the normal at θ=π/4 is x − y = 0.
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