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Find the equation of tangent to curve #y^2(y^2-4)=x^2(x^2-5)# at point #(0,-2)#?

Answer 1

Samantha Adkison

Equation of tangent at #(0,-2)# is #y+2=0#

Let us find the derivative #(dy)/(dx)# for #y^2(y^2-4)=x^2(x^2-5)#, using implicit differentiation. The differential is

#y^2xx2yxx(dy)/(dx)+2yxx(y^2-4)xx(dy)/(dx)=x^2xx2x+2x xx(x^2-5)#

or #2y^3(dy)/(dx)+(2y^3-8y)(dy)/(dx)=2x^3+(2x^3-10x)#

or #(dy)/(dx)=(4x^3-10x)/(4y^3-8y)#

and at #x=0# and #y=-2#, #(dy)/(dx)=0#

As slope of tangent is #0# and it passes through #(0,-2)#

the equation of tangent is #(y+2)=0(x-0)# or #y+2=0#

graph{(y+2)(y^2(y^2-4)-x^2(x^2-5))=0 [-10, 10, -5, 5]}

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Answer 2

Charles Anctil

The equation of the tangent to the curve y^2(y^2-4)=x^2(x^2-5) at the point (0,-2) is y = -2.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.