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

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

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Answer 1 · 2016-10-21T12:38:30

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

Explanation:

Let us find the derivative $\frac{d y}{d x}$ $(dy)/(dx)$ for $y^{2} (y^{2} - 4) = x^{2} (x^{2} - 5)$ $y^2(y^2-4)=x^2(x^2-5)$ , using implicit differentiation. The differential is

$y^{2} \times 2 y \times \frac{d y}{d x} + 2 y \times (y^{2} - 4) \times \frac{d y}{d x} = x^{2} \times 2 x + 2 x \times (x^{2} - 5)$ $y^2xx2yxx(dy)/(dx)+2yxx(y^2-4)xx(dy)/(dx)=x^2xx2x+2x xx(x^2-5)$

or $2 y^{3} \frac{d y}{d x} + (2 y^{3} - 8 y) \frac{d y}{d x} = 2 x^{3} + (2 x^{3} - 10 x)$ $2y^3(dy)/(dx)+(2y^3-8y)(dy)/(dx)=2x^3+(2x^3-10x)$

or $\frac{d y}{d x} = \frac{4 x^{3} - 10 x}{4 y^{3} - 8 y}$ $(dy)/(dx)=(4x^3-10x)/(4y^3-8y)$

and at $x = 0$ $x=0$ and $y = - 2$ $y=-2$ , $\frac{d y}{d x} = 0$ $(dy)/(dx)=0$

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

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

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

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

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Explanation:

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Find the equation of tangent to curve y2(y2−4)=x2(x2−5)y^2(y^2-4)=x^2(x^2-5) at point (0,−2)(0,-2)?

1 Answer

Explanation:

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