[tex]\underline{\underline{\large{\red{\mathcal{✒GIVEN:}}}}}[/tex]
A curve has an equation
[tex]\bullet \: \: \rm{x^{3}−4xy+y3=0}[/tex]
[tex]\underline{\underline{\large{\red{\mathcal{REQUIRED:}}}}}[/tex]
The equation of the tangent to the curve at the point (0, -3)
[tex]\underline{\underline{\large{\red{\mathcal{SOLUTION:}}}}}[/tex]
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1. To find the equation of the tangent to the curve at the point (0, -3) , we first need to find dy/dx using implicit differentiation:
[tex]\small{\tt{ {3x}^{2} - 4x \dfrac{d}{y} - 4y + 3 {y}^{2} \dfrac{dy}{dx} = 0}}[/tex]
[tex]\tt{(3 {y}^{2} - 4x) \dfrac{dy}{dx} = 4y - {3x}^{2} }[/tex]
[tex]\tt{ \dfrac{dy}{dx} = \dfrac{4y - {3x}^{2} }{3 {y}^{2} - 4x } }[/tex]
2. Now we evaluate [tex]\rm{\dfrac{dy}{dx}}[/tex] at the point (0, -3):
[tex]\tt{ \dfrac{dy}{dx} = \dfrac{4( - 3) - 3(0 {)}^{2} }{3( - 3 {)}^{2} - 4(0) } }[/tex]
[tex]\tt{ \dfrac{dy}{dx} = - \dfrac{12}{27} }[/tex]
[tex]\tt{ \dfrac{dy}{dx} = - \dfrac{4}{9} }[/tex]
3. The slope of the tangent at (0, -3) is -4/9.
The equation of the tangent line is given by:
[tex]\small{\boxed{ \bm{{ \red{y - y_{1} = m(x - x_{1}) }}}}}[/tex]
[tex]\tt{y - ( - 3) = - \dfrac{4}{9} (x - 0)}[/tex]
[tex]\tt{y + 3 = - \dfrac{4}{9} x}[/tex]
[tex]\boxed{ \tt{ \purple{ \large{y = - \dfrac{4}{9} x - 3}}}}[/tex]
Final Answer:
Thus, the equation of the tangent line to the curve at the point (0,−3) is
[tex] \rm{ \purple{ \large{D. \: y = - \dfrac{4}{9} x - 3}}}[/tex].
