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EleoNora [17]
3 years ago
11

Find the coordinates of the point in the first quadrant at which the tangent line to the curve (x)^3-xy+(y)^3 =0 is parallel to

the x axis.
Mathematics
1 answer:
Oksanka [162]3 years ago
8 0
<span>Differentiate implicitly:
</span>3x^2-y-xy'+3y^2y'=0
<span>
Solve for y
</span>y'(3y^2-x)=y-3x^2&#10;\\&#10;\\y'={y-3x^2\over3y^2-x}

<span>When the tangent is parallel to the x-axis we have y'=0, so we must solve
</span>y'={y-3x^2\over3y^2-x}=0\implies y=3x^2

<span>To find the actual value of x we plug this expression for y into the original equation
</span>x^3-3x^3+27x^6=0&#10;\\&#10;\\x^3(27x^3-2)=0\implies x=\{0,{\sqrt[3]2\over3}\}

<span>Plugging this into the formula for y above gives the points
</span>(0,0)\text{ and }({\sqrt[3]2\over3},{\sqrt[3]4\over3})

<span>which is where our tangent will be parallel to the x-axis.</span>
<span>

</span>
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