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Flauer [41]
3 years ago
7

Find an equation of the tangent line to the curve 2(x^2+y^2)2=25(x^2−y^2) (a lemniscate) at the point (3,1)

Mathematics
1 answer:
Sholpan [36]3 years ago
6 0
\bf 2[x^2+y^2]^2=25(x^2-y^2)\qquad \qquad 
\begin{array}{lllll}
&x_1&y_1\\
%   (a,b)
&({{ 3}}\quad ,&{{ 1}})\quad 
\end{array}\\\\
-----------------------------\\\\
2\left[ x^4+2x^2y^2+y^4 \right]=25(x^2-y^2)\qquad thus
\\\\\\
2\left[ 4x^3+2\left[ 2xy^2+x^22y\frac{dy}{dx} \right]+4y^3\frac{dy}{dx} \right]=25\left[2x-2y\frac{dy}{dx}  \right]
\\\\\\
2\left[ 4x^3+2\left[ 2xy^2+x^22y\frac{dy}{dx} \right]+4y^3\frac{dy}{dx} \right]=50\left[x-y\frac{dy}{dx}  \right]
\\\\\\


\bf \left[ 4x^3+2\left[ 2xy^2+x^22y\frac{dy}{dx} \right]+4y^3\frac{dy}{dx} \right]=25\left[x-y\frac{dy}{dx}  \right]
\\\\\\
4x^3+4xy^2+4x^2y\frac{dy}{dx}+4y^3\frac{dy}{dx}+25y\frac{dy}{dx}=25x
\\\\\\
\cfrac{dy}{dx}[4x^2y+4y^3+25y]=25x-4x^3+4xy^2
\\\\\\
\cfrac{dy}{dx}=\cfrac{25x-4x^3+4xy^2}{4x^2y+4y^3+25y}\impliedby m=slope

notice... a derivative is just the function for the slope

now, you're given the point 3,1, namely x = 3 and y = 1

to find the "m" or slope, use that derivative, namely f'(3,1)=\cfrac{25x-4x^3+4xy^2}{4x^2y+4y^3+25y}

that'd give you a value for the slope

to get the tangent line at that point, simply plug in the provided values
in the point-slope form

\bf y-{{ y_1}}={{ m}}(x-{{ x_1}})\qquad
\begin{cases}
x_1=3\\
y_1=1\\
m=slope
\end{cases}\\ \qquad \uparrow\\
\textit{point-slope form}

and then you solve it for "y", I gather you don't have to, but that'd be the equation of the tangent line at 3,1

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