Mathematics, 29.01.2020 04:04 claythememe
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)
Answers: 2
![\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] \\\\\\ ](/tpl/images/0480/1986/5699d.png)
![\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](/tpl/images/0480/1986/f824e.png)
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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 poin...
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