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Find an equation of the tangent line to the curve 2(x2+y2)2=25(x2−y2) (a lemniscate) at the point (−3,1). An equation of the tangent line to the lemniscate at the given point is

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LammettHash

[tex]2(x^2+y^2)^2=25(x^2-y^2)[/tex]

Let [tex]y=y(x)[/tex], so that differentiating both sides wrt [tex]x[/tex] gives

[tex]4(x^2+y^2)\left(2x+2y\dfrac{\mathrm dy}{\mathrm dx}\right)=25\left(2x-2y\dfrac{\mathrm dy}{\mathrm dx}\right)[/tex]

If [tex]x=-3[/tex] and [tex]y=1[/tex], the above reduces to

[tex]40\left(-6+2\dfrac{\mathrm dy}{\mathrm dx}\right)=25\left(-6-2\dfrac{\mathrm dy}{\mathrm dx}\right)\implies\dfrac{\mathrm dy}{\mathrm dx}=\dfrac9{13}[/tex]

This is the slope of the tangent line, which has equation

[tex]y-1=\dfrac9{13}(x+3)\implies\boxed{y=\dfrac{9x+40}{13}}[/tex]

Ver imagen LammettHash
abidemiokin

The equation of the tangent line to the curve is expressed as [tex]y-1= 1/5(x+3)[/tex]

The formula for calculating the equation of a line in point-slope form is expressed as y-y0 = m(x-x0)

m is the slope

(x0, y0) is the point on the line

Given the equation [tex]2(x^2+y^2)^2=25(x^2-y^2)[/tex]

To get the required slope, we will need to differentiate the equation implicitly with respect to x as shown:

[tex]4(x^2+y^2)*2x+2y\frac{dy}{dx}=25(2x-2y\frac{dy}{dx} ) \\\ 4(x^2+y^2)*2x+2y\frac{dy}{dx}}=50x-50y\frac{dy}{dx}[/tex]

Substituting the coordinate point (-3, 1)

[tex]4(9+1)*(-6)+2\frac{dy}{dx}}=50(-3)-50\frac{dy}{dx}\\40-12\frac{dy}{dx}=-150-50 \frac{dy}{dx}\\190=-38\frac{dy}{dx}\\ \frac{dy}{dx}=-\frac{1}{5}[/tex]

Write the expression in point slope form:

[tex]y-1= 1/5(x+3)[/tex]

Hence the equation of the tangent line to the curve is expressed as [tex]y-1= 1/5(x+3)[/tex]

Learn more here: https://brainly.com/question/6617153

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