Question

4. Find the equation of the tangent line to curve 3 (x2 + y2) ? = 25 (x2 - y2),= 1 at (2,1).

Answer 1

we have

3.22 + y)? 25(x2 - y2

$\frac{d}{dx}3(x^2+y^2)^2=\frac{d}{dx}25(x^2-y^2)$

$3\cdot 2(x^2+y^2)\left (2x+2y\frac{dy}{dx} \right )=25\left (2x-2y\frac{dy}{dx} \right )$

$12x(x^2+y^2)+12y(x^2+y^2)\frac{dy}{dx}=50x-50y\frac{dy}{dx}$

$12y(x^2+y^2)\frac{dy}{dx}+50y\frac{dy}{dx}=50x-12x(x^2+y^2)$

$(12y(x^2+y^2)+50y)\frac{dy}{dx}=50x-12x(x^2+y^2)$

$\frac{dy}{dx}=\frac{50x-12x(x^2+y^2)}{12y(x^2+y^2)+50y}$

$\frac{dy}{dx}=\frac{25x-6x(x^2+y^2)}{6y(x^2+y^2)+25y}$

the slope of the tangent line is,

dy m de

$m=\frac{25x-6x(x^2+y^2)}{6y(x^2+y^2)+25y}$

slope at (2,1),

$m=\frac{25(2)-6(2)(2^2+1^2)}{6(1)(2^2+1^2)+25(1)}$

$m=\frac{50-12(5)}{6(5)+25}$

$m=\frac{50-60}{30+25}$

$m=-\frac{10}{55}$

$m=-\frac{2}{11}$

the equation of the tangent line is,

Or - 2) = 0h - h

put the x₀ = 2, y₀ = 1 and m = -2/11

$y-1=-\frac{2}{11}(x-2)$

$y=-\frac{2}{11}x+\frac{4}{11}+1$

$y=-\frac{2}{11}x+\frac{15}{11}$