Question

2(x²+y²)²=25(x²-y²) at(3,1),Find the equation of the tangent to given curves at given point.

Answer 1

curve , 2(x² + y²)²= 25(x² - y²)
differentiate curve with respect to x,

4(x² + y²)(2x + 2y.dy/dx) = 25(2x - 2ydy/dx)

4(x² + y²)(x + y.dy/dx) = 25(x - y.dy/dx)

at (3,1) slope of tangent of curve => 4(3² + 1²)(3 + 1.dy/dx) = 25(3 - 1.dy/dx)

=> 4 × 10 × (3 + dy/dx) = 25(3 - dy/dx)

=> 40 × 3 + 40.dy/dx = 75 - 25.dy/dx

=> 45 + 65.dy/dx = 0

=> dy/dx = -9/13

now, equation of tangent of curve :
(y - 1) = -9/13(x - 3)

13(y - 1) + 9(x - 3) = 0

13y - 13 + 9x - 27 = 0

9x + 13y - 40 = 0

Answer 2

Dear Student:

Given: 2(x²+y²)²=25(x²-y²) at(3,1)

$x_{0} =3,y_{0} =1$

$m=\frac{dy}{dx} =slope of tangent$

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

$4(x^{2} +y^{2})(2x+2y\frac{dy}{dx})=25*(2x-2y\frac{dy}{dx})$

At, $x_{0} =3,y_{0} =1$

$4(3^{2} +1^{2})(2*3+2*1\frac{dy}{dx})=25*(2*3-2*1\frac{dy}{dx})$

$m=\frac{dy}{dx} =slope of tangent=-9/13$

Equation of tangent:

$(y-y_{0})=m(x-x_{0})$

$(y-1)=-9/13(x-3)$

Hope it helps

Thanks

With Regards