Equation of the tangent to the curve `2x^(2)+3y^(2)-5=0` at (1, 1) is

Similar Questions

Explore conceptually related problems

The equation of tangent to the curve 2x^(2)+3y^(2)-5=0 at (1,1) is

Find the equation of the tangent to the curve x^(2/3)+y^(2/3)=2 at (1,1).

The equation of the tangent to the curve: x^2-y^2-8x+2y+11=0 at (2,1) is :

The equation of the tangent to the curve 2x^2-xy+3y^2=18 at (3,1) is

The equation of normal to the curve 2x^(2)+3y^(2)-5=0 at (1,1) is

Find the equation of the tangent to the curve. y=x^(3)+3x^(2)-5 and which is perpendicular to 2x-6y+1=0

The equation of the tangent to the curve y = e^(2x) at (0,1) is

Find the equation of the tangent to the curve (x^2)/(a^2)+(y^2)/(b^2)=1 at (x_1,\ y_1) on it.

Find the equation of the tangent of the curve y=3x^(2) at (1, 1).