# Find The Equation Of The Tangent Line To The Curve

 AnonymousFrom: -Posts: -Votes: - Find the equation of the tangent line to the curve (x^2 +y^2)^2 = 4x^2yat the point (1,1).

 AnonymousFrom: -Posts: -Votes: - I think that y = is not meant to be there. So I'm going to answer the question "Find the equation of the tangent line to the curve x² + 4xy + y² = 13 at the point (2, 1)." Are you familiar with implicit differentiation? If we differentiate both sides of this equation with respect to x, the first term gives 2x we need the product formula to differentiate 4xy: It's 4*derivative of x * y + 4*x*derivative of y = 4y + 4xy' using y' to denote dy/dx we use the chain rule to differentiate y²: It's 2y*y'. And the derivative of 13 is 0. So we get 2x + 4y + 4xy' +2yy' = 0 Therefore 4xy' + 2yy' = -2x - 4y 2(2x + y)y' = -2(x + 2y) y' = -(x + 2y)/(2x + y) after cancelling the common factor 2. At the point (2, 1), y' = -(2 + 2)/(4 + 1) i.e. gradient of tangent = -4/5 Using y - y1 = m(x - x1) we get the equation of the tangent y - 1 = -(4/5)(x - 2) 5y - 5 = -4x + 8 4x + 5y - 13 = 0 THERE IS A SHORTCUT: For equations like this, containing only first and second powers of x and y, we can get the equation of the tangent at the point x, y by replacing x² in the equation by xx1, y² by yy1, x by (x + x1)/2, y by (y + y1)/2, and xy by (xy1 + x1y)/2. In this case, that gives 2x + 2(x +2y) + y = 13 leading to the same equation as found above. CONFIRMATION. If you solve the simultaneous equations x² + 4xy + y² = 13 and 4x + 5y - 13 = 0 by eliminating y, we get the solution (x - 2)² = 0, the double root confirming that the line is a tangent to the curve at x = 2

 AnonymousFrom: -Posts: -Votes: - Slope at (1,1) 2(x^2+y^2)(2x+2y dy/dt)=8xy+4x^2 dy/dt =>2(2)(2+2dy/dt)=8+4dy/dt =>dy/dt=0 so the slope of tangent -infinity equation of tangent=> (X-1=0)

 EliteMath1From: USPosts: 5Votes: 20 (x^2+y^2)*(2x+2ydy/dx)=8*x =4+4dy/dx=8 dy/dx=1 y-1=1*(x-1) y=x

 AnonymousFrom: -Posts: -Votes: - For equations like this, containing only first and second powers of x and y, we can get the equation of the tangent at the point x, y by replacing x² in the equation by xx1, y² by yy1, x by (x + x1)/2, y by (y + y1)/2, and xy by (xy1 + x1y)/2. In this case, that gives 2x + 2(x +2y) + y = 13 leading to the same equation as found above. CONFIRMATION. If you solve the simultaneous equations x² + 4xy + y² = 13 and 4x + 5y - 13 = 0 by eliminating y, we get the solution (x - 2)² = 0, the double root confirming that the line is a tangent to the curve at x = 2

 AnonymousFrom: -Posts: -Votes: - Slope at (1,1) 2(x^2+y^2)(2x+2y dy/dt)=8xy+4x^2 dy/dt =>2(2)(2+2dy/dt)=8+4dy/dt =>dy/dt=0 so the slope of tangent -infinity equation of tangent=> (X-1=0)

 AnonymousFrom: -Posts: -Votes: - (x^2+y^2)*(2x+2ydy/dx)=8*x =4+4dy/dx=8 dy/dx=1 y-1=1*(x-1) y=x

 AnonymousFrom: -Posts: -Votes: - Answer: y =1 we need two independent information in order to determine the tangent line, in this problem, ==>we have (1)a point that the tangent line passes: (1,1) ->(x0, y0) ==>we can find (2) the slope of the tangent line by computing y' at (1,1)->m then, we just substitute these in y = y0 + m (x - x0) = 1 + m(x -1) = 1-m + mx ---------------- let us find (2): use implicit differentiation to find y': 2(x2 +y2)(2x + 2yy') = 4(2xy + x2y') =>(x2 +y2)(x + yy') = (2xy + x2y') =>(x2 +y2)(x) - 2xy = x2y' - (x2 +y2)(yy') =>(x2 +y2)(x) - 2xy = {x2 - (x2 +y2)y} y' =>y' = {(x2 +y2)(x) - 2xy} / {x2 - (x2 +y2)y} to find the slope at (1,1), we substitute: y' = {(12 +12)(1) - 2(1)(1)} / {12 - (12 +12)(1)} y' = {2 - 2} / {1 - 2} = 0 -------------------- y = 1-m + mx = 1 y = 1