the points P and Q on the graph of y^(2) -xy -5y +10 =0 have the same x coordina

Question

the points P and Q on the graph of y^(2) -xy -5y +10 =0 have the same x coordinate x=2. Find the point of intersection of the tangent lines to the graph at P and Q

I got this far...my equations are 5/3x + 5/3= y and y=-2/3x +10/3 and i found that my x is 5/7....

Explanation / Answer

y^(2) -xy -5y +10 =0 Differentiating we get 2ydy/dx - y - xdy/dx -5dy/dx = 0 =>dy/dx = y/(2y-x-5) And given the points P and Q on the graph of y^(2) -xy -5y +10 =0 have the same x coordinate x=2. =>y^2-2y-5y+10 = 0 =>y(y-2)-5(y-2) = 0 =>y = 2,5 So points are (2,2) and (2,5) First point tangent: y-2 = (-2/3)(x-2) =>3y-6 = -2x+4 =>2x+3y = 10.......(1) Second point tangent: y-5 = 5/3(x-2) =>3y-15 = 5x-10 =>5x-3y = -5......(2) Solve (1) and (2) to get x and y. Add (1) and (2) => 7x = 5 => x = 5/7 5x-3y = -5 =>3y = 25/7 + 5 = 60/7 =>y = 20/7 So point of intersection is (5/7,20/7)

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