Find the equation of the normal to the curve `x^2+2y^2-4x-6y+8=0`at the point whose abscissa is 2.

Find the equation of the normal to the curve `x^2+2y^2-4x-6y+8=0`at th

1.	Find the equation of the normal to the curve `x^2+2y^2-4x-6y+8=0`at the point whose abscissa is 2.
Answer» Let `m` is the slope of tangent of the curve and `n` is the slope of normal to the curve. Then, `m*n = -1=> n = -1/m`. Now, equation of the curve, `x^2+2y^2-4x-6y+8 = 0` Differentiating it w.r.t. `x`, `=>2x+4ydy/dx - 4 -6dy/dx = 0` `=>dy/dx = (4-2x)/(4y-6)` `:. m = (4-2x)/(4y-6)`. Now, we have to find tyhe equation of the normal at a point whose abscissa is `2`. `:. x = 2` `=> m = (4-2(2))/(4y-6) = 0` `:.` Slope of tangent is `0`. `:. n = -1/0` Now, putting `x = 2`, in the equation of the curve. `=>4+2y^2-8-6y+8 = 0` `=>2y^2-6y+4 = 0` `=>2y^2-4y-2y+4 = 0` `=>(2y-2)(y-2) = 0` `=>y = 1 and y = 2` So, we will find equation of normal at points `(2,1)` and `(2,2)`. `:.` Equation of normal to the curve at `(2,1)`. `=>y-1 = -1/0(x-2) => x = 2` Equation of normal to the curve at `(2,2)`. `=>y-2 = -1/0(x-2) => x = 2` `:. x = 2` is the required equation.

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Find the equation of the normal to the curve `x^2+2y^2-4x-6y+8=0`at the point whose abscissa is 2.

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