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

Explanation

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)`.

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