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.