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

Abscissa means the horizontal co-ordiante of a point.

Given that abscissa `=2`.

i.e., `x=2`

`x^{2}+2 y^{2}-4 x-6 y+8=0 ldots(1)`

Differentiating both sides w.r.t.x ,


`2 x+4 y frac{d y}{d x}-4-6 frac{d y}{d x}=0 `

`Rightarrow frac{d y}{d x}(4 y-6)=4-2 x `

...