The normal to the curve, x^2 + 2xy - 3y^2 = 0, at (1,1)

Question

The normal to the curve,  ${x^2} + 2xy - 3{y^2} = 0$, at $(1,1)$

Accepted Answer

Given curve is

${x^2} + 2xy - 3{y^2} = 0$

Difference $w.r.t.x,$

$$2x + 2x{{dy} \over {dx}} + 2y - 6y{{dy} \over {dx}} = 0$$

$${\left( {{{dy} \over {dx}}} \right)_{\left( {1,1} \right)}} = 1$$

Equation of normal at $(1,1)$ is

$y=2-x$

Solving eq. $(1)$ and $(2),$ we get $x=1,3$

Point of intersection $$\left( {1,1} \right),\left( {3, - 1} \right)$$

Normal cuts the curve again in $4$th quadrant.

Solutions · 1