Q.The normal to the curve $x^2 + 2xy - 3y^2 = 0$ at $(1, 1)$ (A) Meets the curve again in second quadrant. (B) Does not meet the curve again. (C) Meets the curve again in third quadrant. (D) Meets the curve again in fourth quadrant.

The derivative dy/dx evaluated at a point $(x_1,y_1)$ on a curve $y=f(x)$ gives the slope $m$ of the tangent line to the curve at that point — this is the geometric meaning of the derivative. The tangent is the straight line that just touches the curve at $(x_1,y_1)$ and has the same instantaneous direction as the curve there; its equation is $y-y_1=m(x-x_1)$. The normal is the line through the same point that is perpendicular to the tangent; since perpendicular lines have slopes that are negative reciprocals of each other, the normal's slope is $m'=-1/m$ (valid when $m\ne0$), and its equation is $y-y_1=m'(x-x_1)$. When the curve is given implicitly (an equation in $x$ and $y$ that cannot easily be solved for $y$) or parametrically (both $x$ and $y$ given in terms of a parameter such as $\theta$ or $t$), the same idea applies: differentiate implicitly using the chain and product rules, or use $dy/dx=(dy/d\theta)/(dx/d\theta)$ for parametric curves, evaluate the resulting slope at the given point, and substitute into the point-slope forms above. Special cases: a horizontal tangent (slope $0$) gives a vertical normal, and a vertical …