NIMCET 2014 — Mathematics PYQ

Concept:

• The slope of the normal at $(x_1,y_1)$ is given by $m=-{\left(\frac{dx}{dy}\right)}_{(x_1,y_1)}$.

• Equation of a line passing through $(x_1,y_1)$ and $(x_2,y_2)$ is $\frac{y-y_1}{x-x_1}=\frac{y_2-y_1}{x_2-x_1}$.

• Perpendicular distance $D$ of a point $(p,q)$ from line $ax+by+c=0$ is $D=\left|\frac{ap+bq+c}{\sqrt{a^2+b^2}}\right|$.

Calculation:

Given curve: $x^2=4y$

Differentiating with respect to $x$:

Let the point where the normal cuts the curve be $(x_1,y_1)$.

Slope of normal $m=-\frac{dx}{dy}=-\frac{2}{x_1}$.

The normal passes through $(x_1,y_1)$ and $(1,2)$, so the slope is also:

Since $(x_1,y_1)$ lies on the curve, $y_1=\frac{x_1^2}{4}$.

The normal passes through $(2,1)$ and $(1,2)$. Its equation is:

Distance from origin $(0,0)$ to the normal $x+y-3=0$: