NIMCET 2014 — Mathematics PYQ

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Question

NIMCET 2014 — Mathematics PYQ

NIMCET | Mathematics | 2014

A normal to the curve $x^{2} = 4y$ passes through the point $(1, 2)$ . The distance of the origin from the normal is:

Choose the correct answer:

Explanation

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 $x$ :

2x = 4 \frac{dy}{dx} \Rightarrow \frac{dy}{dx} = \frac{x}{2}

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:

\frac{2 - y_{1}}{1 - x_{1}} = -\frac{2}{x_{1}}

Since $(x_{1}, y_{1})$ lies on the curve, $y_{1} = \frac{x_{1}^{2}}{4}$ .

\Rightarrow \left( 2 - \frac{x_{1}^{2}}{4} \right)x_{1} = -2(1 - x_{1})

\Rightarrow 2x_{1} - \frac{x_{1}^{3}}{4} = -2 + 2x_{1}

\Rightarrow \frac{x_{1}^{3}}{4} = 2 \Rightarrow x_{1}^{3} = 8 \Rightarrow x_{1} = 2

y_{1} = \frac{2^{2}}{4} = 1

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

\frac{y - 1}{x - 2} = \frac{2 - 1}{1 - 2} \Rightarrow \frac{y - 1}{x - 2} = -1

y - 1 = -x + 2 \Rightarrow x + y - 3 = 0

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

D = \left| \frac{1(0) + 1(0) - 3}{\sqrt{1^{2} + 1^{2}}} \right|

D = \left| \frac{-3}{\sqrt{2}} \right| = \frac{3}{\sqrt{2}}