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UPSC NDA 2 2016 Mathematics Solved Paper

Question Numbers: 86-87

Consider the parabola

y = x^{2} + 7x + 2 and the straight line y

= 3x – 3.

Question : 87 of 120

Marks: +1, -0

What is the shortest distance from the above point on the parabola to the line?

Solution:

Concept:

The shortest distance from a point

(x_1, y_1)

to a line

ax + by + c = 0

is given by

d = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}}

Explanation:

From the previous solution, the point on the parabola closest to the given line is

(-2, -8)

The equation of the line is

3x - y - 3 = 0

Substitute

x_1 = -2

and

y_1 = -8

into the distance formula:

d = \frac{|3(-2) - (-8) - 3|}{\sqrt{3^2 + (-1)^2}} = \frac{|-6 + 8 - 3|}{\sqrt{9 + 1}} = \frac{|-1|}{\sqrt{10}} = \frac{1}{\sqrt{10}}

Answer:

\frac{1}{\sqrt{10}}

(Option C).

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