Test Index

UPSC NDA 2 2016 Mathematics Solved Paper

Consider the following for the next two (02) items that follow:

Consider the parabola

y=x^{2}+7x+2

and the straight line

y=3x-3

[NDA II 2016]

Question : 86 of 120

Marks: +1, -0

Solution:

Parabola Equation:

y=x^{2}+7x+2

Line equation:

y=3x-3

Since all the points given in the options lie on the parabola

Thus we will calculate the distance from the given line to these points:

for

(0,2)

: distance

=\;\frac{\left|3(0)-2-3\right|}{\sqrt{(3)^{2}+(-1)^{2}}}=\;\frac{5}{\sqrt{10}}

for

(-2,-8)

: distance

=\;\frac{\left|3(-2)-(-8)-3\right|}{\sqrt{10}}=\;\frac{1}{\sqrt{10}}

for

(-7,2)

: distance

=\;\frac{\left|3(-7)-2-3\right|}{\sqrt{10}}=\;\frac{26}{\sqrt{10}}

for

(1,10)

: distance

=\;\frac{\left|3(1)-10-3\right|}{\sqrt{10}}=\;\frac{10}{\sqrt{10}}

\therefore(-2,-8)

is the given point.

Go to Question:

Prev Question

Next Question