Math, asked by shaktisingh7060, 4 months ago


17. Intense atmosphere between the countries, it is important to remain vigilant on our borders. And our solliers are
ever ready for that. A soldier is sitting at the place representing the point (3.2) and notices on enemy jet. He notices
the path and calculates it to be a parabola symmetrical to the yaxis and vertex of parabola is at point (0, 2) with
path represented by y = x + a. For the above situation answer the following:
(1) the value of a in path is
(a) 3
(c) 2
(ir) The distance of jet from any point on path, at a given time, from soldier is
(a) (-3) + (y + 2) (b) Vix - 3)2 + (x + 2) (c) 768-3)*+-2) (d) V-2) + (y - 3)
(iii) The expression for distance in terms of only is
(a) +x+1
(h) Vir-3)* +6+4)
(d) - 2)² + (x - 1)
(w) The value of such that the distance between jet and soldier is least is
1
1
6
() Minimum distance is
(@) V5 units
(b) 5 units
(©)
units
(d)
units
b

Answers

Answered by julieanah14567
5

Answer:

i) (c) 2

ii) (c) root (x-3)^2 - (x-2)^2

iii) (c) root (x-3)^2 + x^4

iv) (a) x=1

v) (a) root 5

Answered by amitnrw
6

Given : A soldier is sitting at the place representing the point (3.2) and notices on enemy jet to be a parabola symmetrical to the y axis and vertex of parabola is at point (0, 2) with path represented by y = x² + a

To find :   value of a in path

The distance of jet from any point on path, at a given time, from soldier

The expression for distance in terms of only x

The value of x such that the distance between jet and soldier is least

Minimum distance

Solution:

y = x² + a

Vertex is  ( 0 , 2)

=> 2 =  0² + a

=> a = 2

Value of a is 2

hence y = x² + 2

The distance of jet from any point on path, at a given time, from soldier ( 3 , 2) is

√{( y - 2)² + ( x - 3)²}

= √{( x² + 2 - 2)² + ( x - 3)²}

= √{( x² )² + ( x - 3)²}

= √{x⁴ + ( x - 3)²}

distance between jet and soldier is least

=> √{x⁴ + ( x - 3)²}  is least

=> x⁴ + ( x - 3)² is least

f(x) = x⁴ + ( x - 3)²

f'(x) = 4x³  + 2(x - 3)  

4x³  + 2x  - 6     = 0

2x³ + x -  3 = 0

=> (x - 1)(2x²  + 2x  + 3) = 0

2x²  + 2x  + 3 gives imaginary roots

Hence x = 1

f''(x) = 12x² + 2 > 0  Hence distance is least

Least distance =  √{x⁴ + ( x - 3)²}

= √{1⁴ + ( 1 - 3)²}

= √5

value of a in path  = 2

The distance of jet from any point on path, at a given time, from soldier = √{( y - 2)² + ( x - 3)²}

The expression for distance in terms of only x   is √{x⁴ + ( x - 3)²}

The value of x such that the distance between jet and soldier is least x = 1

Minimum distance √5

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