Question

These days Chinese and Indian troops are engaged in aggressive melee, face-offs skirmishes at locations near the disputed Pangong Lake in Ladakh. One day a helicopter of enemy is flying along the curve represented by y = x.x + 7. A soldier placed at (3,7) wants to shoot down the helicopter when it is nearest to him. Based on the above information answer the following questions: (1). If (x1, y1) represents the position of helicopter on the curve y = X.X + 7, when the distance D from soldier placed at S(3,7) is minimum what is relationship between x1 and y1? (ii). The soldier at S wants to know when the enemy helicopter is nearest to soldier, find value Y1? (iii). Find the distance 'D' expressed as a function of x1 (iv), When the enemy helicopter is nearest to Soldier, then find the value of D. (v). Find the nearest position of the helicopter from the Soldier.​

Answer 1

Answer:

The nearest option is 5 as we can see.

Answer 2

The answers are y1 = x1² + 7 , 8 , $D^2 = x_1^2 + x_1^4 - 6x_1 +9$ , √5 and point(1,8) respectively.

Given,

Equation of the curve: y = x² + 7

Coordinates of the point = (3,7)

To Find,

(i). If (x1, y1) represents the position of a helicopter on the curve when the distance D from the soldier placed at S(3,7) is minimum what is the relationship between x1 and y1 =?

(ii). The soldier at S wants to know when the enemy helicopter is nearest to the soldier, find value Y1 =?

(iii). Find the distance 'D' expressed as a function of x1 =?

(iv). When the enemy helicopter is nearest to the soldier, then find the value of D=?

(v). Find the nearest position of the helicopter from the Soldier =?

Solution,

(i) Is (x1,y1) lies on the equation of the curve then the relationship will be:

y1 = x1² + 7

(ii) we have $D^2 = x_1^2 + x_1^4 - 6x_1 +9$

$(\frac{d(D^2)}{dx_1} ) = 2x_1 + 4x_1^3 +6$

For the minimum value of D,

$(\frac{d(D^2)}{dx_1} ) = 2x_1 + 4x_1^3 - 6 = 0\\4x_1^2(x_1 - 1)+ 4x_1(x_1-1) + 6(x_1 - 1) = 0\\(x_1 - 1) (4x_1^2 + 4x_1 + 6) = 0\\x_1 -1= 0\\x_1 = 1$

Therefore, y1 = 1² + 7

y1 = 8

(iii) Distance between (x1,y1) and (3,7)

$D = \sqrt{(x1 - 3)^2 + (y1 - 7)^2}$

We know,  y1 = x1² + 7

$D^2 = (x_1 - 3)^2 + (x_1^2 + 7 - 7)^2\\D^2 = x_1^2 + x_1^4 -6x_1 +9$

(iv) for the the minimum value of D, x1 = 1

$D^2 = x_1^2 + x_1^4 -6x_1 + 9\\D^2 = 1^2 + 1^4 -6*1 + 9\\D^2 = 1 + 1 - 6 +9\\D^2 = 11 - 6\\D^2 = 5\\D = \sqrt{5}$

(v) For the minimum value of D the nearest points are (1,8)

Hence,  y1 = x1² + 7 , 8 , $D^2 = x_1^2 + x_1^4 - 6x_1 +9$ , √5  and point(1,8) are the solutions of part (i), (ii), (iii),(iv) and (v) respectively.

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