$\begin{array}{1 1} (A) d_x = 134, \; d_y = 84\\ (B) d_x = 142.6, \; d_y = 92.6 \\(C) d_x = 182.6, \; d_y = 132.6 \\ (D) d_x = 132.4, \; d_y = 82.4 \end{array}$

Want to ask us a question? Click here

Browse Questions

Ad |

0 votes

Hint: If $d_x$ and $d_y$ are the distances from the sub to X and Y and $d_x - d_y = 50$, then the ship must be on a hyperbola with foci at X and Y.

$\begin{array}{1 1} (A) d_x = 134, \; d_y = 84\\ (B) d_x = 142.6, \; d_y = 92.6 \\(C) d_x = 182.6, \; d_y = 132.6 \\ (D) d_x = 132.4, \; d_y = 82.4 \end{array}$

0 votes

Given: Hint: If $d_x$ and $d_y$ are the distances from the sub to X and Y and $d_x - d_y = 50$, then the ship must be on a hyperbola with foci at X and Y, as shown below:

The equation for the hyperbola is $\large\frac{x^2}{a^2} $$ - \large\frac{y^2}{b^2}$$ = 1$

Here $a = \large\frac{1}{2}$$50 = 25$, the distance from the center to each vertex and $c = 100$, distance from center to foci X or Y. .

$\Rightarrow b = \sqrt (c^2 - a^2) = \sqrt (100^2 - 25^2) = \sqrt 9375$

$\Rightarrow $ The equation for the hyperbola is $\large\frac{x^2}{625} $$ - \large\frac{y^2}{9375}$$ = 1$

Substitute, $y = 60, \rightarrow \large\frac{x^2}{625} $$ - \large\frac{3600}{9375}$$ = 1$

$\Rightarrow x^2 = \large\frac{9375+3600}{9375}$$\times 625 = 865$

$\Rightarrow x = \sqrt 865 \approx 29.4$

Therefore, we can now calculate $d_x$ and $d_y$ as follows:

$d_x = \sqrt (29.41+100)^2 \approx 142.6$ Km

$d_y = \sqrt (29.41 -100)^2 \approx 92.6$ Km

Ask Question

Take Test

x

JEE MAIN, CBSE, NEET Mobile and Tablet App

The ultimate mobile app to help you crack your examinations

...