Question

Find the distance between the points (-a cos theta, b sin theta) and ( a sin theta, -b coz theta)

Answer 1

Answer:

The answer of the question is ==√a²+b²+2sin∅cos∅(a²+b²)

Step-by-step explanation:

Distance between two points (x1,y1) and (x2,y2) applying distance formula

distance d=√(x2-x1)^2 +(y2-y1)^2

In the above question X1=-acos∅

x2=asin∅;  y1=bsin∅; y2=-bcos∅

substituting the values in the above equation of distance formula.

d=√(asin∅+acos∅)^2+(bsin∅+bcos∅)^2

=√a²sin∅²+a²cos∅²+b²sin∅²+b²cos∅²+2a²sin∅cos∅+2b²sin∅cos∅

=√a²+b²+2sin∅cos∅(a²+b²)

Thus the solution of the above question can be found

Concepts involved: Distance formula between two points

                          Basic trignometric formulas to simplify the equations

Answer 2

Answer:

The distance between given two points is

$\sqrt{(a^2+b^2)(1+sin2\theta)}$

Step-by-step explanation:

Distance formula:

The distance between two points $(x_1,y_1)\:and\:(x_2,y_2)$ is

$d= \sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$

sin2A=2 sinA cosA

Given points are $(-acos\theta,bsin\theta)\:and\:(asin\theta, -bcos\theta)$

Now,

$d= \sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\\\\d= \sqrt{(-acos\theta-asin\theta)^2+(bsin\theta+bcos\theta)^2}\\\\d= \sqrt{(acos\theta+asin\theta)^2+(bsin\theta+bcos\theta)^2}$

$d= \sqrt{[a^2cos^2\theta+a^2sin^2\theta+2.a^2.sin\theta.cos\theta]+[b^2sin^2\theta+b^2cos^2\theta+2.b^2.sin\theta.cos\theta]}$

$d= \sqrt{a^2(cos^2\theta+sin^2\theta+2.sin\theta.cos\theta)+b^2(sin^2\theta+cos^2\theta+2.sin\theta.cos\theta)}$

$d= \sqrt{a^2(1+2.sin\theta.cos\theta)+b^2(1+2.sin\theta.cos\theta)}$

$d= \sqrt{(a^2+b^2)(1+2.sin\theta.cos\theta)}\\\\d= \sqrt{(a^2+b^2)(1+sin2\theta)}$