Question

find the distance between two points(a sin alpha,-b cos alpha),(-a cos alpha, b sin alpha)..........

Answer 1

Answer:

$d=\sqrt{(a^2+b^2)(1+2sin2\alpha)}$

Step-by-step explanation:

The two points are

$(a\sin\alpha,-b\cos\alpha), (-a\cos\alpha,b\sin\alpha)$

The distance formula is given by

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

Substituting the known values

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

Now, apply the formula $(a+b)^2=a^2+b^2+2ab$

$d=\sqrt{a^2\cos^2\alpha+a^2\sin^2\alpha+2a^2\sin\alpha</p><p>\cos\alpha+b^2\sin^2\alpha+b^2cos^2\alpha)^2+2b^2\sin\alpha\cos\alpha}\\\\d=\sqrt{a^2(\sin^2\alpha+\cos^2\alpha)+2(a^2+b^2)(\sin\alpha\cos\alpha)+b^2(\sin^2\alpha+\cos^2\alpha)}$

Now using the formula, $\sin^2\alpha+\cos^2\alpha=1$

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

Factor out $a^2+b^2$

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

Answer 2

Step-by-step explanation:

i hope this will help you guys