Question

need answer for this question​

Answer 1

Answer:

B

Step-by-step explanation:

The given points are, (a sin(α), -b cos(α)) and ( -a cos(α), b sin(α))

From the distance formula, we have,

d=√{(x2-x1)²+ (y2-y1)²}

Now,

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

Taking in common (sin(α)+cos(α))²

we get,

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

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

we know that, sin(α)+cos(α)=sin(α+π/4)

and, √(x²)=|x|,

so,

$= \sqrt{2( {a}^{2} + {b}^{2} )} . | \sin( \alpha + \frac{\pi}{4} ) |$

$= \sqrt{2}. \sqrt{ {a}^{2} + {b}^{2} } . | \sin( \alpha + \frac{\pi}{4} ) |$

Hope you get this...!

Answer 2

Answer:

I love it and then delete this communication