Question

a point moves so that the sum of uts distance from ie.0 and -ae, 0 is 2q prove that locus of the point is​

Answer 1

Answer:

Mass of neutron = 1.0087 a.m.u. Mass of proton = 1.0073 a.m.u. Mass of helium nucleus = 4.0015 a.m.u. As we know that, in a helium nucleus there are 2 neutrons and 2 protons

Answer 2

We have to find Area Enclosed by ellipse

X'+ Since Ellipse is symmetrical about both x-axis and y-axis.

$\sf{Area \: of \: ellipse = 4 \times Area \: of \: OAB}$

$\sf{Area \: of \: ellipse = 4 \times \int^{a}_0y \: dx }$

We know that

$\: \: \: \: \: \implies\sf \dfrac{ {x}^{2} }{ {a}^{2} } + \dfrac{ {y}^{2} }{ {b}^{2} } = 1$

$\: \: \: \: \: \implies\sf \dfrac{ {y}^{2} }{ {b}^{2} } = 1 - \dfrac{ {x}^{2} }{ {a}^{2} }$

${ \: \: \: \: \: \implies\sf \dfrac{ {y}^{2} }{ {b}^{2} } = \dfrac{ {a}^{2} - {x}^{2} }{ {a}^{2} } }$

${ \: \: \: \: \: \implies\sf {y}^{2} = \dfrac{ {b}^{2}( {a}^{2} - {x}^{2} )}{ {a}^{2} } }$

${ \: \: \: \: \: \implies\sf y = \pm\sqrt{ \dfrac{ {b}^{2}( {a}^{2} - {x}^{2} )}{ {a}^{2} } }}$

${ \: \: \: \: \: \implies\sf y = \pm \dfrac{b}{a} \sqrt{ ( {a}^{2} - {x}^{2} ) }}$

Since OAB is in 1st

Quadrant, value of y is positive

${ \: \: \: \: \: \implies\sf y = \dfrac{b}{a} \sqrt{ {a}^{2} - {x}^{2} }}$

$\sf{Area \: of \: ellipse = 4 \times \int^{a}_0y \: dx }$

• Putting the value of y

$\sf{Area \: of \: ellipse = 4 \times \int^{a}_0 \dfrac{b}{a} \sqrt{ {a}^{2} - {x}^{2} }\: dx }$

$\sf{Area \: of \: ellipse = \dfrac{4b}{a} \times \int^{a}_0 \sqrt{ {a}^{2} - {x}^{2} }\: dx }$

$\bf it \: is \:of \: form \: \: \: \: \: \\ \boxed{ \sf \sqrt{ {a}^{2} - {x}^{2} } \: dx = \frac{1}{2} x \sqrt{ {a}^{2} - {x}^{2} } + \dfrac{ {a}^{2} }{2} {\sin }^{ - 1} \dfrac{x}{a} + c }$

$\sf{Area \: of \: ellipse = \dfrac{4b}{a} \bigg [ \dfrac{x}{2} \sqrt{ {a}^{2} - {x}^{2} } + \dfrac{ {a}^{2} }{2} {\sin }^{ - 1} \frac{x}{a} \bigg ]^{a} _0 }$

$\sf{Area \: of \: ellipse = \dfrac{4b}{a} \bigg [ \bigg(\dfrac{a}{2} \sqrt{ {a}^{2} - {a}^{2} } + \dfrac{ {a}^{2} }{2} {\sin }^{ - 1} \dfrac{a}{a} \bigg) - \bigg(\dfrac{0}{2} \sqrt{ {a}^{2} - 0 } + \dfrac{ {a}^{2} }{2} {\sin }^{ - 1} (0) \bigg) \bigg ]^{a} _0 }$

$\sf{Area \: of \: ellipse = \dfrac{4b}{a} \bigg [ 0 + \dfrac{ {a}^{2} }{2} \sin ^{ - 1} (1) - 0 - 0 \bigg ] }$

$\sf{Area \: of \: ellipse = \dfrac{4b}{a} \times \dfrac{ {a}^{2} }{2} \sin ^{ - 1}(1) }$

$\sf{Area \: of \: ellipse = 2ab \times \sin^{ - 1} (1) }$

$\sf{Area \: of \: ellipse = 2ab \times \dfrac{\pi}{2} }$

$~~\underbrace{\boxed{\bf{Area \: of \: ellipse = \pi ab}}}$

Hence, the Required Area = πab units².