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  • A particle motion along the ellipse . when particle is at point (3,8) its x-component of veloc

A particle motion along the ellipse \mathrm{x^2+8 y^2=4}. when particle is at point (3,8) its x-component of velocity is 12, what will be the y-component of velocity.

Option: 1

\mathrm{\frac{-9}{128}}


Option: 2

\mathrm{\frac{-9}{25}}


Option: 3

\frac{-9}{5}


Option: 4

\frac{-9}{16}


Answers (1)

best_answer

Ellipse equation is , \mathrm{x^2+8 y^2=4} --------(1)

Differentation with both sides -

\mathrm{2 x \frac{d x}{d t}+16 y \frac{d y}{d t}=0} -----------(2)

 Divide both sides factor of 2 -

\mathrm{x \frac{d x}{d t}+8 y \frac{d y}{d t}=0}

Now, At point (3,8) , \mathrm{\frac{d x}{d t}=12} we get \mathrm{\frac{d y}{d t}=\text { ? }}

\mathrm{\begin{aligned} 3 \times 12+8 \times 8 \frac{d y}{d t} & =0 \\ 64 \frac{d y}{d t} & =-36 \\ \frac{d y}{d t} & =-\frac{36}{64}=-\frac{9}{16} \end{aligned}}

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