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A particle motion along a ellipse \mathrm{x^{2}+4 y^{2}=0}
when particle is at point At (2,1) its \mathrm{x}- component of velocity is 6 , then the \mathrm{y} -component of velocity is.

Option: 1

4


Option: 2

3


Option: 3

-3


Option: 4

-2


Answers (1)

best_answer

Given, Ellipse equation is given by -
\mathrm{x^{2}+4 y^{2}=0}

Differenalition both sides -
\mathrm{2 \times \frac{d x}{d t}+4 \times 2 y \frac{d y}{d t}=0 }
\mathrm{x \frac{d x}{d t}+4 y \frac{d y}{d t}=0\quad-(1) }

Now,
point (2,1) and \mathrm{\frac{d x}{d t}=6}, \mathrm{\frac{d y}{d t}= ?}

from \mathrm{\varepsilon q^{n}(1)}.
\mathrm{2 \times 6+4 \times 1 \cdot \frac{d y}{d t} =0 }
\mathrm{4 \frac{d y}{d t} =-12 }
\mathrm{\frac{d y}{d t} =-3}

Posted by

Irshad Anwar

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