Question

Will give brainliest!  pts to the correct answer! Are quadratic equations already in rectangular form? I'm asked to eliminate the parameter t from these two questions to get it into the rectangular equation with x and y. Is it possible?
x(t) = 0.5t^2+0.4t-87.30
y(t)= 0.4t^2+0.1t +29.44

Answer 1

$\bf x(t)=0.5t^2+0.4t-87.30\implies x(t)=\cfrac{1}{2}t^2+\cfrac{2}{5}t-\cfrac{873}{10} \\\\\\ x=\cfrac{1}{2}\left( t^2+\cfrac{4}{5}t+\boxed{?}^2 \right)-\cfrac{873}{10}$

so, as you already know, we'll borrow from our good friend Mr Zero, 0, to get that missing value for the perfect square trinomial,

$\bf x=\cfrac{1}{2}\left[ t^2+\cfrac{4}{5}t+\left( \cfrac{2}{5} \right)^2-\left( \cfrac{2}{5} \right)^2 \right]-\cfrac{873}{10} \\\\\\ x=\cfrac{1}{2}\left[t^2+\cfrac{4}{5}t+\cfrac{4}{25}-\cfrac{4}{25} \right]-\cfrac{873}{10} \\\\\\ x=\cfrac{1}{2}\left[t^2+\cfrac{4}{5}t+\left( \cfrac{2}{5} \right)^2\right]-\cfrac{2}{25}-\cfrac{873}{10} \\\\\\ x=\cfrac{1}{2}\left[t^2+\cfrac{4}{5}t+\left( \cfrac{2}{5} \right)^2\right]-\cfrac{4369}{50}$

$\bf x=\cfrac{1}{2}\left( t+\frac{2}{5} \right)^2-\cfrac{4369}{50}\implies x+\cfrac{4369}{50}=\cfrac{1}{2}\left( t+\frac{2}{5} \right)^2 \\\\\\ 2x+\cfrac{4369}{25}=\left( t+\frac{2}{5} \right)^2\implies \sqrt{2x+\cfrac{4369}{25}}=t+\cfrac{2}{5} \\\\\\ \sqrt{2x+\cfrac{4369}{25}}-\cfrac{2}{5}=t$

$\bf -------------------------------\\\\ y(t)=0.4t^2+0.1t+29.44\implies y=\cfrac{2}{5}t^2+\cfrac{1}{10}t+\cfrac{736}{25} \\\\\\ y=\cfrac{2}{5}\left( \sqrt{2x+\cfrac{4369}{25}}-\cfrac{2}{5} \right)^2+\cfrac{1}{10}\left( \sqrt{2x+\cfrac{4369}{25}}-\cfrac{2}{5} \right)+\cfrac{736}{25}$