Hello, i need help trying to solve these problems, dont understand how to arrive to answers... thak you very much in advance..

Solve-
1)4x^2+9y^2-16x+18y-11=0

2) y^2= -20x vertex foci

3) Sin^2x-sinx= 0

2. ## complete the quare

$\displaystyle 4x^2+9y^2-16x+18y-11=0$

this is an ellipse and you will need to complete the square to solve it

the Ellipse Equation is
$\displaystyle \frac{\left(x - h\right)^2}{a^2} + \frac{\left(y - k\right)^2}{b^2} = 1$

rewrite

$\displaystyle 4x^2-16x +9y^2+18y = 11$

factor out

$\displaystyle 4\left(x^2-4x \right)+ 9\left(y^2+2y \right) = 11$

now add in so squares can be construted

$\displaystyle 4\left(x^2-4x + 4 \right)+ 9\left(y^2+2y + 1 \right) = 11+16 +9 = 36$

factor the squares

$\displaystyle 4\left(x^2-2\right)^2+ 9\left(y^2+1\right)^2 = 36$

divide thru with 36 and simplify

$\displaystyle \frac{\left(x^2-2\right)^2}{3^2}+ \frac{\left(y^2+1\right)^2}{2^2} = 1$
foci:$\displaystyle \left(h \pm c,k\right)$
vertices:$\displaystyle \left(h \pm a,k\right)$
axis$\displaystyle \left(h,k \pm b\right)$

3. $\displaystyle y^2 = -20x$ vertex foci

Use 4p for this.

$\displaystyle \sin^2{x} - \sin{x}= 0$

Treat this like an algebra problem....

$\displaystyle \sin{x}(\sin{x} - 1) = 0$

Then think of values where

$\displaystyle \sin{x} = 0$ and $\displaystyle \sin{x} = 1$