Question

Due inu um 102019 IT.Dy pri Show Intro/Instructions 7. Given the conic section r = cay find the x and y intercept(s) and the focus (foci). Identify the type of conic section and then graph the conic section. Intercepts: 0.67 0.47 Foci: 0.21 -0.8-0.6-0.4-22 0.2 0.4 0.6 0.8 -0.4+ Be sure to graph the conic section in your work. Preview Intercepts as ordered pairs: Preview Foci as ordered pairs:

Answer 1

$r=\frac{3}{6+4sin\theta}$

$6r+4rsin\theta=3$

$6r=3-4rsin\theta$

$36r^2=(3-4rsin\theta)^2$

convert to cartesian form

$x=rcos\theta$

$y=rsin\theta$

$x^2+y^2=r^2(sin^2\theta+cos^2\theta)=r^2$

put these in polar equation

$36(x^2+y^2)=(3-4y)^2$

$36(x^2+y^2)=9+16y^2-24y$

$36x^2+20y^2+24y-9=0$

$36x^2+20(y^2+\frac{6}{5}y)-9=0$

$36x^2+20(y^2+\frac{6}{5}y+({\frac{3}{5}})^2-({\frac{3}{5}})^2)-9=0$

$36x^2+20((y+{\frac{3}{5}})^2-({\frac{9}{25}})-9=0$

$36x^2+20((y+{\frac{3}{5}})^2={\frac{9*26}{25}}$

$36x^2+20((y+{\frac{3}{5}})^2={\frac{234}{25}}$

$\frac{x^2}{\frac{234}{26*36}}+\frac{(y+{\frac{3}{5}})^2}{\frac{234}{500}}=1$

$\frac{x^2}{0.25}+\frac{(y+{\frac{3}{5}})^2}{0.468}=1$

this is ellipse with major axis parallel to y axis

intercepts

major axes vertices 0,0.3 and 0,-1.5

$\\c^2=0.468-0.25=0.218\\c=0.4669$

foci are $(0,-0.6\pm0.4669)$

$(0,-1.0669),(0,-0.1331)$

-0.4(0.0.3) 9.00 -1 -0.9 -0.8 -0.7 -0.6 0.1 0.2 0.3 0.4 0.6 0.7 0.8 0.9 0.5 -0.4 -0.3 -0.2 -0.1 0 --0:17 Q -0:3 -0.4 -0.5 --0:6 -0.7 -0:84 -1-1 i (0, -1.5)