Question

The conic F is defined as 9x²-36x+4y² = 0

a) State the nature of F

b)Determine its characteristic elements

c) Sketch F

Answer 1

a)

$$F: 9x^2-36x+4y^2 = 0$$

$$9x^2-36x+36+4y^2=36$$

$$9(x-2)^2+4y^2=36$$

$$\dfrac{(x-2)^2}{4}+\dfrac{y^2}{9}=1$$

A conic (F) is an ellipse. Standard form

$$\dfrac{y^2}{9}+\dfrac{(x-2)^2}{4}=1$$

Major axis is vertical.

b)

$$h=2, k=0, a=3, b=2$$

$$c^2=a^2-b^2=9-4=5, c=\sqrt{5}$$

Center: $(h, k)=(2,0)$

Vertices: $(h, k\pm a), (2, -3), (2, 3)$

Covertices: $(h\pm b, k), (0, 0), (4, 0)$

Foci: $(h, k\pm c), (2, -\sqrt{5}), (2, \sqrt{5})$

The equations of the directrices are $y=k±a^2/c$

$y=-\dfrac{9\sqrt{5}}{5},y=\dfrac{9\sqrt{5}}{5}$

$x=0, \dfrac{y^2}{9}+\dfrac{(0-2)^2}{4}=1=>y=0$

The graph passes through the origin.

$y=0, \dfrac{(0)^2}{9}+\dfrac{(x-2)^2}{4}=1, x_1=0, x_2=4$

c)