Question

Given the conic section -3.6x²+1.6y²+72x+6.4y = 58/5

a) State the nature of the conic section

b) Find the characteristic elements of the conic section

c) Sketch the conic section indicating all the elements.

Answer 1

a)

$$-3.6x^2+1.6y^2+72x+6.4y = 58/5$$

$$-3.6(x^2-20x+100)+360+1.6(y^2+4y+4)-6.4=11.6$$

$$3.6(x-10)^2-1.6(y+2)^2=342$$

$$\dfrac{(x-10)^2}{95}-\dfrac{(y+2)^2}{213.75}=1$$

A conic is a hyperbola. Standard form

$$\dfrac{(x-10)^2}{95}-\dfrac{(y+2)^2}{213.75}=1$$

Horizontal hyperbola.

b)

$$h=10, k=-2, a=\sqrt{95}, b=\sqrt{213.75}=1.5\sqrt{95}$$

$$c^2=a^2+b^2=95+213.75=308.75, c=\sqrt{308.75}$$

Center: $(h, k)=(10,-2)$

Vertices: $(h\pm a,k), (10-\sqrt{95}, -2), (10+\sqrt{95}, -2)$

Covertices: $(h, k\pm b), (10, -2-1.5\sqrt{95}), (10, -2+1.5\sqrt{95})$

Foci: $(h\pm c,k), (10-\sqrt{308.75}, -2), (10+\sqrt{308.75}, -2)$

The equations of the asymptotes are 

$$y=\dfrac{b}{a}(x-h)+k,y=-\dfrac{b}{a}(x-h)+k$$ $$y=1.5(x-10)-2,y=-1.5(x-10)-2$$

$x=0,\dfrac{(0-10)^2}{95}-\dfrac{(y+2)^2}{213.75}=1$

$$(y+2)^2=11.24$$

$$y=-2\pm1.5\sqrt{5}$$

$$(0, -2-1.5\sqrt{5}), (0, -2+1.5\sqrt{5})$$

$y=0,\dfrac{(x-10)^2}{95}-\dfrac{(0+2)^2}{213.75}=1$

$$(x-10)^2=\dfrac{839}{9}$$

$$x=10\pm\dfrac{\sqrt{839}}{3}$$

$$(10-\dfrac{\sqrt{871}}{3},0), (10+\dfrac{\sqrt{871}}{3}, 0)$$

c)