Question

Show that for eccentricity equal to one in Equation 13.10 for conic sections, the path is a parabola. Do this by substituting Cartesian coordinates, x and y, for the polar coordinates, r and θ, and showing that it has the general form for a parabola, x=ay2+by+c.

(a) ( {Parabola equation is satisfied} )
(b) ( {Hyperbola equation is satisfied} )
(c) ( {Ellipse equation is satisfied} )
(d) ( {Circle equation is satisfied} )

Answer 1

When the eccentricity of a conic is 1, the resulting shape is a parabola. Converting the polar equation of a conic section to Cartesian coordinates confirms that for e=1, the equation simplifies to the general form of a parabola x = ay^2 + by + c, with specific coefficients.

Step-by-step explanation:

When the eccentricity (e) of a conic section is exactly one, the resulting path is a parabola. To show this in Cartesian coordinates, we start with the polar form of a conic section, r = (ed)/(1 + ecos(θ)), where d is the distance from the directrix to the focus and the origin is the focus. For a parabola, e is 1, so the equation simplifies to r = d/(1 + cos(θ)).

To convert this to Cartesian coordinates, we use the identities x = rcos(θ) and y = rsin(θ) and obtain x = (dcos(θ))/(1 + cos(θ)) and y = (dsin(θ))/(1 + cos(θ)). By multiplying the denominator on the x-coordinate to the numerator, we get x = d and y = dtan(θ)/2. Squaring both sides of the equation for y gives us y2 = (d2tan2(θ))/4 = 4x2/d2. Solving for x yields x = ay2, where a is a constant, showing that the path described by the equation is a parabola, which has the general parabolic form of x = ay2 + by + c, with b = 0 and c = 0.