First, we rewrite the conic in standard form by multiplying the numerator and denominator by the reciprocal of 2, which is
$\text{\hspace{0.17em}}\frac{1}{2}.$
Because
$\text{\hspace{0.17em}}e=\frac{3}{2},e>1,$ so we will graph a
hyperbola with a focus at the origin. The function has a
$\text{\hspace{0.17em}}\mathrm{sin}\text{}\theta \text{\hspace{0.17em}}$ term and there is a subtraction sign in the denominator, so the directrix is
$\text{\hspace{0.17em}}y=-p.$
First, we rewrite the conic in standard form by multiplying the numerator and denominator by the reciprocal of 5, which is
$\text{\hspace{0.17em}}\frac{1}{5}.$
Because
$\text{\hspace{0.17em}}e=\frac{4}{5},e<1,$ so we will graph an
ellipse with a
focus at the origin. The function has a
$\text{\hspace{0.17em}}\text{cos}\text{\hspace{0.17em}}\theta ,$ and there is a subtraction sign in the denominator, so the
directrix is
$\text{\hspace{0.17em}}x=-p.$
Deﬁning conics in terms of a focus and a directrix
So far we have been using polar equations of conics to describe and graph the curve. Now we will work in reverse; we will use information about the origin, eccentricity, and directrix to determine the polar equation.
Given the focus, eccentricity, and directrix of a conic, determine the polar equation.
Determine whether the directrix is horizontal or vertical. If the directrix is given in terms of
$\text{\hspace{0.17em}}y,$ we use the general polar form in terms of sine. If the directrix is given in terms of
$\text{\hspace{0.17em}}x,$ we use the general polar form in terms of cosine.
Determine the sign in the denominator. If
$\text{\hspace{0.17em}}p<0,$ use subtraction. If
$\text{\hspace{0.17em}}p>0,$ use addition.
Write the coefficient of the trigonometric function as the given eccentricity.
Write the absolute value of
$\text{\hspace{0.17em}}p\text{\hspace{0.17em}}$ in the numerator, and simplify the equation.
Finding the polar form of a vertical conic given a focus at the origin and the eccentricity and directrix
Find the polar form of the
conic given a
focus at the origin,
$\text{\hspace{0.17em}}e=3\text{\hspace{0.17em}}$ and
directrix$\text{\hspace{0.17em}}y=-2.$
The directrix is
$\text{\hspace{0.17em}}y=-p,$ so we know the trigonometric function in the denominator is sine.
Because
$\text{\hspace{0.17em}}y=\mathrm{-2},\mathrm{\u20132}<0,$ so we know there is a subtraction sign in the denominator. We use the standard form of
Finding the polar form of a horizontal conic given a focus at the origin and the eccentricity and directrix
Find the
polar form of a conic given a
focus at the origin,
$\text{\hspace{0.17em}}e=\frac{3}{5},$ and
directrix$\text{\hspace{0.17em}}x=4.$
Because the directrix is
$\text{\hspace{0.17em}}x=p,$ we know the function in the denominator is cosine. Because
$\text{\hspace{0.17em}}x=4,4>0,$ so we know there is an addition sign in the denominator. We use the standard form of
I think the formula for calculating algebraic is the statement of the equality of two expression stimulate by a set of addition, multiplication, soustraction, division, raising to a power and extraction of Root. U believe by having those in the equation you will be in measure to calculate it
the Cayley hamilton Theorem state if A is a square matrix and if f(x) is its characterics polynomial then f(x)=0 in another ways evey square matrix is a root of its chatacteristics polynomial.