To calculate: The equation

usagirl007A
2021-01-15
Answered

To calculate: The equation

You can still ask an expert for help

timbalemX

Answered 2021-01-16
Author has **108** answers

Step 1Formula: The general equation of the hyperbola is $\frac{{x}^{2}}{{a}^{2}}\text{}-\text{}\frac{{y}^{2}}{{b}^{2}}=1$ where coordinate of the focus $(\pm \text{}c,\text{}0)\text{}\text{and}\text{}{c}^{2}={a}^{2}\text{}+\text{}{b}^{2}$ Step 2Calculation:Consider the equation of the conic sections$100{y}^{2}\text{}+\text{}4x={x}^{2}\text{}+\text{}104$ That is$100{y}^{2}\text{}-\text{}{x}^{2}\text{}+\text{}4x=104$

$100{y}^{2}\text{}-\text{}({x}^{2}\text{}-\text{}4x)=104$

$100{y}^{2}\text{}-\text{}({x}^{2}\text{}-\text{}4x\text{}+\text{}4)=104\text{}-\text{}4$

$100{y}^{2}\text{}-\text{}{(x\text{}-\text{}2)}^{2}=100$ Or,$\frac{{y}^{2}}{1}\text{}-\text{}\frac{{(x\text{}-\text{}2)}^{2}}{{10}^{2}}=1$ Therfore the standard form of the conic section is$\frac{{y}^{2}}{1}\text{}-\text{}\frac{{(x\text{}-\text{}2)}^{2}}{{10}^{2}}=1$ And since, general equation of the hyperbola is $\frac{{x}^{2}}{{a}^{2}}\text{}-\text{}\frac{{y}^{2}}{{b}^{2}}=1$ , hence, it's a hyperbola.

asked 2021-08-09

To calculate: The simplified form of the expression , $\frac{8}{\sqrt{15}-\sqrt{11}}$

asked 2021-08-06

Evaluate each expression.

P(14,0)

P(14,0)

asked 2021-01-24

The polar equation of the conic with the given eccentricity and directrix and focus at origin:
$r=41\text{}+\text{}\mathrm{cos}\theta$

asked 2020-11-24

Decide if the equation defines an ellipse, a hyperbola, a parabola, or no conic section at all.

asked 2021-08-12

Graph the lines and conic sections $r=\frac{1}{(1+\mathrm{cos}\theta )}$

asked 2022-04-22

Formula for analytical finding ellipse and circle intersection points if exist

I need a formula that will give me all points of random ellipse and circle intersection (ok, not fully random, the center of circle is laying on ellipse curve)

I need step by step solution (algorithm how to find it) if this is possible.

I need a formula that will give me all points of random ellipse and circle intersection (ok, not fully random, the center of circle is laying on ellipse curve)

I need step by step solution (algorithm how to find it) if this is possible.

asked 2022-04-13

Is there a parametrization of a hyperbola ${x}^{2}-{y}^{2}=1$ by functions x(t) and y(t) both birational?

Consider the hyperbola${x}^{2}-{y}^{2}=1$ . I am aware of some parametrizations like:

1.$(x\left(t\right),y\left(t\right))=(\frac{{t}^{2}+1}{2t},\frac{{t}^{2}-1}{2t})$

2.$(x\left(t\right),y\left(t\right))=(\frac{{t}^{2}+1}{{t}^{2}-1},\frac{2t}{{t}^{2}-1})$

3.$(x\left(t\right),y\left(t\right))=(\text{cosh}t,\text{sinh}t)$

4.$(x\left(t\right),y\left(t\right))=(\mathrm{sec}\left(t\right),\mathrm{tan}\left(t\right))$

The first and the second are by rational functions x(t) and y(t). But the functions are not birational(i.e. with rational inverse of each).

Is there a parametrization where:

- x(t) is rational with inverse also rational, and

- y(t) is rational with inverse also rational?

Is possible, to find a parametrization where both are rational and at least one of the has inverse rational?

Consider the hyperbola

1.

2.

3.

4.

The first and the second are by rational functions x(t) and y(t). But the functions are not birational(i.e. with rational inverse of each).

Is there a parametrization where:

- x(t) is rational with inverse also rational, and

- y(t) is rational with inverse also rational?

Is possible, to find a parametrization where both are rational and at least one of the has inverse rational?