How do you name the curve given by the conic r=41+cos⁡θ?

Convert to the General Cartesian Form: Ax2+Bxy+Cy2+Dx+Ey+F=0 Compute the determinant: Δ=B2−4AC If Δ&lt;0, then it is an ellipse or a circle. If B=0and A=C, then it is a circle. Otherwise, it is an ellipse. If Δ=0, then it is a parabola. If Δ&gt;0, them it is a hyperbola. Given: r=41+cos⁡(θ) r+rcos⁡(θ)=4 Substitute r=x2+y2and rcos⁡(θ)=x: x2+y2+x=4 x2+y2=4−x x2+y2=x2−8x+16 y2+8x−16=0 Please observe that, for the the above equation, the coefficients of the General Cartesian Form are, A=B=E=0,C=1,D=8,and F=−16 Δ=02−4(0)(1)=0 It is a parabola.

"How do you name the curve given by..." solved and explained

dohtarjev510

Answered question

2022-02-10

How do you name the curve given by the conic

r = \frac{4}{1 + \cos θ}

Answer & Explanation

s3au1d8wh

Beginner2022-02-11Added 11 answers

Convert to the General Cartesian Form:
$A x^{2} + B x y + C y^{2} + D x + E y + F = 0$
Compute the determinant:
$Δ = B^{2} - 4 A C$
If $Δ < 0$ , then it is an ellipse or a circle. If $B = 0 and A = C$ , then it is a circle. Otherwise, it is an ellipse.
If $Δ = 0$ , then it is a parabola.
If $Δ > 0$ , them it is a hyperbola.
Given: $r = \frac{4}{1 + \cos (θ)}$
$r + r \cos (θ) = 4$
Substitute $r = \sqrt{x^{2} + y^{2}} and r \cos (θ) = x$ :
$\sqrt{x^{2} + y^{2}} + x = 4$
$\sqrt{x^{2} + y^{2}} = 4 - x$
$x^{2} + y^{2} = x^{2} - 8 x + 16$

$y^{2} + 8 x - 16 = 0$
Please observe that, for the the above equation, the coefficients of the General Cartesian Form are, $A = B = E = 0, C = 1, D = 8, and F = - 16$
$Δ = 0^{2} - 4 (0) (1) = 0$
It is a parabola.

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