Polar coordinate equation 2 ρ cos ^ 2 θ / 2 = 5 is a circle B ellipse C hyperbola D parabola

Polar coordinate equation 2 ρ cos ^ 2 θ / 2 = 5 is a circle B ellipse C hyperbola D parabola

2ρcos^2 θ/2=ρ(cos θ+1)=ρcos θ+ρ=x+ρ,
So x + ρ = 5, ρ = 5-x, square: x ^ 2 + y ^ 2 = (5-x) ^ 2, simplify the parabola y ^ 2 = 25-10x

How to prove the arc length formula of polar coordinate equation?
One of the methods I already know is to turn it into a parametric equation of x = R (θ) cos (θ), y = R (θ) sin (θ)
Is there a direct way?
What is the root cause of the error of DL = R (θ) d θ? Is it related to the radius of curvature? Is it correct to replace R (θ) with radius of curvature? (although it is very complicated)
If you want to tell me which book to look for, I only have Tongji's advanced mathematics and a balang's AP calculus
In addition, I think it's not beautiful to show Greek letters with such small characters, so please don't use letters such as φ except for θ, PI, etc

The fundamental reason for the error of DL = R (θ) d θ is that dl-r (θ) d θ does not get the higher-order infinitesimal of D θ, but the infinitesimal of the same order. By comparing polar coordinates with rectangular coordinates as shown in the figure, we can see that the reason why the curve integral in rectangular coordinates can not directly integrate DX is that there is a lot of difference between DX and DL. Similarly, there is a lot of difference between DL and R (θ) d θ

On hyperbola
It is known that in the triangle ABC, B (- 5,0) C (5,0), the motion of point a satisfies the trajectory equation of SINB sinc = 3 / 5sina

According to the sine theorem, sinc = 2R * | ab |, SINB = 2R * | AC |,
sinA=2R*|BC| ,
So the original equation is: 2R * | AC | - 2R | BC | = 3 / 5 * 2R | ab|
Remove 2R,
The results are as follows: | AC | - | BC | = 3 / 5 | ab | = 3 / 5 * 10 = 6
So the trajectory of point a is the right branch of hyperbola!

Mathematical problems of hyperbolic eccentricity in Senior Two
Let the hyperbola C: x ^ 2 / A ^ - y ^ = 1 (a > 0) intersect with the straight line L: x + y = 1 and two different points a and B to find the value range of the eccentricity of the hyperbola
Just give me an idea

Combine the above two equations: (1-A ^ 2) y ^ 2-2y + 1-A ^ 2 = 0
(1) 1-A ^ 2 is not equal to 0 (2) 4-4 (1-A ^ 2) ^ 2 > 0
Solve a ^ 2, and then according to: 1 + B ^ 2 / A ^ 2 = e under the root sign