We need to verify theorem 5 Focus-Directrix definition:0 < e < 1 text{and} c=d(e - 2 - 1)

EunoR

EunoR

Answered question

2020-11-27

We need to verify theorem 5 Focus-Directrix definition: 0 < e < 1 and c=d(e  2  1)

Answer & Explanation

liannemdh

liannemdh

Skilled2020-11-28Added 106 answers

Step 1 A conic section is the set of all points in a plane whose distances from a particular point (focus) and a particular line (directrix) in the plane have a constant ratio. If P is a point of a cinic section, F is the conic's focus and D is the point of the directrix closest to P, then e=PFPD and PF=e × PD, where e is a constant and the eccentricity of the conic. Step 2 Formula: Assume P(x, y) is a point on the ellipse and PF=(x  c)2 + y2 so the equation PF=ePD will be written as. Step 3 Calculation: If is given that, c=d(e  2  1) and we know that a=ce. Assume the coordinates of the focus F=(c, 0) with vertical directrix x=ae, then x=d + c=c(e  2  1) + c=ce  2=(ae)e  2=ae
(x  c)2 + y2=e(x  ae)
(x  c)2 + y2

=e2(x  ae)2x2  2cx + c2 + y2

=e2x2  2aex + a2(1  e2)x2 + y2

=2cx  2aex + a2  c2   (since c=ae)
(1  e2)x2 + y2

=2(ae)x  2aex + a2  (ae)2(1  e2)x2 + y2

=a2(1  e2)
(1  e2)x2a2(1  e2) + y2a2(1  e2)

=a2(1e2)a2(1e2)(since b2)

(a2c2a2a2e2)x2a2+y2b2

=1

Thus, the theorem 5 Focus-Directrix definition is verified.

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