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

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We need to verify theorem 5 Focus-Directrix definition: 0 &amp;lt; e &amp;lt; 1 and c=d(e − 2 − 1)

liannemdh · Accepted Answer

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&#039;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(1−e2)a2(1−e2)(since b2)  (a2−c2−a2−a2e2)x2a2+y2b2  =1  Thus, the theorem 5 Focus-Directrix definition is verified.

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

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