Question

Draw a parabola ,when the distance of the focus from the directrix is 50 mm​

Answer 1

• Draw directrix DD. At any point C on it draw axis CA ⊥r to DD.
• Distance between the focus and the directrix is 40mm. So mark F the focus, such that CF=40mm.
• For parabola, e is 1; so construct the right angled ΔCXY such that XY/CX = 1 (X is any point on axis).
• From F draw a 45º line to intersect CY at s.
• From S erect vertical to cut CF at vertex V.Now SV = FV.
• From similar Δs Cthe points 1,2,…,5 at approximately equal interXY and CVS, SV/CV = XY/CX = FV/CV = 1.
• Along the axis CA mark the points 1,2,…,5 at approximately equal intervals.
• Through 1,2,..,5 erect verticals to intersect the line CY (produced if necessary) at 1′,2′,…,5′ respectively.
• With 11′ as radius and F as center, Draw two arcs on either side of the axis to intersect the vertical line drawn through 1 at P1 and P1′.
• Repeat the above and obtain P2 & P2′,P3 & P3′,….,P5&P5′ corresponding to points 2,3,4,5 respectively. Draw a smooth parabola through P5,…,P1,V,P1′,etc.
• Tangent and Normal: Join PF. At F, draw a perpendicular to QP. NM is the normal.

The directrix is the line y=-p. Any point (x,y) on the parabola will be the same distance from the focus as it is from the directrix. That is, if d1 is the distance from the focus to the point on the parabola, and $d_2$is the distance from the directrix to the point on the parabola, then $d_1=d_2.$