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erive the equation of parabola y^2 = 4ax in standard for |
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Answer» Let S be the focus and the linelbe the directrix. Draw SZ perpendicular to the linel. Take O as the midpoint of SZ. Now take O as the origin. Take OS produced as X-axis and the perpendicular to OS at O as Y-axis. Let SZ = 2a. The OS = a. Therefore, S=(a,0) Now let P(x,y) be any point on the curve. Join PS and draw PM perpendicular to the lineland PN perpendicular to x-axis Then by definition of the parabola, PS = PM. By distance formula, PS = rt { (x-a)^2 + y- 0)^2} and PM = NZ = x + a Hence squaring we have (x-a)^2 +y^2 = (x+a)^2 or x^2 -2ax + a^2 + y^2 = x^2 + 2ax + a^2 i.e., y^2 = 4ax Hence the equation of the parabola is y^2 = 4ax |
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