Question

If three normals from a point to the parabola y = 4 a x cut the axis in points
whose distances from the vertex are in A.P., show that the point lies on the curve
27 a y = 2 (x - 2 a)^3​

Answer 1

Step-by-step explanation:

The equation of any normal to the parabola is y = mx – 2am – am3 It passes through the point (h, k) if am3 + m (2a - h) + k = 0 ... (1) The normal cuts the axis of the parabola viz., y = 0 at point where x = 2a + am2 Hence the abscissa of the points in which the normal through (h, k) meet the axis of the parabola are x1 = 2a + am1 2 , x2 = 2a + am2 2 , x3 = 2a + am3 2 Since x1 , x2 , x3 are in A.P. (2a + am21 ) + (2a + am32 ) = 2 (2a + am22 ) m21 + m22 = 2m22 ... (2) Also, from (1), m1 + m2 + m3 = 0. m2m3 + m3m1 + m1m2 = 2a - h/a and m1 m2 m3 = – k/a ... (5)Read more on Sarthaks.com - https://www.sarthaks.com/484003/three-normals-from-point-parabola-4ax-meet-axis-parabola-points-whose-abscissa-find-locus?show=484011#a484011

Answer 2

Answer:

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