JEE Advanced 2016 (3)

Geometry Level 4

The centroid of the triangle formed by the feet of co-normal points on the curve 25 x 2 + 25 y 2 − 250 x − 300 y + 1525 = ( 3 x + 4 y − 12 ) 2 25x^{2}+25y^{2}-250x-300y+1525=(3x+4y-12)^{2} lies on the line

4 x − 3 y + 1 = 0 4x-3y+1=0 None of these 4 x − 3 y − 2 = 0 4x-3y-2=0 4 x − 3 y + 3 = 0 4x-3y+3=0

Shivam Jadhav by Shivam Jadhav , 22, India

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1 solution

Prakhar Bindal
Mar 27, 2016

Nice Problem

There is a well Known property of parabola that the centroid of triangle formed by conormal points lies on the axis of parabola.

The given Equation when transformed Will look like

(x-5)^2 + (y-6)^2 = (3x+4y-12 / 5 )^2

That means that given curve is locus of a point which moves such that distance from a fixed point and line is same which is the basic definition of parabola .

Here focus is (5,6) and directrix is 3x+4y-12 = 0

Axis is a line perpendicular to directrix passing through focus . Hence from 2 conditions we get equation of axis of parabola

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