Maths-
General
Easy
Question
The equations of the normals at the ends of the latus rectum of the parabola y2 = 4ax are given by
- x2 – y2 – 6ax + 9a2 = 0
- x2 – y2 – 6ax – 6ay + 9a2 = 0
- x2 – y2 – 6ay + 9a2= 0
- none of these
Hint:
Find the end points of latus rectum of given parabola ,Also find the slope of normal at that point to form two equations.
The correct answer is: x2 – y2 – 6ax + 9a2 = 0
The co-ordinates of end points of latus rectum is (a, 2a) and (a,-2a) .
Then slope of tangent of y2 = 4ax is
Slope of tangent at (a, 2a) and (a,-2a) are 1 and -1 respectively.
As, Product of Slopes of perpendicular lines is -1.
Then Slope of Normal at (a, 2a) and (a,-2a) are and respectively.
Equations of normal are y - 2a = -1(x-a) => y+ x -3a = 0 and y + 2a = 1(x-a) => x-y -3a =0
Then equation of pair of lines is (y+ x -3a)( x-y -3a ) = 0
The equations of the normals at the ends of the latus rectum of the parabola y2 = 4ax are given by
x2 - y2- 6ax + 9a2 = 0 .
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