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# The equations of the normals at the ends of the latus rectum of the parabola y^{2} = 4ax are given by

- x
^{2} – y^{2} – 6ax + 9a^{2} = 0
- x
^{2} – y^{2} – 6ax – 6ay + 9a^{2} = 0
- x
^{2} – y^{2} – 6ay + 9a^{2}= 0
- none of these

^{2}– y^{2}– 6ax + 9a^{2}= 0^{2}– y^{2}– 6ax – 6ay + 9a^{2}= 0^{2}– y^{2}– 6ay + 9a^{2}= 0Hint:

### 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: x^{2} – y^{2} – 6ax + 9a^{2} = 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

x^{2 }- y^{2}- 6ax + 9a^{2} = 0 .

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