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A normal chord of the parabola y^2=4ax s...

A normal chord of the parabola `y^2=4ax` subtends a right angle at the vertex, find the slope of chord.

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Equation of normal to the parabola `y^(2)=4ax` having slope 'm' is
`y=mx-2am-am^(3)` (1)
This normal meets the parabola at P and Q.
Given that PQ subtends right angle at vertex.
Now, we can get the equation of pair of straight lines OP and OQ using homogenization.
From (1), `1=(mx-y)/(2am+am^(3))`
Homogenizing the equation `y^(2)=4ax`, we get
`y^(2)=4ax((mx-y)/(2am+am^(3)))`
`or4amx^(2)-(2am+am^(3))y^(2)-4axy=0` (2)
This is the equation of pair of straight lines OP and OQ.
Since lines OP and OQ are perpendicular, sum of coefficients of `x^(2)andy^(2)` is zero.
`i.e.," "4am-2am-am^(3)=0`
`or" "m^(2)=2`
`:." "m=pmsqrt(2)`
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