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Three normals are drawn from the point (...

Three normals are drawn from the point (c, 0) to the curve `y^2 = x`. Show that c must be greater than 1/2. One normal is always the axis. Find c for which the other two normals are perpendicular to each other.

Text Solution

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The correct Answer is:
`xgt(1)/(2),c=(3)/(4)`
Equation of normal to parabola `y^(2)=4ax` having slope m is
`y=mx-2am-am^(3)`
For parabola `y^(2)=x,a=(1)/(4)`.
Thus, equation of normal is
`y=mx-(m)/(2)-(m^(3))/(4)`
This, equation passes through (c,0).
`:." "mc-(m)/(2)-(m^(2))/(4)=0` (1)
`rArr" "m[c-(1)/(2)-(m^(2))/(4)]=0`
`rArr" "m=0orm^(2)=4(c-(1)/(2))` (2)
For, m=0 normal is y=0, i.e., x-axis.
For other two normals to be real `m^(2)ge0`
`rArr" "4(c-(1)/(2))ge0orcge(1)/(2)`
For `c=(1)/(2),m=0`.
Therefore, for other real normal, `cgt1//2`.
Now, other two normals are perpendicular to each other.
`:.` Product of their slope =-1
So, from (1),
`4((1)/(2)-c)=-1`
`rArr" "c=(3)/(4)`
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