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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. |
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Answer» Correct Answer - `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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