The number of normals that can be drawn throu | vedclass

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(D) The equation of the parabola is $y^2=4ax$,where $a=1$.Any normal to the parabola $y^2=4ax$ at point $(at^2, 2at)$ has the equation $y = -tx + 2at

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$3$

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(D) The equation of the parabola is $y^2=4ax$,where $a=1$.Any normal to the parabola $y^2=4ax$ at point $(at^2, 2at)$ has the equation $y = -tx + 2at + at^3$.If this normal passes through $(9,6)$,then $6 = -9t + 2(1)t + (1)t^3$.$6 = -7t + t^3 \Rightarrow t^3 - 7t - 6 = 0$.By testing values,$t=-1$ is a root: $(-1)^3 - 7(-1) - 6 = -1 + 7 - 6 = 0$.Dividing by $(t+1)$,we get $(t+1)(t^2 - t - 6) = 0 \Rightarrow (t+1)(t-3)(t+2) = 0$.The roots are $t = -1, 3, -2$.Since there are $3$ distinct real values for $t$,there are $3$ distinct normals that can be drawn from the point $(9,6)$ to the parabola.

The number of normals that can be drawn through the point $(9,6)$ to the parabola $y^2=4x$ is

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