If x + y = k is a normal to the parabola {y^2 | vedclass

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(B) The equation of the parabola is ${y^2} = 12x$,which is of the form ${y^2} = 4ax$,where $a = 3$.The equation of a normal to the parabola ${y^2} = 4

Vedclass · Accepted Answer

$9$

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(B) The equation of the parabola is ${y^2} = 12x$,which is of the form ${y^2} = 4ax$,where $a = 3$.The equation of a normal to the parabola ${y^2} = 4ax$ at point $t$ is given by $y + tx = 2at + at^3$.Substituting $a = 3$,the normal equation becomes $y + tx = 6t + 3t^3$.Comparing this with the given normal $x + y = k$,we have $tx + y = 6t + 3t^3$ and $x + y = k$.Since these represent the same line,the coefficients must be proportional:$\frac{t}{1} = \frac{1}{1} = \frac{6t + 3t^3}{k}$.From $\frac{t}{1} = 1$,we get $t = 1$.Substituting $t = 1$ into the ratio $\frac{1}{1} = \frac{6(1) + 3(1)^3}{k}$,we get $1 = \frac{6 + 3}{k}$,which implies $k = 9$.

If $x + y = k$ is a normal to the parabola ${y^2} = 12x$,then $k$ is

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