For any non-zero real value of m,the equation | vedclass

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(C) The given equation of the line is $m x - y + (10 + m^2) = 0$. Rearranging this as a quadratic equation in $m$: $m^2 + m x + (10 - y) = 0$. Since t

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$x^2=-4(y-10)$

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(C) The given equation of the line is $m x - y + (10 + m^2) = 0$. Rearranging this as a quadratic equation in $m$: $m^2 + m x + (10 - y) = 0$. Since the line is a tangent to the parabola,the discriminant $D$ of this quadratic equation must be zero. For $a m^2 + b m + c = 0$,$D = b^2 - 4ac = 0$. Here,$a = 1$,$b = x$,and $c = (10 - y)$. Substituting these values: $x^2 - 4(1)(10 - y) = 0$. $x^2 - 40 + 4y = 0$. $x^2 = -4y + 40$. $x^2 = -4(y - 10)$. Thus,the correct option is $C$.

For any non-zero real value of $m$,the equation of the parabola to which the line $m x-y+10+m^2=0$ is a tangent,is

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