If 2x + y + lambda = 0 is a normal to the par | vedclass

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(A) Given the parabola equation is $y^{2} = 8x$. Comparing with $y^{2} = 4ax$,we get $a = 2$. The slope of the normal to the parabola $y^{2} = 4ax$ at

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$-24$

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(A) Given the parabola equation is $y^{2} = 8x$. Comparing with $y^{2} = 4ax$,we get $a = 2$. The slope of the normal to the parabola $y^{2} = 4ax$ at point $(x_{1}, y_{1})$ is given by $m = -\frac{y_{1}}{2a}$. Given the normal equation $2x + y + \lambda = 0$,which can be written as $y = -2x - \lambda$. Comparing this with $y = mx + c$,the slope $m = -2$. Equating the slopes: $-2 = -\frac{y_{1}}{2(2)}$ $\Rightarrow -2 = -\frac{y_{1}}{4}$ $\Rightarrow y_{1} = 8$. Since the point $(x_{1}, y_{1})$ lies on the parabola $y^{2} = 8x$,we have $8^{2} = 8x_{1}$ $\Rightarrow 64 = 8x_{1}$ $\Rightarrow x_{1} = 8$. Since the point $(8, 8)$ lies on the line $2x + y + \lambda = 0$,we substitute the values: $2(8) + 8 + \lambda = 0$ $16 + 8 + \lambda = 0$ $24 + \lambda = 0$ $\lambda = -24$.

If $2x + y + \lambda = 0$ is a normal to the parabola $y^{2} = 8x$,then the value of $\lambda$ is

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