The product of the slopes of the common tange | vedclass

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(D) The equation of a tangent to the parabola $y^2 = 8x$ with slope $m$ is $y = mx + \frac{2}{m}$.The condition for the line $y = mx + c$ to be a tang

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$-\frac{1}{4}$

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(D) The equation of a tangent to the parabola $y^2 = 8x$ with slope $m$ is $y = mx + \frac{2}{m}$.The condition for the line $y = mx + c$ to be a tangent to the ellipse $\frac{x^2}{32} + \frac{y^2}{8} = 1$ is $c^2 = a^2m^2 + b^2$,where $a^2 = 32$ and $b^2 = 8$.Substituting $c = \frac{2}{m}$ into the condition: $(\frac{2}{m})^2 = 32m^2 + 8$.$\frac{4}{m^2} = 32m^2 + 8$.Dividing by $4$: $\frac{1}{m^2} = 8m^2 + 2$.$8m^4 + 2m^2 - 1 = 0$.Let $t = m^2$. Then $8t^2 + 2t - 1 = 0$.$(4t - 1)(2t + 1) = 0$.Since $m^2$ must be positive,$m^2 = \frac{1}{4}$.Thus,$m = \frac{1}{2}$ or $m = -\frac{1}{2}$.The product of the slopes is $(\frac{1}{2}) 	imes (-\frac{1}{2}) = -\frac{1}{4}$.

The product of the slopes of the common tangents to the ellipse $\frac{x^2}{32} + \frac{y^2}{8} = 1$ and the parabola $y^2 = 8x$ is:

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