Let L_{1} be a tangent to the parabola y^{2}= | vedclass

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(A) The equation of a tangent to the parabola $y^{2}=4a(x-h)$ is $y=m(x-h)+\frac{a}{m}$.For the parabola $y^{2}=4(x+1)$,we have $a=1$ and $h=-1$. The

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$x+3=0$

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(A) The equation of a tangent to the parabola $y^{2}=4a(x-h)$ is $y=m(x-h)+\frac{a}{m}$.For the parabola $y^{2}=4(x+1)$,we have $a=1$ and $h=-1$. The tangent $L_{1}$ is $y=m(x+1)+\frac{1}{m}$,which simplifies to $y=mx+m+\frac{1}{m}$.For the parabola $y^{2}=8(x+2)$,we have $a=2$ and $h=-2$. The tangent $L_{2}$ is $y=m'(x+2)+\frac{2}{m'}$,which simplifies to $y=m'x+2m'+\frac{2}{m'}$.Since $L_{1}$ and $L_{2}$ intersect at right angles,$m \cdot m' = -1$,so $m' = -\frac{1}{m}$.Substituting $m'$ into the equation for $L_{2}$:$y = -\frac{1}{m}x + 2(-\frac{1}{m}) + \frac{2}{-1/m} = -\frac{1}{m}x - \frac{2}{m} - 2m = -\frac{1}{m}x - 2(m+\frac{1}{m})$.Equating the two expressions for $y$:$mx + m + \frac{1}{m} = -\frac{1}{m}x - 2(m+\frac{1}{m})$.Rearranging the terms:$(m+\frac{1}{m})x + (m+\frac{1}{m}) + 2(m+\frac{1}{m}) = 0$.$(m+\frac{1}{m})(x+3) = 0$.Since $m+\frac{1}{m} 
eq 0$ for real tangents,we must have $x+3=0$.

Let $L_{1}$ be a tangent to the parabola $y^{2}=4(x+1)$ and $L_{2}$ be a tangent to the parabola $y^{2}=8(x+2)$ such that $L_{1}$ and $L_{2}$ intersect at right angles. Then $L_{1}$ and $L_{2}$ meet on the straight line

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