The shortest distance between the curves y^2= | vedclass

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(D) The given parabola is $y^2=8x$,where $4a=8 \Rightarrow a=2$. The equation of the normal to the parabola at point $(at^2, 2at)$ is $y = -tx + 2at +

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$2\sqrt{2}-1$

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(D) The given parabola is $y^2=8x$,where $4a=8 \Rightarrow a=2$. The equation of the normal to the parabola at point $(at^2, 2at)$ is $y = -tx + 2at + at^3$. Alternatively,using slope $m$,the normal is $y = mx - 2am - am^3$. Substituting $a=2$,we get $y = mx - 4m - 2m^3$. The given circle is $x^2+y^2+12y+35=0$,which can be written as $x^2+(y+6)^2=1$. The center is $C(0, -6)$ and radius $r=1$. The shortest distance between the curves is the distance from the center $C$ to the parabola minus the radius $r$. The normal from $C(0, -6)$ to the parabola satisfies $-6 = m(0) - 4m - 2m^3$,so $2m^3 + 4m - 6 = 0$,which simplifies to $m^3 + 2m - 3 = 0$. By inspection,$m=1$ is a root. Thus,$(m-1)(m^2+m+3)=0$. Since $m^2+m+3=0$ has no real roots,$m=1$. The point $P$ on the parabola where the normal passes through $C$ is $(am^2, -2am) = (2(1)^2, -2(2)(1)) = (2, -4)$. The distance $PC = \sqrt{(2-0)^2 + (-4 - (-6))^2} = \sqrt{2^2 + 2^2} = \sqrt{8} = 2\sqrt{2}$. Therefore,the shortest distance is $PC - r = 2\sqrt{2} - 1$.

The shortest distance between the curves $y^2=8x$ and $x^2+y^2+12y+35=0$ is:

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