The slope of the line touching both the parabolas y2 = 4x and x2 = 32y is
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General parabola should be applied;Y²=4ax
therefore we differentiate;dy/dx=2a/y
(x1,y1),dy/dx=2a/y1=m
y1=2a/m & y²=4ax
whereas,(2a/m)²=4ax,so x1=a/m²
equation of a tangent is y-y1=m(x-x1)
y-2a/m=m(x-a/m²)
equate x²=4ay2x=4a dy/dx
dy/dx=x/2a
at(x²,y²),dy/dx=x²/2a=m
x2=2am
(2am)²=4ay2
4a²m²=4ay2 or y²=am²
hence,the equation of a tangent is;y-am²=m(x-2am)
y=mx-am²(equation of a tangent at x²=4ay)
hence the curve are;y²=4x(where a=1) & x²=32y(where a=8)
and the tangent to both curves are same
hence,y(mx+1/m &y=mx-8m²)
hence m³=-1/8
m=-1/2
therefore we differentiate;dy/dx=2a/y
(x1,y1),dy/dx=2a/y1=m
y1=2a/m & y²=4ax
whereas,(2a/m)²=4ax,so x1=a/m²
equation of a tangent is y-y1=m(x-x1)
y-2a/m=m(x-a/m²)
equate x²=4ay2x=4a dy/dx
dy/dx=x/2a
at(x²,y²),dy/dx=x²/2a=m
x2=2am
(2am)²=4ay2
4a²m²=4ay2 or y²=am²
hence,the equation of a tangent is;y-am²=m(x-2am)
y=mx-am²(equation of a tangent at x²=4ay)
hence the curve are;y²=4x(where a=1) & x²=32y(where a=8)
and the tangent to both curves are same
hence,y(mx+1/m &y=mx-8m²)
hence m³=-1/8
m=-1/2
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