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If the parabola y^2=4x and x^2=32y intersect at (16,8) at an angle theta , then theta equals

Answers (1)

D Deependra Verma

Solution:

$y=2sqrtx$

$herefore$ $fracmathrmd ymathrmd x=frac1sqrtxRightarrow m_1=(fracmathrmd ymathrmd x)_(16,8)=1$

$y=fracx^232$

$Rightarrow$ $fracmathrmd ymathrmd x=frac16x Rightarrow m_2 =(fracmathrmd ymathrmd x)_(16,8)=1$

$heta = an^-1[fracfrac14-11+frac14]= an^-1(frac35).$

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