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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

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

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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).

Posted by

Deependra Verma

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