The angle of intersection between the curve \(x^2\) = 32y and \(y^2\) = 4x at point (16, 8) is

Solution :

\(x^2\) = 32y  \(\implies\)  \(dy\over dx\) = \(x\over 16\)  \(\implies\)  \(y^2\) = 4x \(\implies\)  \(dy\over dx\) = \(2\over y\)

\(\therefore\)  at  (16, 8), \((dy\over dx)_1\) = 1, \((dy\over dx)_2\) = \(1\over 4\)

So, required angle = \(tan^{-1}({1 – {1\over 4}\over 1 + 1({1\over 4})})\)

= \(tan^{-1}({3\over 5})\)


Similar Questions

Find the equations of the tangent and the normal at the point ‘t’ on the curve x = \(a sin^3 t\), y = \(b cos^3 t\).

Find the equation of the normal to the curve y = \(2x^2 + 3 sin x\) at x = 0.

Find the angle between the curves xy = 6 and \(x^2 y\) =12.

Check the orthogonality of the curves \(y^2\) = x and \(x^2\) = y.

Find the equation of the tangent to curve y = \(-5x^2 + 6x + 7\)  at the point (1/2, 35/4).

Leave a Comment

Your email address will not be published.