The two parabolas y^2 = 4x and x^2 = 4y inter | vedclass

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(C) Solving $x^2 = 4y$ and $y^2 = 4x$,we get $x = 0, y = 0$ and $x = 4, y = 4$.Since the abscissa is not zero,the point $P$ is $(4, 4)$.The equation o

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The tangents to each curve at $P$ make complementary angles with the $x$-axis

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(C) Solving $x^2 = 4y$ and $y^2 = 4x$,we get $x = 0, y = 0$ and $x = 4, y = 4$.Since the abscissa is not zero,the point $P$ is $(4, 4)$.The equation of the tangent to $y^2 = 4x$ at $(4, 4)$ is $y(4) = 2(x + 4)$,which simplifies to $2x - y + 4 = 0$ (Slope $m_1 = 2$).The equation of the tangent to $x^2 = 4y$ at $(4, 4)$ is $x(4) = 2(y + 4)$,which simplifies to $x - 2y + 4 = 0$ (Slope $m_2 = 1/2$).Let $	heta_1$ and $	heta_2$ be the angles the tangents make with the $x$-axis. Then $	an 	heta_1 = 2$ and $	an 	heta_2 = 1/2$.Since $	an 	heta_1 = \cot 	heta_2 = 	an(90^\circ - 	heta_2)$,we have $	heta_1 = 90^\circ - 	heta_2$,or $	heta_1 + 	heta_2 = 90^\circ$.Thus,the tangents make complementary angles with the $x$-axis.

The two parabolas $y^2 = 4x$ and $x^2 = 4y$ intersect at a point $P$,whose abscissa is not zero,such that

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