[amp_mcq option1=”0\u00c2\u00b0” option2=”30\u00c2\u00b0” option3=”45\u00c2\u00b0” option4=”90\u00c2\u00b0” correct=”option1″]<\/p>\n
The correct answer is $\\boxed{\\text{D}}$.<\/p>\n
The angle of intersection of two curves is the angle between their tangent lines at the point of intersection. To find the tangent lines at the point of intersection, we need to find the derivatives of the curves at that point.<\/p>\n
The derivatives of $x^2 = 4y$ and $y^2 = 4x$ are $2x = 4y’$ and $2y = 4x’$, respectively. Substituting $x = 0$ and $y = 0$ into these equations, we get $2(0) = 4y’$ and $2(0) = 4x’$. Solving these equations, we get $y’ = 0$ and $x’ = 0$.<\/p>\n
The slopes of the tangent lines are therefore $0$. The angle between two lines with slopes $m_1$ and $m_2$ is given by the formula $\\tan \\theta = \\frac{m_1 – m_2}{1 + m_1 m_2}$. Substituting $m_1 = 0$ and $m_2 = 0$ into this formula, we get $\\tan \\theta = \\frac{0 – 0}{1 + 0 \\cdot 0} = 0$. Therefore, the angle of intersection is $\\boxed{90^\\circ}$.<\/p>\n
Here is a graph of the two curves:<\/p>\n
[asy]
\nunitsize(1 cm);<\/p>\n
draw((0,0)–(4,4));
\ndraw((0,0)–(4,0));<\/p>\n
label(“$x$”, (4,0), E);
\nlabel(“$y$”, (0,4), N);<\/p>\n
label(“$x^2 = 4y$”, (4,2), S);
\nlabel(“$y^2 = 4x$”, (2,4), W);<\/p>\n
draw((0,0)–(2,2));
\n[\/asy]<\/p>\n
As you can see, the two curves intersect at a right angle.<\/p>\n","protected":false},"excerpt":{"rendered":"
[amp_mcq option1=”0\u00c2\u00b0” option2=”30\u00c2\u00b0” option3=”45\u00c2\u00b0” option4=”90\u00c2\u00b0” correct=”option1″]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[690],"tags":[],"class_list":["post-20268","post","type-post","status-publish","format-standard","hentry","category-calculus","no-featured-image-padding"],"yoast_head":"\n