جواب کار در کلاس صفحه 42 درس 2 هندسه دوازدهم (آشنایی با مقاطع مخروطی)

\(\begin{array}{l}O(0\;,\;1)\;\,,\,\,r = 3\\\\ \Rightarrow {(x - 0)^2} + {(y - 1)^2} = {3^2}\\\\ \Rightarrow {x^2} + {(y - 1)^2} = 9\end{array}\)

\(\begin{array}{l}O(0\;,\;0)\\\\ \Rightarrow {(x - 0)^2} + {(y - 0)^2} = {r^2}\\\\ \Rightarrow {x^2} + {y^2} = {r^2}\end{array}\)

الف

\(\begin{array}{l}{x^2} + {y^2} - 2x - 6y - 1 = 0\\\\ \Rightarrow a = - 2\;,\;b = - 6\;,\;c = - 1\\\\ \Rightarrow {a^2} + {b^2} - 4c = \\\\4 + 36 + 4 = 44 > 0\\\\O'( - \frac{a}{2}\;,\; - \frac{b}{2}) \Rightarrow O'(1\;,\;3)\\\\r = \frac{{\sqrt {{a^2} + {b^2} - 4c} }}{2} = \frac{{\sqrt {44} }}{2} = \\\\\frac{{2\sqrt {11} }}{2} = \sqrt {11} \end{array}\)

ب

\(\begin{array}{l}{x^2} + {y^2} + 2x + 3y + 4 = 0\\\\ \Rightarrow a = 2\;,\;b = 3\;,\;c = 4\\\\ \Rightarrow {a^2} + {b^2} - 4c = \\\\4 + 9 - 16 = - 3 < 0\end{array}\)

این معادله هیچ نقطه از صفحه را مشخص نمی کند.

ج

\(\begin{array}{l}2{x^2} + 2{y^2} - 3x + 4y - 2 = 0\\\\ \Rightarrow {x^2} + {y^2} - \frac{3}{2}x + 2y - 1 = 0\\\\ \Rightarrow a = - \frac{3}{2}\;,\;b = 2\;,\;c = - 1\\\\ \Rightarrow {a^2} + {b^2} - 4c = \\\\\frac{9}{4} + 4 + 4 = \frac{{41}}{4} > 0\\\\O'( - \frac{a}{2}\;,\; - \frac{b}{2}) \Rightarrow O'(\frac{3}{2},\; - 1)\\\\r = \frac{{\sqrt {{a^2} + {b^2} - 4c} }}{2} = \frac{{\sqrt {\frac{{41}}{4}} }}{2} = \\\\ = \frac{{\frac{{\sqrt {41} }}{2}}}{2} = \frac{{\sqrt {41} }}{4}\end{array}\)