Bizga parametrik tenglamalar to'plami uchun birinchi hosila kerak bo'ladi. Bizga quyidagi hosilalar kerak bo'ladi,

\[\frac{{dx}}{{dt}} = - 6\sin \left( {3t} \right) - 12\cos \left( {3t} \right)\hspace{0.5in}\frac{{dy}}{{dt}} = 18{\sec ^2}\left( {6t} \right)\]

Birinchi hosila shundan iboratki,

\[\frac{{dy}}{{dx}} = \frac{{\displaystyle \,\,\frac{{dy}}{{dt}}\,\,}}{{\displaystyle \,\,\frac{{dx}}{{dt}}\,\,}} = \frac{{18{{\sec }^2}\left( {6t} \right)}}{{ - 6\sin \left( {3t} \right) - 12\cos \left( {3t} \right)}} = \frac{{3{{\sec }^2}\left( {6t} \right)}}{{ - \sin \left( {3t} \right) - 2\cos \left( {3t} \right)}}\]

2-qadam

\(t = \frac{\pi }{2}\) dagi tangens chiziqning qiyaligi u holda,

\[m = {\left. {\frac{{dy}}{{dx}}} \right|_{t = \frac{\pi }{2}}} = \frac{{3{{\left( { - 1} \right)}^2}}}{{ - \left( { - 1} \right) - 2\left( 0 \right)}} = 3\]

\(t = \frac{\pi }{2}\) da parametrik egri chiziq nuqtada,

\[\left. {{x_{t\, = \,\frac{\pi }{2}}}} \right| = 2\left( 0 \right) - 4\left( { - 1} \right) = 4\hspace{0.25in}\,\,\left. {{y_{t\, = \,\frac{\pi }{2}}}} \right| = 3\left( 0 \right) = 0\hspace{0.5in} \Rightarrow \hspace{0.5in}\left( {4,0} \right)\]

3-qadam

Bu muammo uchun (yagona) tangens chizig'i shundan iboratki,

\[y = 0 + 3\left( {x - 4} \right)\hspace{0.25in} \to \hspace{0.5in}\require{bbox} \bbox[2pt,border:1px solid black]{{y = 3x - 12}}\]