Tìm hai điểm M, N trên (P):y^2=x, biết vectơ IM =4× vectơ IN và I (0;2) 26/07/2021 Bởi Valerie Tìm hai điểm M, N trên (P):y^2=x, biết vectơ IM =4× vectơ IN và I (0;2)
Đáp án: \(\left[ \begin{array}{l}\left\{ \begin{array}{l}M(36,6)\\N(9,3)\end{array} \right.\\\left\{ \begin{array}{l}M(4, – 2)\\N(1,1)\end{array} \right.\end{array} \right.\) Giải thích các bước giải: Vì M,N ∈ (P) ->M(m²,m) , N(n²,n) \(\begin{array}{l}\overrightarrow {IM} = 4\overrightarrow {IN} \\ \leftrightarrow \left\{ \begin{array}{l}{m^2} – 0 = 4({n^2} – 0)\\m – 2 = 4(n – 2)\end{array} \right. \leftrightarrow \left\{ \begin{array}{l}{m^2} = 4{n^2}\\m = 4n – 6\end{array} \right.\\ \leftrightarrow \left\{ \begin{array}{l}\left[ \begin{array}{l}m = 2n\\m = – 2n\end{array} \right.\\m = 4n – 6\end{array} \right. \leftrightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}m – 2n = 0\\m – 4n = – 6\end{array} \right.\\\left\{ \begin{array}{l}m + 2n = 0\\m – 4n = – 6\end{array} \right.\end{array} \right. \leftrightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}m = 6\\n = 3\end{array} \right.\\\left\{ \begin{array}{l}m = – 2\\n = 1\end{array} \right.\end{array} \right.\\ \leftrightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}M(36,6)\\N(9,3)\end{array} \right.\\\left\{ \begin{array}{l}M(4, – 2)\\N(1,1)\end{array} \right.\end{array} \right.\\ \end{array}\) Bình luận
Đáp án:
\(\left[ \begin{array}{l}
\left\{ \begin{array}{l}
M(36,6)\\
N(9,3)
\end{array} \right.\\
\left\{ \begin{array}{l}
M(4, – 2)\\
N(1,1)
\end{array} \right.
\end{array} \right.\)
Giải thích các bước giải:
Vì M,N ∈ (P) ->M(m²,m) , N(n²,n)
\(\begin{array}{l}
\overrightarrow {IM} = 4\overrightarrow {IN} \\
\leftrightarrow \left\{ \begin{array}{l}
{m^2} – 0 = 4({n^2} – 0)\\
m – 2 = 4(n – 2)
\end{array} \right. \leftrightarrow \left\{ \begin{array}{l}
{m^2} = 4{n^2}\\
m = 4n – 6
\end{array} \right.\\
\leftrightarrow \left\{ \begin{array}{l}
\left[ \begin{array}{l}
m = 2n\\
m = – 2n
\end{array} \right.\\
m = 4n – 6
\end{array} \right. \leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
m – 2n = 0\\
m – 4n = – 6
\end{array} \right.\\
\left\{ \begin{array}{l}
m + 2n = 0\\
m – 4n = – 6
\end{array} \right.
\end{array} \right. \leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
m = 6\\
n = 3
\end{array} \right.\\
\left\{ \begin{array}{l}
m = – 2\\
n = 1
\end{array} \right.
\end{array} \right.\\
\leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
M(36,6)\\
N(9,3)
\end{array} \right.\\
\left\{ \begin{array}{l}
M(4, – 2)\\
N(1,1)
\end{array} \right.
\end{array} \right.\\
\end{array}\)