Question

Cho (d): 2kx+(k-1).y=2

a) Tim k de (d) song song duong thang y=\(\sqrt{3}x\)

b) Tim k de khoang cach tu goc O den (d) lon nhat

Answer 1

a: (d): 2kx+(k-1)y=2

=>(k-1)y=2-2kx

\(\Leftrightarrow y=x\cdot\dfrac{-2k}{k-1}+\dfrac{2}{k-1}\)

Để hai đường song song thì \(-\dfrac{2k}{k-1}=\sqrt{3}\)

=>\(2k=-\sqrt{3}k+\sqrt{3}\)

=>\(k\left(2+\sqrt{3}\right)=\sqrt{3}\)

=>\(k=\sqrt{3}\left(2-\sqrt{3}\right)\)

b: \(d\left(O;d\right)=\dfrac{\left|0\cdot2k+0\cdot\left(k-1\right)-2\right|}{\sqrt{\left(2k\right)^2+\left(k-1\right)^2}}=\dfrac{2}{\sqrt{\left(4k^2+k^2-2k+1\right)}}\)

Để d lớn nhất thì \(\sqrt{5k^2-2k+1}_{MIN}\)

\(\Leftrightarrow A=5k^2-2k+1_{MIN}\)

A=5(k^2-2/5k+1/5)

=5(k^2-2/5k+1/25+4/25)

=5(k-1/5)^2+4/5>=4/5

Dấu = xảy ra khi k=1/5