Syarat agar garis \(ax + y = 0\) menyinggung lingkaran dengan pusat \((-1, 3)\) dan jari-jari 1 adalah \(a = \dotso\)
Jawaban: B
Gunakan rumus jarak titik \(\color{blue} (x_1, y_1)\) ke garis \(\color{blue}ax + by + c = 0\)
\(\color{blue}r = \left|\dfrac{ax_1 + by_1 + c}{\sqrt{a^2 + b^2}}\right|\)
Jarak titik pusat lingkaran \((-1, 3)\) ke garis singgung \(ax + y = 0\) adalah jari-jari.
\(r = \left|\dfrac{ax + y}{\sqrt{a^2 + 1^2}}\right|\)
\(1= \left|\dfrac{a(-1) + 3}{\sqrt{a^2 + 1^2}}\right|\)
\(1= \left|\dfrac{-a+ 3}{\sqrt{a^2 + 1}}\right|\)
Kali silang
\(\sqrt{a^2 + 1}= -a + 3\)
Kuadratkan kedua ruas,
\(a^2 + 1 = (-a + 3)^2\)
\(a^2 + 1 = a^2 \:-\:6a + 9\)
\(6a = 8\)
\(a = \dfrac{8}{6}\)
\(a = \dfrac{4}{3}\)
Jika \(\tan 2\alpha = 4\sin \alpha \cos \alpha\) untuk \(\dfrac{\pi}{2} < \alpha < \pi\), maka \(\cos \alpha = \dotso\)
Jawaban: D
Rumus sudut rangkap yang diperlukan:
\(\color{blue} \sin 2 \alpha = 2 \sin \alpha \cos \alpha\)
\(\color{blue} \cos 2 \alpha = 2\cos^2 \alpha\:-\:1\)
\(\tan 2\alpha = 2 \cdot 2\sin \alpha \cos \alpha\)
\(\tan 2\alpha = 2 \cdot \sin 2\alpha\)
\(\tan 2\alpha \:-\: 2 \cdot \sin 2\alpha = 0\)
\(\dfrac{\sin 2\alpha }{\cos 2\alpha } \:-\: 2 \cdot \sin 2\alpha = 0\)
\(\sin 2 \alpha \left(\dfrac{1}{\cos 2\alpha }\:-\:2\right) = 0\)
\(\dfrac{1}{\cos 2\alpha }\:-\:2 = 0\)
\(\dfrac{1}{\cos 2\alpha } = 2\)
\(\cos 2\alpha = \dfrac{1}{2}\)
\(2\cos^2 \alpha\:-\:1 =\dfrac{1}{2}\)
\(2\cos^2 \alpha = \dfrac{3}{2}\)
\(\cos^2 \alpha = \dfrac{3}{4}\)
\(\cos \alpha = \pm\sqrt{\dfrac{3}{4}}\)
\(\cos \alpha = \pm \dfrac{1}{2}\sqrt{3}\)
Karena \(\dfrac{\pi}{2} < \alpha < \pi\) (kuadran II) maka ambil nilai cosinus yang negatif
\(\cos \alpha =- \dfrac{1}{2}\sqrt{3}\)
Vektor \(\vec{u} = (x, y, 1)\) sejajar dengan \(\vec{v} = (-1, 3, z)\). Jika \(\vec{u}\) tegak lurus \((3, -2, 3)\), maka \(y = \dotso\)
Jawaban: B
\(\vec{u}\) sejajar dengan \(\vec{v}\) maka berlaku:
\(\color{blue}\vec{u} = k\cdot \vec{v}\)
\( (x, y, 1) = k\cdot (-1, 3, z)\)
\(x = -k\)
\(y = 3k\)
\(\vec{u}\) tegak lurus \((3, -2, 3)\) maka berlaku:
\(\color{blue}\vec{u} \cdot (3, -2, 3) = 0\)
\((x, y, 1) \cdot (3, -2, 3) = 0\)
\(3x\:-\:2y + 3 = 0\)
Substitusikan nilai \(x = -k\) dan \(y = 3k\)
\(3(-k)\:-\:2(3k) + 3 = 0\)
\(-3k\:-\:6k + 3 = 0\)
\(-9k + 3 = 0\)
\(-9k = -3\)
\(k = \dfrac{3}{9} = \dfrac{1}{3}\)
\(y = 3k\)
\(y = \cancel{3}\cdot \dfrac{1}{\cancel{3}}\)
\(y = 1\)
