Dua garis yang saling tegak lurus menyinggung elips 2x² + 3y² + 4x – 12y – 36 = 0. Jika salah satu garis mempunyai kemiringan –(3/2). Tentukan titik potong kedua garis singgung.

Jawaban : (–1 ± (5/13)√(57) ± (10/13)√(10), 2 ± (10/39)√(57) ± (20/39)√(10) ± (65/39)√(10)) Pembahasan : Konsep : Persamaan elips : (x-p)²/a² + (y-q)²/b² = 1 dimana (p, q) adalah pusat Persamaan garis singgung elips: (y - q) = m(x - p) ± √(a²m²+b²) 2x² + 3y² + 4x – 12y – 36 = 0 2x² + 4x + 2 + 3y² – 12y + 12 – 36 – 2 – 12 = 0 2(x² + 2x + 1) + 3(y² – 4y + 4) – 50 = 0 2(x + 1)² + 3(y – 2)² = 50 (bagi 50) (x + 1)²/(50/2) + (y – 2)²/(50/3) = 1 50/2 > 50/3 maka p = -1 q = 2 a² = (50/2) >> a = √(50/2) b² = (50/3) >> b = √(50/3) persamaan garis singgung : (y - q) = m(x - p) ± √(a²m²+b²) (y - 2) = m(x - (-1)) ± √((50/2)m²+50/3) (y - 2) = m(x + 1) ± √((50/2)m²+50/3) .......................... (0) Dua garis yang saling tegak lurus menyinggung elips. Salah satu garis mempunyai kemiringan –(3/2) m1 = –(3/2) ............................... (1) tegak lurus maka m1.m2 = –1 m2 = –1/m1 m2 = –1/(–(3/2)) m2 = (2/3) ............................... (2) substitusi (1) ke (0) (y – 2) = –(3/2)(x + 1) ± √((50/2)(–(3/2))²+50/3) (y – 2) = –(3/2)(x + 1) ± √((50/2)(9/4)+50/3) (y – 2) = –(3/2)(x + 1) ± √(225/4+50/3) (y – 2) = –(3/2)(x + 1) ± √(675/12+200/12) (y – 2) = –(3/2)(x + 1) ± √(475/12) (y – 2) = –(3/2)(x + 1) ± √(5700/144) (y – 2) = –(3/2)(x + 1) ± √(57(100/144)) (y – 2) = –(3/2)(x + 1) ± (10/12)√(57) (y – 2) = –(3/2)(x + 1) ± (5/6)√(57) ................. (3) substitusi (2) ke (0) (y – 2) = (2/3)(x + 1) ± √((50/2)(2/3)²+50/3) (y – 2) = (2/3)(x + 1) ± √((50/2)(4/9)+50/3) (y – 2) = (2/3)(x + 1) ± √(100/9+150/9) (y – 2) = (2/3)(x + 1) ± √(250/9) (y – 2) = (2/3)(x + 1) ± √(10(25/9)) (y – 2) = (2/3)(x + 1) ± (5/3)√(10) ................. (4) (3) dan (4) (y – 2) = (y – 2) (2/3)(x + 1) ± (5/3)√(10) = –(3/2)(x + 1) ± (5/6)√(57) (2/3)(x + 1) + (3/2)(x + 1) = ± (5/6)√(57) ± (5/3)√(10) (4/6)(x + 1) + (9/6)(x + 1) = ± (5/6)√(57) ± (5/3)√(10) (13/6)(x + 1) = ± (5/6)√(57) ± (5/3)√(10) (x + 1) = (6/13)(± (5/6)√(57) ± (5/3)√(10)) (x + 1) = ± (5/13)√(57) ± (10/13)√(10) x = –1 ± (5/13)√(57) ± (10/13)√(10) .............................. (5) substitusi (5) ke (4) (y – 2) = (2/3)(x + 1) ± (5/3)√(10) (y – 2) = (2/3)(–1 ± (5/13)√(57) ± (10/13)√(10) + 1) ± (5/3)√(10) (y – 2) = (2/3)(± (5/13)√(57) ± (10/13)√(10)) ± (5/3)√(10) (y – 2) = (± (10/39)√(57) ± (20/39)√(10)) ± (5/3)√(10) (y – 2) = ± (10/39)√(57) ± (20/39)√(10) ± (65/39)√(10) y = 2 ± (10/39)√(57) ± (20/39)√(10) ± (65/39)√(10) ................ (6) Jadi, titik potongnya adalah (–1 ± (5/13)√(57) ± (10/13)√(10), 2 ± (10/39)√(57) ± (20/39)√(10) ± (65/39)√(10))