Irisan kerucut: Perbedaan antara revisi
Konten dihapus Konten ditambahkan
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| Lingkaran || colspan=2 align="center"| <math>y = mx \pm r\sqrt{1+m^2} |
| Lingkaran || colspan=2 align="center"| <math>y = mx \pm r\sqrt{1+m^2}</math> |
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| Parabola || <math>y = mx - p m |
| Parabola || <math>y = mx - p m</math> || <math>y = mx + \frac{p}{m}</math> |
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| Elips || <math>y = mx \pm \sqrt{b^2 + a^2m^2} |
| Elips || <math>y = mx \pm \sqrt{b^2 + a^2m^2}</math> || <math>y = mx \pm \sqrt{a^2m^2 + b^2}</math> |
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| Hiperbola || <math>y = mx \pm \sqrt{b^2 - a^2m^2} |
| Hiperbola || <math>y = mx \pm \sqrt{b^2 - a^2m^2}</math> || <math>y = mx \pm \sqrt{a^2m^2 - b^2}</math> |
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! !! colspan=2 align="center"| Titik pusat (h,k) |
! !! colspan=2 align="center"| Titik pusat (h,k) |
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| Lingkaran || colspan=2 align="center"| <math>(y - k) = m(x - h) \pm r\sqrt{1+m} |
| Lingkaran || colspan=2 align="center"| <math>(y - k) = m(x - h) \pm r\sqrt{1+m}</math> |
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| Parabala || <math>(y - k) = m(x - h) - p m |
| Parabala || <math>(y - k) = m(x - h) - p m</math> || <math>(y - k) = m(x -h) + \frac{p}{m}</math> |
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| Elips || <math>(y - k) = m(x - h) \pm \sqrt{b^2 + a^2m^2} |
| Elips || <math>(y - k) = m(x - h) \pm \sqrt{b^2 + a^2m^2}</math> || <math>(y - k) = m(x - h) \pm \sqrt{a^2m^2 + b^2}</math> |
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| Hiperbola || <math>y = mx \pm \sqrt{b^2 - a^2m^2} |
| Hiperbola || <math>y = mx \pm \sqrt{b^2 - a^2m^2}</math> || <math>y = mx \pm \sqrt{a^2m^2 - b^2}</math> |
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| Baris 214: | Baris 214: | ||
| Lingkaran || colspan=2 align="center"| <math>x x_1 + y y_1 = r^2</math> |
| Lingkaran || colspan=2 align="center"| <math>x x_1 + y y_1 = r^2</math> |
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| Parabola || <math>x x_1 = 2py + 2py_1 |
| Parabola || <math>x x_1 = 2py + 2py_1</math> || <math>y y_1 = 2px + 2px_1</math> |
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| Elips || <math>\frac{x x_1}{b^2} + \frac{y y_1}{a^2} = 1</math> || <math>\frac{x x_1}{a^2} + \frac{y y_1}{b^2} = 1</math> |
| Elips || <math>\frac{x x_1}{b^2} + \frac{y y_1}{a^2} = 1</math> || <math>\frac{x x_1}{a^2} + \frac{y y_1}{b^2} = 1</math> |
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! !! colspan=2 align="center"| Titik pusat (h,k) |
! !! colspan=2 align="center"| Titik pusat (h,k) |
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| Lingkaran || colspan=2 align="center"| <math>(x - h)(x_1 - h) + (y - k)(y_1 - k) = r^2 |
| Lingkaran || colspan=2 align="center"| <math>(x - h)(x_1 - h) + (y - k)(y_1 - k) = r^2</math> atau <br><math> x x_1 + y y_1 + \frac{1}{2} A x + \frac{1}{2} A x_1 + \frac{1}{2} B y + \frac{1}{2} B y_1 + C = 0</math> |
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| Parabola || <math>(x - h)(x_1 - h) = 2p(y - k) + 2p(y_1 - k) |
| Parabola || <math>(x - h)(x_1 - h) = 2p(y - k) + 2p(y_1 - k)</math> || <math>(y - k)(y_1 - k) = 2p(x - h) + 2p(x_1 - h)</math> |
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| Elips || <math>\frac{(x - h)(x_1 - h)}{b^2} + \frac{(y - k)(y_1 - k)}{a^2} = 1 |
| Elips || <math>\frac{(x - h)(x_1 - h)}{b^2} + \frac{(y - k)(y_1 - k)}{a^2} = 1</math> || <math>\frac{(x - h)(x_1 - h)}{a^2} + \frac{(y - k)(y_1 - k)}{b^2} = 1</math> |
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| Hiperbola || <math>\frac{(x - h)(x_1 - h)}{b^2} - \frac{(y - k)(y_1 - k)}{a^2} = 1 |
| Hiperbola || <math>\frac{(x - h)(x_1 - h)}{b^2} - \frac{(y - k)(y_1 - k)}{a^2} = 1</math> || <math>\frac{(x - h)(x_1 - h)}{a^2} - \frac{(y - k)(y_1 - k)}{b^2} = 1</math> |
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| Baris 309: | Baris 309: | ||
; Titik pusat (h,k) |
; Titik pusat (h,k) |
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* Tentukan persamaan garis singgung <math>y^2 - 6y - 8x + 9 = 0 |
* Tentukan persamaan garis singgung <math>y^2 - 6y - 8x + 9 = 0</math> melalui persamaan yang tegak lurus <math>y - 2x - 5 = 0</math>! |
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jawab: |
jawab: |
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ubah ke bentuk sederhana |
ubah ke bentuk sederhana |
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:<math>y^2 - 6y - 8x + 9 = 0 |
:<math>y^2 - 6y - 8x + 9 = 0</math> |
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:<math>y^2 - 6y + 9 = 8x |
:<math>y^2 - 6y + 9 = 8x</math> |
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:<math>(y - 3)^2 = 8x |
:<math>(y - 3)^2 = 8x</math> |
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cari gradien persamaan <math>y - 2x - 5 = 0 </math> |
cari gradien persamaan <math>y - 2x - 5 = 0 </math> |
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:<math>y - 2x - 5 = 0 |
:<math>y - 2x - 5 = 0</math> |
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:<math>y = 2x + 5 |
:<math>y = 2x + 5</math> |
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gradien (<math>m_1 |
gradien (<math>m_1</math>) = 2 karena tegak lurus menjadi <math>m_2 = - \frac{1}{2}</math> |
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cari <math>p |
cari <math>p</math> |
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:<math>(y - 3)^2 = 8x -> (y - 3)^2 = 4 (2x) jadi p = 2 |
:<math>(y - 3)^2 = 8x -> (y - 3)^2 = 4 (2x) \text { jadi } p = 2</math> |
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:<math>y = - \frac{1}{2}x + \frac{2}{- \frac{1}{2}} -> y = - \frac{1}{2}x - 4 |
:<math>y = - \frac{1}{2}x + \frac{2}{- \frac{1}{2}} -> y = - \frac{1}{2}x - 4</math> |
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* Tentukan persamaan garis singgung <math>y^2 - 6y - 8x + 9 = 0 |
* Tentukan persamaan garis singgung <math>y^2 - 6y - 8x + 9 = 0</math> yang berordinat 6! |
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jawab: |
jawab: |
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ubah ke bentuk sederhana |
ubah ke bentuk sederhana |
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:<math>y^2 - 6y - 8x + 9 = 0 |
:<math>y^2 - 6y - 8x + 9 = 0</math> |
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:<math>y^2 - 6y + 9 = 8x |
:<math>y^2 - 6y + 9 = 8x</math> |
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:<math>(y - 3)^2 = 8x |
:<math>(y - 3)^2 = 8x</math> |
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cari absis dimana ordinat 6 |
cari absis dimana ordinat 6 |
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:<math>(y - 3)^2 = 8x |
:<math>(y - 3)^2 = 8x</math> |
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:<math>(6 - 3)^2 = 8x |
:<math>(6 - 3)^2 = 8x</math> |
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:<math>9 = 8x |
:<math>9 = 8x</math> |
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:<math>x = \frac{9}{8} |
:<math>x = \frac{9}{8}</math> |
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:<math>(y - 3)^2 = 8x -> (y - 3)^2 = 4 (2x) \text { jadi } p = 2</math> |
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dengan cara bagi adil |
dengan cara bagi adil |
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:<math>(y - k)(y_1 - k) = 2px + 2px_1 |
:<math>(y - k)(y_1 - k) = 2px + 2px_1</math> |
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:<math>(y - 3)(6 - 3) = |
:<math>(y - 3)(6 - 3) = 2(2)x + 2(2)(\frac{9}{8})</math> |
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:<math>(y - 3)3 = 4x + \frac{9}{2} |
:<math>(y - 3)3 = 4x + \frac{9}{2}</math> |
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:<math>3y - 9 = 4x + \frac{9}{2} |
:<math>3y - 9 = 4x + \frac{9}{2}</math> |
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:<math>3y = 4x + \frac{27}{2} |
:<math>3y = 4x + \frac{27}{2}</math> |
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:<math>y = \frac{4}{3}x + \frac{27}{6} |
:<math>y = \frac{4}{3}x + \frac{27}{6}</math> |
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* Tentukan persamaan garis singgung yang melalui (1,6) terhadap <math>y^2 - 6y - 8x + 9 = 0 </math>! |
* Tentukan persamaan garis singgung yang melalui (1,6) terhadap <math>y^2 - 6y - 8x + 9 = 0 </math>! |
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| Baris 350: | Baris 352: | ||
:<math>y^2 - 6y - 8x + 9 = 0 </math> |
:<math>y^2 - 6y - 8x + 9 = 0 </math> |
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:<math>y^2 - 6y + 9 = 8x </math> |
:<math>y^2 - 6y + 9 = 8x </math> |
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:<math>(y - 3)^2 = 8x |
:<math>(y - 3)^2 = 8x</math> |
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:<math>(y - 3)^2 = 8x -> (y - 3)^2 = 4 (2x) \text { jadi } p = 2</math> |
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:<math>(y - 3)^2 - 8x = 0 maka masukkan lah (1,6) (6 - 3)^2 - 8 (1) = 9 - 8 = 1 > 0 |
:<math>(y - 3)^2 - 8x = 0 \text { maka masukkan lah (1,6) } (6 - 3)^2 - 8 (1) = 9 - 8 = 1 > 0</math> (luar) |
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dengan cara bagi adil |
dengan cara bagi adil |
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:<math>(y - k)(y_1 - k) = 2px + 2px_1 |
:<math>(y - k)(y_1 - k) = 2px + 2px_1</math> |
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:<math>(y - 3)(6 - 3) = |
:<math>(y - 3)(6 - 3) = 2(2)x + 2(2)(1)</math> |
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:<math>(y - 3)3 = 4x + 4 |
:<math>(y - 3)3 = 4x + 4</math> |
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:<math>3y - 9 = 4x + 4 |
:<math>3y - 9 = 4x + 4</math> |
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:<math>3y = 4x + 13 |
:<math>3y = 4x + 13</math> |
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:<math>y = \frac{4}{3}x + \frac{13}{3} |
:<math>y = \frac{4}{3}x + \frac{13}{3}</math> |
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masukkan lah <math>(y - 3)^2 = 8x |
masukkan lah <math>(y - 3)^2 = 8x</math> |
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:<math>(\frac{4}{3}x + \frac{13}{3} - 3)^2 = 8x |
:<math>(\frac{4}{3}x + \frac{13}{3} - 3)^2 = 8x</math> |
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:<math>(\frac{4}{3}x + \frac{4}{3})^2 = 8x |
:<math>(\frac{4}{3}x + \frac{4}{3})^2 = 8x</math> |
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:<math>\frac{16}{9}x^2 + \frac{32}{9}x + \frac{16}{9} - 8x = 0 |
:<math>\frac{16}{9}x^2 + \frac{32}{9}x + \frac{16}{9} - 8x = 0</math> |
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:<math>\frac{16}{9}x^2 - \frac{40}{9}x + \frac{16}{9} = 0 |
:<math>\frac{16}{9}x^2 - \frac{40}{9}x + \frac{16}{9} = 0</math> (dibagi 8/9) |
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:<math>2x^2 + 5x + 2 = 0 |
:<math>2x^2 + 5x + 2 = 0</math> |
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maka kita mencari nilai x |
maka kita mencari nilai x |
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:<math>x = \frac{- b \pm \sqrt{b^2 - 4ac}}{2a} |
:<math>x = \frac{- b \pm \sqrt{b^2 - 4ac}}{2a}</math> |
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:<math>x = \frac{5 \pm \sqrt{25 - 16}}{4} |
:<math>x = \frac{5 \pm \sqrt{25 - 16}}{4}</math> |
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:<math>x = \frac{5 \pm \sqrt{9}}{4} |
:<math>x = \frac{5 \pm \sqrt{9}}{4}</math> |
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:<math>x_1 = \frac{5 + \sqrt{9}}{4} = 2 |
:<math>x_1 = \frac{5 + \sqrt{9}}{4} = 2</math> atau <math>x_2 = \frac{5 - \sqrt{9}}{4} = \frac{1}{2}</math> |
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maka kita mencari nilai y |
maka kita mencari nilai y |
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: untuk <math>x_1 |
: untuk <math>x_1</math> |
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:<math>y_1 = \frac{4}{3} (2) + \frac{13}{3} = \frac{8}{3} + \frac{13}{3} = 7</math> |
:<math>y_1 = \frac{4}{3} (2) + \frac{13}{3} = \frac{8}{3} + \frac{13}{3} = 7</math> |
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jadi <math>(2, 7) |
jadi <math>(2, 7)</math> |
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: untuk <math>x_2 |
: untuk <math>x_2</math> |
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:<math>y_2 = \frac{4}{3} (\frac {1}{2}) + \frac{13}{3} = \frac{2}{3} + \frac{13}{3} = 5</math> |
:<math>y_2 = \frac{4}{3} (\frac {1}{2}) + \frac{13}{3} = \frac{2}{3} + \frac{13}{3} = 5</math> |
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jadi <math>(\frac{1}{2}, 5) |
jadi <math>(\frac{1}{2}, 5)</math> |
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kembali dengan cara bagi adil |
kembali dengan cara bagi adil |
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: untuk persamaan singgung pertama |
: untuk persamaan singgung pertama |
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:<math>(y - k)(y_1 - k) = 2px + 2px_1 |
:<math>(y - k)(y_1 - k) = 2px + 2px_1</math> |
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:<math>(y - 3)(7 - 3) = |
:<math>(y - 3)(7 - 3) = 2(2)x + 2(2)(2)</math> |
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:<math>(y - 3)4 = 4x + 8 |
:<math>(y - 3)4 = 4x + 8</math> |
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:<math>4y - 12 = 4x + 8 |
:<math>4y - 12 = 4x + 8</math> |
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:<math>4y = 4x + 20 </math> (dibagi 4) |
:<math>4y = 4x + 20 </math> (dibagi 4) |
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:<math>y = x + 5 |
:<math>y = x + 5</math> |
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: untuk persamaan singgung kedua |
: untuk persamaan singgung kedua |
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:<math>(y - k)(y_2 - k) = 2px + 2px_2 |
:<math>(y - k)(y_2 - k) = 2px + 2px_2</math> |
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:<math>(y - 3)(5 - 3) = |
:<math>(y - 3)(5 - 3) = 2(2)x + 2(2)(\frac{1}{2})</math> |
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:<math>(y - 3)2 = 4x + 2 |
:<math>(y - 3)2 = 4x + 2</math> |
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:<math>2y - 6 = 4x + 2 |
:<math>2y - 6 = 4x + 2</math> |
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:<math>2y = 4x + 8 |
:<math>2y = 4x + 8</math> (dibagi 2) |
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:<math>y = 2x + 4 |
:<math>y = 2x + 4</math> |
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== Referensi == |
== Referensi == |
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