Hallar la ecuación de la parábola cuyo eje es paralelo al eje Y y pasa por 3 puntos (0;3) , (3;4), (4;11) PORFAVOOOR
Respuestas
Respuesta dada por:
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Como el eje es paralelo al eje Y entonces buscas la ecuación de una parábola vertical cuya ecuación es:
![(x-h)^2=4p(y-k) (x-h)^2=4p(y-k)](https://tex.z-dn.net/?f=%28x-h%29%5E2%3D4p%28y-k%29)
Ahora debes sustituir los puntos que te dan en la ecuación:
![(0-h)^2=4p(3-k)\to h^2=12p-4pk \\ (3-h)^2=4p(4-k)\to 9-6h+h^2=16p-4pk\\(4-h)^2=4p(11-k)\to 16-8h+h^2=44p-4pk \\ \\ 9-6h+h^2-h^2=16p-4pk-(12p-4pk) \\ 9-6h=4p\\\\16-8h+h^2-h^2=44p-4pk-(12p-4pk)\\16-8h=32p\\2-h=4p \\ \\ \to 9-6h=2-h\\9-2=6h-h\\ \boxed{7/5=h} \\ \\ 2-h=4p\\2-7/5=3/5=4p\\ \boxed{p=3/20} \\ \\ h^2=12p-4pk \\ (7/5)^2=12(3/20)-4(3/20)k \\ 49/25-36/20=-(12/20)k\\4/25=-(3/5)k \\ (4/25)(-5/3)=k\\ \boxed{-4/15=k} (0-h)^2=4p(3-k)\to h^2=12p-4pk \\ (3-h)^2=4p(4-k)\to 9-6h+h^2=16p-4pk\\(4-h)^2=4p(11-k)\to 16-8h+h^2=44p-4pk \\ \\ 9-6h+h^2-h^2=16p-4pk-(12p-4pk) \\ 9-6h=4p\\\\16-8h+h^2-h^2=44p-4pk-(12p-4pk)\\16-8h=32p\\2-h=4p \\ \\ \to 9-6h=2-h\\9-2=6h-h\\ \boxed{7/5=h} \\ \\ 2-h=4p\\2-7/5=3/5=4p\\ \boxed{p=3/20} \\ \\ h^2=12p-4pk \\ (7/5)^2=12(3/20)-4(3/20)k \\ 49/25-36/20=-(12/20)k\\4/25=-(3/5)k \\ (4/25)(-5/3)=k\\ \boxed{-4/15=k}](https://tex.z-dn.net/?f=%280-h%29%5E2%3D4p%283-k%29%5Cto+h%5E2%3D12p-4pk+%5C%5C++%283-h%29%5E2%3D4p%284-k%29%5Cto+9-6h%2Bh%5E2%3D16p-4pk%5C%5C%284-h%29%5E2%3D4p%2811-k%29%5Cto+16-8h%2Bh%5E2%3D44p-4pk+%5C%5C++%5C%5C+9-6h%2Bh%5E2-h%5E2%3D16p-4pk-%2812p-4pk%29+%5C%5C++9-6h%3D4p%5C%5C%5C%5C16-8h%2Bh%5E2-h%5E2%3D44p-4pk-%2812p-4pk%29%5C%5C16-8h%3D32p%5C%5C2-h%3D4p+%5C%5C++%5C%5C+%5Cto+9-6h%3D2-h%5C%5C9-2%3D6h-h%5C%5C+%5Cboxed%7B7%2F5%3Dh%7D+%5C%5C++%5C%5C+2-h%3D4p%5C%5C2-7%2F5%3D3%2F5%3D4p%5C%5C+%5Cboxed%7Bp%3D3%2F20%7D+%5C%5C++%5C%5C++h%5E2%3D12p-4pk+%5C%5C++%287%2F5%29%5E2%3D12%283%2F20%29-4%283%2F20%29k+%5C%5C++49%2F25-36%2F20%3D-%2812%2F20%29k%5C%5C4%2F25%3D-%283%2F5%29k+%5C%5C++%284%2F25%29%28-5%2F3%29%3Dk%5C%5C+%5Cboxed%7B-4%2F15%3Dk%7D+)
La ecuación es:
![(x-7/5)^2=4(3/20)(y+4/15)\\ \boxed{(x- \frac{7}{5} )^2= \frac{3}{5}(y+ \frac{4}{15} ) } (x-7/5)^2=4(3/20)(y+4/15)\\ \boxed{(x- \frac{7}{5} )^2= \frac{3}{5}(y+ \frac{4}{15} ) }](https://tex.z-dn.net/?f=+%28x-7%2F5%29%5E2%3D4%283%2F20%29%28y%2B4%2F15%29%5C%5C+%5Cboxed%7B%28x-+%5Cfrac%7B7%7D%7B5%7D++%29%5E2%3D+%5Cfrac%7B3%7D%7B5%7D%28y%2B+%5Cfrac%7B4%7D%7B15%7D++%29++%7D)
Saludos!
Ahora debes sustituir los puntos que te dan en la ecuación:
La ecuación es:
Saludos!
Adjuntos:
![](https://es-static.z-dn.net/files/dc1/d6bc2ef26ff3814988b7e97146c08fc5.png)
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