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let's consider a point which belongs to our parabola, namely P=(x_0,y_0)
then its distance d from the directrix is: \[d = 3 - 1 = ...?\]
what is d?
noe the distance d', between point P and the focus of the parabola is: \[\Large d' = \sqrt {{{\left( {{x_0} - 1} \right)}^2} + {{\left( {{y_0} - 3} \right)}^2}} \] where I have applied the formula of the distance between 2 generic points: \[\Large d = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} \]
now in order to find the requested equation we have to start from this equation: \[\sqrt {{{\left( {{x_0} - 1} \right)}^2} + {{\left( {{y_0} - 3} \right)}^2}} = 2\] which expresses the condition d=d' please square both sides, what equation do you get?
sorry I have made an error, the distance d between the point P and the directrix is: \[\left| {y - 1} \right|\] so we have to start from this equation: \[\sqrt {{{\left( {{x_0} - 1} \right)}^2} + {{\left( {{y_0} - 3} \right)}^2}} = \left| {y - 1} \right|\] squaring both sides, we get: \[{\left( {{x_0} - 1} \right)^2} + {\left( {{y_0} - 3} \right)^2} = {\left( {{y_0} - 1} \right)^2}\]
now, please compute the squares of the three binomial, then simplify, and you will get the requested equation
binomials*
yes!
hint: \[\Large {x_0}^2 + 1 - 2{x_0} + {y_0}^2 + 9 - 6{y_0} = {y_0}^2 + 1 + 2{y_0}\] now simplify similar terms, please
after a simplification, we get: \[\Large x_0^2 - 2{x_0} + 9 = 8{y_0}\] now divide both sides by 8, what do you get?
sorry, we have to rewrite the left side using the completing square method
in order to that we have: \[\Large x_0^2 - 2{x_0} + 9 = x_0^2 - 2{x_0} + 1 + 8 = {\left( {{x_0} - 1} \right)^2} + 8\]
so we can write: \[\Large {\left( {{x_0} - 1} \right)^2} + 8 = 8{y_0}\]
now, please divide both sides by 8, what do you get?
it is: \[\Large {y_0} = \frac{1}{8}{\left( {{x_0} - 1} \right)^2} + 1\]
yes! since we have found the subsequent values: a=1/8, h=1 and k=1
yes!
Thanks! :)
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