Question

Derive the equation of a parabola with a focus of 6,2 and a directrix of y=1

Answer 1

Answer:

Let ($x_{0},y_{0}$) be any point on the parabola.

then, find the distance between ($x_{0},y_{0}$) and the focus (6,2). Also, find the distance between($x_{0},y_{0}$) and directrix (i.e y=1).  

Using distance formula: let ($x_{1},y_{1 }$) and ($x_{2},y_{2}$) be any point then,

distance = $\sqrt{(y_{2}-y_{1})^2+(x_{2}-x_{1})^2}$.

The distance between ($x_{0},y_{0}$) and (6,2) is, $\sqrt{(x_{0}-6)^2+(y_{0}-2)^2}$

Similarly,

The distance between ($x_{0},y_{0}$)  and the directrix, y=1 is, $|y_{0}-1|$.

Equate the two distance expression:

$\sqrt{(x_{0}-6)^2+(y_{0}-2)^2}$=$|y_{0}-1|$.

Now, Squaring both sides we have;

$(x_{0}-6)^2+(y_{0}-2)^2 = (y_{0}-1)^2$

Simplify and bring all the terms to one side:

$x_{0}^2-16x_{0}-2y_{0}+39=0$  or we can write this as;

$y_{0}=\frac{1}{2}( x_{0}^2-16x_{0}+39)$

therefore, it is true for all values of ($x_{0},y_{0}$) on parabola and hence we can rewrite with (x,y) .

therefore, the equation of parabola with focus (6,2) and a directrix of y=1 is,

$y=\frac{1}{2}( x^2-16x+39)$.

Answer 2

Answer:

f(x)= 1/2 (x-6)^2 + 3/2

Step-by-step explanation: