What is the vertex form of the equation of the parabola with a focus at (29,37) and a directrix of y=34 ?

Jul 24, 2018

The vertex form of equation is $y = \frac{1}{6} {\left(x - 29\right)}^{2} + 35.5$

Explanation:

Focus is at $\left(29 , 37\right)$and directrix is $y = 34$. Vertex is at midway

between focus and directrix. Therefore vertex is at

$\left(29 , \frac{37 + 34}{2}\right) \mathmr{and} \left(29 , 35.5\right)$ The vertex form of equation of

parabola is y=a(x-h)^2+k ; (h.k) ; being vertex.

$h = 29 \mathmr{and} k = 35.5$, so the equation of parabola is

$y = a {\left(x - 29\right)}^{2} + 35.5$. Distance of vertex from directrix is

$d = 35.5 - 34 = 1.5$, we know $d = \frac{1}{4 | a |}$

$\therefore 1.5 = \frac{1}{4 | a |} \mathmr{and} | a | = \frac{1}{1.5 \cdot 4} = \frac{1}{6}$. Here the directrix is

below the vertex , so parabola opens upward and $a$ is positive.

$\therefore a = \frac{1}{6}$ . The equation of parabola is $y = \frac{1}{6} {\left(x - 29\right)}^{2} + 35.5$

graph{1/6(x-29)^2+35.5 [-160, 160, -80, 80]} [Ans]

${\left(x - 29\right)}^{2} = 6 \left(y - \frac{71}{2}\right)$

Explanation:

The vertex of parabola having focus at $\left(29 , 37\right)$ & directrix $y = 34$ will lie on the axis of symmetry in the middle of focus & directrix whose coordinates ate given as

$V e r t e x \setminus \equiv \left(\setminus \frac{29 + 29}{2} , \setminus \frac{37 + 34}{2}\right)$

$\setminus \equiv \left(29 , \frac{71}{2}\right)$

Now, the equation of vertical parabola with vertex at $\left({x}_{1} , {y}_{1}\right) \setminus \equiv \left(29 , \frac{71}{2}\right)$will be

${\left(x - {x}_{1}\right)}^{2} = 4 a \left(y - {y}_{1}\right)$

${\left(x - 29\right)}^{2} = 4 a \left(y - \frac{71}{2}\right) \setminus \ldots \ldots \ldots . . \left(1\right)$

The equation of directrix of above parabola:

$y - \frac{71}{2} = - a$

$y = \frac{71}{2} - a$

But equation of directrx is given $y = 34$ hence

$\frac{71}{2} - a = 34$

$a = \frac{71}{2} - 34$

$a = \frac{3}{2}$

setting value of $a$ in (1), we get equation of parabola:

${\left(x - 29\right)}^{2} = 4 \setminus \cdot \frac{3}{2} \left(y - \frac{71}{2}\right)$

${\left(x - 29\right)}^{2} = 6 \left(y - \frac{71}{2}\right)$