we know that
Te equation of a vertical parabola is
![y=A(x-h)^2+k](https://img.qammunity.org/2023/formulas/mathematics/college/wocki4w264tajt0jv2jhvn2vkjf4k2rkgl.png)
where
A is the leading coefficient
(h,k) is the vertex
In this problem
(h,k)=(0,0)
so
![y=Ax^2](https://img.qammunity.org/2023/formulas/mathematics/college/qbzg43ykylyvym05syy53ukpk21ltk2u4m.png)
we have that
For x=a, y=20 -----> is given
substitute
![\begin{gathered} 20=A(a)^2 \\ A=(20)/(a^2) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/ft8ktifo04gjplg8rbm1lge0grpdsirw7x.png)
therefore
the equation is
![y=((20)/(a^2))x^2](https://img.qammunity.org/2023/formulas/mathematics/college/zv0vpfuitzf5ke1ed02yo4do7xm8sjxvnz.png)
Remember that
the equation of the parabola in standard form is equal to
(x − h) 2 = 4 p (y − k)
vertex is (0,0)
so
x^2=4py
where
p is the distance between the vertex and the focus
in this problem
p=5
substitute
x^2=4(5)y
x^2=20y
y=(1/20)x^2
Compare this equation with the previous equation
so
![(1)/(20)=(20)/(a^2)](https://img.qammunity.org/2023/formulas/mathematics/college/zmd4xyyqetkmyse6ffbadbte24ruvxjhxn.png)
solve for a
![\begin{gathered} a^2=20^2 \\ a=20 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/4ctzls9cns982tl49yucy3ausxnphl81ma.png)
therefore
the wide of the lamp is 2a
so
2(20)=40 cm
the answer is 40 cm