We know that the origin of the coordinate plane is taken at the vertex of the parabolic reflector, where the axis of the reflector is along the positive x – axis.

We know that the equation of the parabola is of the form y^{2} = 4ax (as it is opening to the right)

Since, the parabola passes through point A(10, 5),

y^{2} = 4ax

10^{2} = 4a(5)

100 = 20a

a = \( \dfrac{100}{20}\)

= 5

The focus of the parabola is (a, 0) = (5, 0), which is the mid – point of the diameter.

Hence, the focus of the reflector is at the mid-point of the diameter.

Answered by Aaryan | 2 years agoAn equilateral triangle is inscribed in the parabola y^{2} = 4ax, where one vertex is at the vertex of the parabola. Find the length of the side of the triangle.

A man running a racecourse notes that the sum of the distances from the two flag posts from him is always 10 m and the distance between the flag posts is 8 m. Find the equation of the posts traced by the man.

Find the area of the triangle formed by the lines joining the vertex of the parabola x^{2} = 12y to the ends of its latus rectum.

A rod of length 12 cm moves with its ends always touching the coordinate axes. Determine the equation of the locus of a point P on the rod, which is 3 cm from the end in contact with the x-axis.

An arch is in the form of a semi-ellipse. It is 8 m wide and 2 m high at the centre. Find the height of the arch at a point 1.5 m from one end.