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

Consider the equation of an ellipse with coordinate axes as its major and minor axes: x^2/a^2 + y^2/b^2 = 1

Let a > b.
Here, a = 8/2 = 4 m and b = 2 m.
Hence, the equation becomes x^2/4^2 + y^2/2^2 = 1.
i.e. x^2/16 + y^2/4 = 1
i.e. x^2 + 4y^2 = 16 (1)

So we are required to find the ordinate of a point on the ellipse whose abscissa is a - 1.5 = 4 - 1.5 = 2.5 m.

So, put x = 2.5 in equation (1)
(5/2)^2 + 4y^2 = 16
25 + 16y^2 = 64
16y^2 = 39
y = √(39)/4, -√(39)/4

But since we are required to find height which is positive quantity so the required height of the point is √(39)/4 = 1.56 metres

1.56 Should be your answer.

answer: c

because if you look at the chart you can see a pattern and compare it to the question he simplify out your

but i think the formula is f(x)=6x

1,2,4,5,10,11,20,22,44,55,110,220