An arch is in the form of a semiellipse. 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.

Explanation

Let equation of ellipse be `(x^(2))/(a^(2))+y^(2)/(b^(2))=1`

Here, `2a=8rArra=4`
and b=2
`:.(x^(2))/(16)+(y^(2))/(4)=1`
`rArr" "x^(2)+4y^(2)=16`
Given that, `AP=1.5rArrOP=OA-AP=4-1.5=2.5m`
Let PM = k
`:.` Coordinates of `M-=(2.5,k)`
Put in equation (1)
`(2.5)^(2)+4k^(2)=16`
`rArr" "4k^(2)=16-6.25=9.75`
`rArr" "k^(2)=(9.75)/(4)=(39)/(16)`
`rArr" "k=(sqrt(39))/(4)=1.56m`