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4.   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.

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The equation of the semi-ellipse will be of the form 

\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\:,y>0

Now, According to the question,

the length of major axis = 2a = 8  \Rightarrow a=4

The length of the semimajor axis =2\Rightarrow b=2

Hence the equation will be,

\frac{x^2}{4^2}+\frac{y^2}{2^2}=1\:,y>0

\frac{x^2}{16}+\frac{y^2}{4}=1\:,y>0

Now, at point 1.5 cm from the end, the x coordinate is 4-1.5 = 2.5

So, the height at this point is 

\frac{(2.5)^2}{16}+\frac{y^2}{4}=1\Rightarrow y=\sqrt{4(1-\frac{2.5^2}{16})}

y\approx 1.56m

Hence the height of the required point is 1.56 m.

Posted by

Pankaj Sanodiya

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