$\begin {array} {1 1} (A)\;1.56 \: m & \quad (B)\;3.12 \: m \\ (C)\;0.78 \: m & \quad (D)\;2.43 \: m \end {array}$

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- Equation of an ellipse whose major axis is along the x - axis is given by $ \large\frac{x^2}{a^2}$$+\large\frac{y^2}{b^2}$$=1$

Step 1 :

It is given that the height of the arc from the centre is 2m.

The width of the arc from the centre is 8 m

The length of the major axis is 8 m and length of the semi - minor axis is 2m.

Let the origin of the coordinates plane be the centre of the ellipse.

The major axis is along the x - axis.

The semi - ellipse is as shown in the fig.

Step 2 :

Since the major axis is along the x - axis, the equation of the semi ellipse is $ \large\frac{x^2}{a^2}$$+\large\frac{y^2}{b^2}$$=1$. Where a is the semi major axis.

$ \therefore 2a = 8$

$ \Rightarrow a = 4$ and b = 2

$ \therefore $ Equation of the semi ellipse is

$ \large\frac{x^2}{16}$$+\large\frac{y^2}{4}$$=1$

$ y \geq 0$

Step 3 :

Let us take a point B on x - axis.

Such that AB = 1.5 m

$AC \perp 0B$

$ \therefore $ 0A = 4-1.5

= 2.5 m

$ \therefore $ The x - coordinates of point C is 2.5 .

Substituting the value of x with 2.5 in the equation of semi ellipse we get,

$ \large\frac{(2.5)^2}{16}$$+ \large\frac{y^2}{4}$$=1$

(i.e) $\large\frac{y^2}{4}$$=1- \large\frac{6.25}{16}$

$ \therefore y^2 = 4 \bigg( \large\frac{16-6.25}{16} \bigg)$

$ = \large\frac{9.75}{4}$

$ \therefore y^2 = 2.4375$

$ \Rightarrow y = 1.56$ ( approx )

$ \therefore AC = 1.56 m$

Thus the length of the arc at a point 1.5 m from one end is approximately 1.56 m

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