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Questions  >>  CBSE XI  >>  Math  >>  Conic Sections
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Q)

An arch is in the form of a parabola with its axis vertical. The arch is 10 m high and 5 m wide at the base. How wide is it 2 m from the vertex of the parabola?

$\begin {array} {1 1} (A)\;1.118\: m & \quad (B)\;2.23\: m \\ (C)\;1\: m & \quad (D)\;2\: m \end {array}$

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A)
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  • If the focus is on the positive y axis, then the parabola is open upwards, Hence the equation is $ x^2=4ay$
Step 1 :
Let us take the origin of the coordinates plane as the vertex of the arch.
Also the vertical axis is along the positive y - axis.
This is as shown in the adjacent figure.
The equation is of the form $x^2=4ay$
Step 2 :
The equation passes through the point $ \bigg( \large\frac{5}{2}$$, 10 \bigg)$
$ \bigg( \large\frac{5}{2} \bigg)^2$$ = 4a(10)$
$ \Rightarrow \large\frac{25}{4}$$ = 40a$
$ \therefore a = \large\frac{25}{160}$$ = \large\frac{5}{32}$
Hence the arch is in the form of a parabola whose equation is
Step 3 :
$x^2= 4 \times \large\frac{5}{32}$$ \times y$
$ = x^2 = \large\frac{5}{8}$$y$
When $ y = 2m$
$x^2=\large\frac{5}{8}$$(2)$
$ \Rightarrow x^2 = \large\frac{5}{4}$
$ \therefore x \pm \sqrt{\large\frac{5}{4}} m$
$ \therefore AB = 2 \times \sqrt{\large\frac{5}{4}} m$
$ = 2 \times 1.118 m$
$ = 2.23 m$ approximately.
$ \therefore $ When the arch is 2m from the vertex of the parabola, its width is approximately 2.23 m
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