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Exercise 12.3 · Q2

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

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✓ Free question

Model the arc as a downward-opening parabola with vertex at the top; use the base data to fix the parabola, then find the width 2 m below the vertex.

Parabola with vertex at origin, axis vertical, opening downward (measuring yy as distance dropped from the vertex): x2=4ayx^2=4ay.

  1. Place the vertex (top of the arc) at the origin, with yy measured as the vertical drop from the vertex and xx the horizontal half-width at that drop.
  2. At the base, the drop is y=10y=10 m (the full height of the arc) and the half-width is x=52=2.5x=\dfrac{5}{2}=2.5 m (base width 5 m).
  3. Substitute (2.5,10)(2.5,10) into x2=4ayx^2=4ay: (2.5)2=4a(10)⇒6.25=40a⇒a=6.2540=532(2.5)^2=4a(10)\Rightarrow6.25=40a\Rightarrow a=\dfrac{6.25}{40}=\dfrac{5}{32}.
  4. Parabola: x2=4(532)y=58yx^2=4\left(\dfrac{5}{32}\right)y=\dfrac{5}{8}y.
  5. At y=2y=2 m from the vertex: x2=58(2)=54⇒x=54=52x^2=\dfrac{5}{8}(2)=\dfrac{5}{4}\Rightarrow x=\sqrt{\dfrac54}=\dfrac{\sqrt5}{2} m.
  6. Full width at that level =2x=2×52=5≈2.236=2x=2\times\dfrac{\sqrt5}{2}=\sqrt5\approx2.236 m.
  7. Self-check: at y=10y=10, x2=58(10)=6.25⇒x=2.5x^2=\dfrac58(10)=6.25\Rightarrow x=2.5 ✓, matching the base half-width.
✓Final answer

Width of the arc 2 m from the vertex =5≈2.236=\sqrt5\approx2.236 m.

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