At a water fountain, water attains a maximum height of $4\,\text m$ at horizontal distance of $0.5\,\text m$ from its origin. If the path of water is a parabola, find the height of water at a horizontal distance of $0.75\,\text m$ from the point of origin.

At a water fountain, water attains a maximum height of 4text m at horizontal distance of 0.5text m from its or

Exercise 5.5 · Q3

4\,\text m

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Concept understanding — Real-life Applications of Conics

Each conic's reflective (or orbital, or structural) property explains why it is the shape engineers and nature actually choose:

Parabola. A ray from the focus reflects parallel to the axis, and conversely — headlight bulbs and dish receivers sit at the focus (car headlights, satellite dishes, field microphones, solar cookers). Parabolic arches (river bridges, the Eiffel Tower) are chosen additionally for structural stability.

Ellipse. By Kepler's laws, planets orbit the Sun on ellipses with the Sun at one focus (comets like Halley's, $e\approx0.97$ , too). A ray from one focus reflects to the other focus — whispering-gallery ceilings, and lithotripters (a shock wave from one focus is focused onto a kidney stone at the other, avoiding surgery). Earth itself is an oblate spheroid, an ellipse revolved about its minor axis.

Hyperbola. Some comets travel on hyperbolic (non-returning) paths. A ray directed at one focus reflects to the other — used to locate a ship or aircraft from time-difference-of-arrival signals at two stations (one branch of a hyperbola with the stations as foci), and in two-mirror telescopes, where a parabolic primary and hyperbolic secondary share a focus so light is relayed to the hyperbola's second focus behind the primary mirror.

Worked-example pattern. Real-life problems translate physical measurements into $a$ , $b$ , $c$ (or the vertex/focus): a bridge's arch height and width become the parabola's $a$ ; a tunnel's width and peak height become the ellipse's $a,b$ ; a signal time-difference becomes $2a$ for a hyperbola (with the stations' separation giving $2c$ ). Once $a^2,b^2$ (or $a$ alone, for a parabola) are pinned down, the specific question — a height at a given point, an eccentricity, a diameter — is read off the standard-form equation exactly as in the algebra sections.

Tip

The very first step in every applied problem is choosing WHERE to put the origin (vertex, or centre, or the midpoint of two foci/stations) — pick it to match the curve's standard form as closely as possible, and the rest of the problem becomes ordinary substitution.

The maximum height point of a parabolic path is its vertex; the water's starting point (the origin) is a second known point, used to find $a$ .

Step 1. Set up. Vertex $(0.5,4)$ , opens down: $(x-0.5)^2=-4a(y-4)$ .

Step 2. Use the origin $(0,0)$ to find $a$ .

$(0-0.5)^2=-4a(0-4) \Rightarrow 0.25=16a \Rightarrow a=\dfrac{0.25}{16}=\dfrac1{64}$ .

Step 3. Find the height at $x=0.75$ .

$(0.75-0.5)^2=-4\left(\dfrac1{64}\right)(y-4) \Rightarrow (0.25)^2=-\dfrac1{16}(y-4) \Rightarrow 0.0625=-\dfrac1{16}(y-4)$

$\Rightarrow y-4=-0.0625\times16=-1 \Rightarrow y=3$ .

✓Final answer

3\,\text m

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2019Set ANNUAL2 marks
$\dfrac{\pi}{3}$
›Reveal solutionSolution
Figure — A rightward-opening parabola with vertex V at the origin and axis along the positive x-axis; mark th

Using the focus-polar equation of a parabola $r=\dfrac{l}{1-\cos\theta}$ with the given $(r,\theta)$ point gives the semi-latus rectum, hence the orbit's Cartesian equation $y^2=80x$ and a closest approach of 20 million km.

Model the orbit as a parabola with the sun $S$ at the focus. Take $S$ as pole and the axis of the parabola as the initial line, oriented so the curve opens away from the sun (comet arrives from far away and swings past the sun).

The polar equation of a parabola referred to its focus is $r=\dfrac{l}{1-\cos\theta}$ , where $l$ is the semi-latus rectum ( $l=2a$ for $y^2=4ax$ ).

Substitute the given data $r=80$ , $\theta=\pi/3$ : $\cos(\pi/3)=\tfrac12$ , so $80=\dfrac{l}{1-\tfrac12}=\dfrac{l}{\tfrac12}=2l$ .

Hence $l=40$ (million km), so $a=\dfrac{l}{2}=20$ .

With the vertex $V$ taken as the Cartesian origin and the axis along the x-axis, the orbit is $y^2=4ax=80x$ .

The focus (sun) is at $(a,0)=(20,0)$ , so the vertex — the point of the orbit nearest the sun — is at distance $a=20$ million km from the sun; this is the comet's closest approach.

Sketch: a rightward-opening parabola with vertex $V$ at the origin, axis $VS$ along the positive x-axis, sun $S$ marked at $(20,0)$ ; the comet's point $P$ on the curve at $80$ million km from $S$ , with $\angle PSV$ -extended $=\pi/3$ .

✓Final answer
$y^2=80x$

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Q.At a water fountain, water attains a maximum height of 4 m4\,\text m4m at horizontal distance of 0.5 m0.5\,\text m0.5m from its origin. If the path of water is a parabola, find the height of water at a horizontal distance of 0.75 m0.75\,\text m0.75m from the point of origin.

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Q.At a water fountain, water attains a maximum height of $4\,\text m$ at horizontal distance of $0.5\,\text m$ from its origin. If the path of water is a parabola, find the height of water at a horizontal distance of $0.75\,\text m$ from the point of origin.

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