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Exercise 5.5 · Q6

Q.Cross section of a Nuclear cooling tower is in the shape of a hyperbola with equation x2302−y2442=1\dfrac{x^2}{30^2}-\dfrac{y^2}{44^2}=1. The tower is 150 m150\,\text m tall and the distance from the top of the tower to the centre of the hyperbola is half the distance from the base of the tower to the centre of the hyperbola. Find the diameter of the top and base of the tower.

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Translate 'top distance is half the base distance' and 'total height 150150' into two numbers tt (top) and 2t2t (base) that add to 150150, then substitute each as yy into the hyperbola equation to get the corresponding xx (half-diameter) at that height.

Step 1. Find the two distances from the centre. Let the top-to-centre distance be tt and the base-to-centre distance be 2t2t (top is half the base, as stated). Since the tower's total height spans both distances (centre lying between top and base):

t+2t=150⇒3t=150⇒t=50t+2t=150 \Rightarrow 3t=150 \Rightarrow t=50.

So top distance =50 m=50\,\text m, base distance =100 m=100\,\text m.

Step 2. Diameter at the top (y=50y=50).

x2900−25001936=1⇒x2900=1+25001936≈1+1.2913=2.2913\dfrac{x^2}{900}-\dfrac{2500}{1936}=1 \Rightarrow \dfrac{x^2}{900}=1+\dfrac{2500}{1936}\approx1+1.2913=2.2913

⇒x2≈900(2.2913)≈2062.2⇒x≈45.41\Rightarrow x^2\approx900(2.2913)\approx2062.2 \Rightarrow x\approx45.41. …

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