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Exercises · Q21

Q.For the hyperbola x29−y216=1\dfrac{x^2}{9}-\dfrac{y^2}{16}=1, find aa, bb, and the eccentricity ee.

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Comparing x29−y216=1\dfrac{x^2}{9}-\dfrac{y^2}{16}=1 with the standard hyperbola form x2a2−y2b2=1\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1:

a2=9⇒a=3,b2=16⇒b=4a^2=9 \Rightarrow a=3, \qquad b^2=16 \Rightarrow b=4

For a hyperbola, c2=a2+b2c^2=a^2+b^2 (note the plus sign, unlike the ellipse):

c2=9+16=25⇒c=5c^2=9+16=25 \quad \Rightarrow \quad c=5

Eccentricity: e=ca=53≈1.667e=\dfrac{c}{a}=\dfrac53 \approx 1.667.

Figure 8 — Hyperbola x²/9 − y²/16 = 1 showing both branches with vertices at (3, 0) and (−3, 0), and foci at (5, 0) and (−5, 0)
Figure 8 — Hyperbola x²/9 − y²/16 = 1 showing both branches with vertices at (3, 0) and (−3, 0), and foci at (5, 0) and (−5, 0)
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