Question
If the conic is a hyperbola and the origin lies on the hyperbola then its eccentricity is(a) $\frac{\sqrt{17}}{4}$(b) $\frac{\sqrt{17}}{3}$(c) $\frac{9}{\sqrt{17}}$(d) $\frac{16}{\sqrt{15}}$
Step 1
Since the origin lies on the hyperbola, the equation of the hyperbola can be written as: $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ Show more…
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