Decoding Hyperbolas: Finding the Equation with Given Center and Points

In this blog post, we'll unravel the mystery of hyperbolas by exploring how to find their equations when the center and two specific points are provided. By diving into the world of algebraic curves, we'll decipher the unique properties of hyperbolas and learn the step-by-step process to determine their equations accurately. Join us on this mathematical journey as we unravel the secrets of hyperbolic geometry!

Step 1: Formulate the Equation with Center and Transverse Axis

Given center \((h, k) = (2, -4)\) and the standard equation for a hyperbola with a horizontal transverse axis:

\(\left(\frac{{(x - 2)^2}}{{a^2}}\right) - \left(\frac{{(y + 4)^2}}{{b^2}}\right) = 1 \quad \text{(equation 1)}\)

Step 2: Substitute the Point \((-3, -4)\) to Solve for \(a\)

Substituting \((-3, -4)\) into equation 1:

\(\left(\frac{{((-3) - 2)^2}}{{a^2}}\right) - \left(\frac{{((-4) + 4)^2}}{{b^2}}\right) = 1\)

Simplifying, we get:

\(a^2 = 25 \implies a = 5\)

Step 3: Substitute the Point \((14, 0)\) to Solve for \(b\)

Substituting \((14, 0)\) into equation 1:

\(\left(\frac{{(12)^2}}{{a^2}}\right) - \left(\frac{{(4)^2}}{{b^2}}\right) = 1\)

Solving for \(b^2\) using a=5:

\(\frac{{144}}{{25}} - 1 = \frac{{16}}{{b^2}} \implies \frac{{119}}{{25}} = \frac{{16}}{{b^2}} \implies b^2 = \left(\frac{{400}}{{119}}\right) = \left(\frac{{20\sqrt{{119}}}}{{119}}\right)^2\)

Step 4: Write the Final Equation of the Hyperbola

Substituting the values of \(a\) and \(b^2\) into equation 1:

\(\left(\frac{{(x - 2)^2}}{{25}}\right) - \left(\frac{{(y + 4)^2}}{{\left(\frac{{20\sqrt{{119}}}}{{119}}\right)^2}}\right) = 1\)

Conclusion:

Therefore, the equation of the hyperbola is:

\(\left(\frac{{(x - 2)^2}}{{25}}\right) - \left(\frac{{(y + 4)^2}}{{\left(\frac{{20\sqrt{{119}}}}{{119}}\right)^2}}\right) = 1\)

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