Answer to Question #163330 in Calculus for sars

Question

Write the equation of each conic section with the given properties:

A hyperbola with foci at (15,0) and (-15,0) and asymptotes y = 3/4x and y = -3/4x

Answer 1

Solution:

$\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$

The info about the asymptotes indicates that $\frac{b}{a}$ = $\frac{3}{4}$.

The distance from either focus to either asymptote is "b,"

so I examine the line through (15,0), perpendicular to y = ($\frac{3}{4}$)x.

Its equation will be y = -($\frac{3}{4}$)(x - 15) = 20 - ($\frac{4}{3}$)x. It intersects y = ($\frac{3}{4}$ )x where 20 = ($\frac{25}{12}$)x, or x = $\frac{48}{5}$ , y = $\frac{36}{5}$ .

The distance from that point to (15,0) is

sqrt[($\frac{27}{5}$)² + ($\frac{36}{5}$)²] = 3*$\sqrt{(\frac{3}{5})^2+(\frac{4}{5})^2}=3$.

So the equation of the hyperbola is

$\dfrac{x^2}{4}-\dfrac{y^2}{3}=1$ .