Question

Find an equation in standard form for the hyperbola with vertices at (0,+-6) and asymptote at Y= +- 3/7x

Answer 1

Answer:

$\frac{y^2}{36}-\frac{x^2}{196}=1$

Step-by-step explanation:

From the question we are told that:

Vertices of hyperbola at $(0,\pm6)$

Asymptotes at $Y= \pm 3/7x$

Generally the expression for Vertices of hyperbola is mathematically given by

 $(0,\pm a)$

Therefore

 $(0,\pm a)=(0,\pm6)$

Comparing variables

 $a=6$

Generally the expression for Asymptotes of hyperbola is mathematically given by

 $y=\pm \frac{a}{b}x$

Therefore

 $\pm \frac{a}{b}x= \pm \frac{3}{7}x$

Therefore

 $\frac{6}{b}= \frac{3}{7}$

 $b=\frac{42}{3}$

 $b=14$

Generally the equation for Hyperbola of hyperbola is mathematically given by

 $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$

 $\frac{y^2}{6^2}-\frac{x^2}{14^2}=1$

 $\frac{y^2}{36}-\frac{x^2}{196}=1$