Find an equation in standard form for the hyperbola with vertices at (0, ±10) and asymptotes at y = ±five divided by four times x..

Answer:The equation of the hyperbola in standard form is[tex]\frac{y^{2}}{100}-\frac{4x^{2}}{625}=1[/tex]Step-by-step explanation:* We will take about the standard form equation of the hyperbola- If the given coordinates of the vertices (0 , a) and (0 , -a) ∴ The transverse axis is the y-axis. (because x = 0)- If the given asymptotes at y = ± (b/a) x∴  Use the standard form  ⇒ y²/a² - x²/b² = 1* Lets use this to solve our problem∵ The vertices are (0 , 10) and (0 , -10)∴ a = ±10∴ a² = 100∵ The asymptotes at y = ± 5/4 x∴ ± 5/4 = ± b/a∵ a = ± 10∴ ± 5/4 = ± b/10 ⇒ using cross multiplication∴± (4b) = ± (5 × 10) = ± 50 ⇒ divide both sides by 4∴ b = ± 25/2∴ b² = 625/4* Now Lets write the equation* y²/100 - x²/(625/4) = 1∵ x² ÷ 625/4 = x² × 4/625 = (4x²/625)∴ y²/100 - 4x²/625 = 1* The equation of the hyperbola in standard form is   [tex]\frac{y^{2}}{100}-\frac{4x^{2}}{625}=1[/tex]