y = -2 x² ⇒⇒⇒ x² = -y/2 →(1)33 x² + y² = 27 → (2) By substitution from (1) at (2) with the value of x² ∴ 33 (-y/2) + y² = 27 ∴ y² - 16.5 y - 27 = 0 a = 1 , b = -16.5 , c = -27 ∴ [tex]y= \frac{-b \pm \sqrt{b^2-4ac} }{2a} = \frac{16.5 \pm \sqrt{(-16.5)^2-4*1*(-27)} }{2*1}[/tex] ∴ y = 18 or y = -3/2 By substitution from at (1) with the value of y for y = 18 ⇒⇒⇒ x² = -18/2 = -9 (unacceptable) for y = -3/2 ⇒⇒⇒ x² = -(-3/2)/2 = 3/4 ∴ [tex]x= \pm \sqrt{ \frac{3}{4} } = \pm \frac{ \sqrt{3} }{2}
[/tex] The correct options are 2 , 7 Solution of the system of equations is [tex]( \frac{ \sqrt{3} }{2} , \frac{-3}{2} ) [/tex] and [tex]( -\frac{\sqrt{3} }{2} , \frac{-3}{2} ) [/tex] ================================== The second problem: The general equation of the hyperbole is [tex] \frac{x^2}{a^2} - \frac{y^2}{b^2} =1
Transverse axis is horizontal
The equation if the asymptotes are y = \pm \frac{b}{a}x [/tex] For the given equation: [tex] \frac{x^2}{225} - \frac{y^2}{36} = 1[/tex] a² = 225 ⇒⇒⇒ a = √225 = 15 b² = 36 ⇒⇒⇒ b = √36 = 6 ∴ the slope of the asymptotes = b/a and -b/a b/a = 6/15 = 2/5 -b/a = -6/15 = -2/5 ∴ m = 2/5 and m = -2/5
