Question

What are the answers to these 2 question 4&5 PLZ HELP

Answer 1

The first problem:

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
∴ $y= \frac{-b \pm \sqrt{b^2-4ac} }{2a} = \frac{16.5 \pm \sqrt{(-16.5)^2-4*1*(-27)} }{2*1}$
∴ 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
∴ $x= \pm \sqrt{ \frac{3}{4} } = \pm \frac{ \sqrt{3} }{2}$
The correct options are 2 , 7
Solution of the system of equations is 
$( \frac{ \sqrt{3} }{2} , \frac{-3}{2} )$
and   
$( -\frac{\sqrt{3} }{2} , \frac{-3}{2} )$
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The second problem:
The general equation of the hyperbole is 
$\frac{x^2}{a^2} - \frac{y^2}{b^2} =1 Transverse axis is horizontal The equation if the asymptotes are y = \pm \frac{b}{a}x$
For the given equation:
$\frac{x^2}{225} - \frac{y^2}{36} = 1$
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