Question

x/y= 2/3 than 2x2 +3y2 /2x2 -3y2?​

Answer 1

2/3

Step-by-step explanation:

$X^2+3xy+y^2=5. at (1,1)</p><p>2x+3y+3xdy/dx+2ydy/dx=0</p><p>dy/dx(3x+2y)=-(2x+3y)</p><p>dy/dx=-(2x+3y)÷(3x+2y)</p><p>dy/dx(1,1)=-(2+3)÷(3+2)</p><p>dy/dx=-1</p><p></p><p>eq of the tangent</p><p></p><p>(x-1)=-1(y-1)</p><p>(x-1)=-y+1</p><p>X+y=2</p><p></p><p></p><p>eq of normal</p><p>(x-1)=1(y-1)</p><p>x-1=y-1$

Answer 2

Answer:

x/y=2/3

x^2/y^2=4/9

2x^2+3y^2/2x^2-3y^2

divide numerator and denominator by y^2

(2x^2/y^2+3y^2/y^2)/(2x^2/y^2-3y^2/y^2)

=(2.4/9+3)/(2.4/9-3)

=(8/9+3)/(8/9-3)

=(8+27)/9/(8-27)/9

=35/ -19

= -35/19

= -1 16/19