If 2 x2-3 x y +y2+x+2 y-8=0, then (d y/d x) is equal to

Question

If 2 x2-3 x y +y2+x+2 y-8=0, then (d y/d x) is equal to (A) (3 y-4 x-1/2 y-3 x+2) (B) (3 y+4 x+1/2 y+3 x+2) (C) (3 y-4 x+1/2 y-3 x-2) (D) (3 y-4 x+1/2 y+3 x+2)

Tardigrade Admin · Accepted Answer

We have, 2 x2-3 x y+ y2+x+2 y-8=0 On differentiating w.r.t. x, we get 4 x-3 x (d y/d x)-3 y+2 y (d y/d x)+1+(2 d y/d x)-0=0 ⇒ 4 x+(-3 x+2 y+2) (d y/d x)-3 y+1=0 ⇒ (d y/d x)=(3 y-4 x-1/2 y-3 x+2)

Q. If $2 x^{2} - 3 x y + y^{2} + x + 2 y - 8 = 0$ , then $\frac{d y}{d x}$ is equal to

$\frac{3 y - 4 x - 1}{2 y - 3 x + 2}$

$\frac{3 y + 4 x + 1}{2 y + 3 x + 2}$

$\frac{3 y - 4 x + 1}{2 y - 3 x - 2}$

$\frac{3 y - 4 x + 1}{2 y + 3 x + 2}$

Q. If 2x2−3xy+y2+x+2y−8=0, then dxdy​ is equal to

2y−3x+23y−4x−1​

2y+3x+23y+4x+1​

2y−3x−23y−4x+1​

2y+3x+23y−4x+1​

We have, 2x2−3xy+y2+x+2y−8=0 On differentiating w.r.t. x, we get 4x−3xdxdy​−3y+2ydxdy​+1+dx2dy​−0=0 ⇒4x+(−3x+2y+2)dxdy​−3y+1=0 ⇒dxdy​=2y−3x+23y−4x−1​

Q. If $2 x^{2} - 3 x y + y^{2} + x + 2 y - 8 = 0$ , then $\frac{d y}{d x}$ is equal to

$\frac{3 y - 4 x - 1}{2 y - 3 x + 2}$

$\frac{3 y + 4 x + 1}{2 y + 3 x + 2}$

$\frac{3 y - 4 x + 1}{2 y - 3 x - 2}$

$\frac{3 y - 4 x + 1}{2 y + 3 x + 2}$