Myanmar Matriculation 2019 Math No 14

(Myanmar Examboard Matriculation, 2019 No 14 (b))

Find the normals to the curve $x y+2 x-y=0$ that are parallel to the line  $2 x+y=0 .$  [5 Marks]

Solution:  

$$\begin{eqnarray*}x y+2 x-y&=&0 \\y(x-1) &=&-2 x \\y &=&\dfrac{-2 x}{x-1} =-2-\frac{2}{x-1}=-2-2(x-1)^{-1}\\\dfrac{d y}{d x} &=&0+2(x-1)^{-2} =\dfrac{2}{(x-1)^{2}}\end{eqnarray*}$$

 Since gradient of normal is $-2,\dfrac{d y}{d x}=\dfrac{1}{2}.$

Thus,

$$\begin{eqnarray*}\frac{2}{(x-1)^{2}} &=&\frac{1}{2} \\(x-1)^{2} &=&4 \\x-1 &=&\pm 2 \\x &=&-1 \text { or } x=3 . \end{eqnarray*}$$

If  $x=-1, y=-1.$ 

At $(-1,-1)$ the normal equation  is $y+1=-2(x+1) .$  

If  $x=3, y=-3.$

At $(3,-3)$ the normal equation is $y+3=-2(x-3).$