All solutions here are SUGGESTED. Mr. Teng will hold no liability for any errors. Comments are entirely personal opinions.

(i)
Differentiating the given equation by x,

\Rightarrow 2x - 2y \frac{dy}{dx} + 2x \frac{dy}{dx} + 2y = 0

(x-y)\frac{dy}{dx} = -y-x

\frac{dy}{dx} = \frac{x+y}{y-x}

(ii)
Tangent parallel to x-axis \Rightarrow \frac{dy}{dx} = 0

Then, y= -x

\Rightarrow x^2 - (-x)^2 + 2x (-x)+4=0

x^2 = 2

x = \pm \sqrt{2}

y = -\sqrt{2}, ~\sqrt{2}

Therefore, coordinates are (\sqrt{2}, -\sqrt{2}) and (-\sqrt{2}, \sqrt{2}).

KS Comments:

Quite straight forward with the implicit differentiations at (i). Students should understand what it means when tangents are parallel to x-axis. Lastly, since the question wants coordinates, students are expected to give in coordinates instead of merely the x and y values!

Leave a Reply