A vertical tangent forms a square · source reference

Printed question

The equation of a curve is given by $4x^3+3x^2y=y^3-2$. Find $\frac{dy}{dx}$ in terms of $x$ and $y$, simplifying your answer. [2]

The curve meets the line $y=-x$ at point $P$. Find (i) the coordinates of $P$ [2] and (ii) the equation of the tangent at $P$. [2]

The tangent to the curve at $P$ cuts the $x$-axis at $R$ and the normal to the curve at $P$ cuts the $y$-axis at $Q$. If $O$ denotes the origin, use the results above to give a geometrical description of the quadrilateral $OQPR$. [1]

[2012/NYJC/Prelim/I/3]

Printed working

$4x^3+3x^2y=y^3-2$. Differentiating with respect to $x$:

$$12x^2+6xy+3x^2\frac{dy}{dx}=3y^2\frac{dy}{dx},$$ $$(3x^2-3y^2)\frac{dy}{dx}=-12x^2-6xy,$$ $$\frac{dy}{dx}=\frac{4x^2+2xy}{y^2-x^2}=\frac{2x(2x+y)}{(y-x)(y+x)}.$$

(i) Curve meets $y=-x$ when $4x^3+3x^2(-x)=-x^3-2$, so $2x^3=-2$, $x=-1$, $y=1$. Thus $P=(-1,1)$.

(ii) At $(-1,1)$, $dy/dx$ is undefined. Thus tangent at $P$ is $x=-1$. The normal is $y=1$; $Q=(0,1)$ and $R=(-1,0)$, giving the square $OQPR$.

Teacher notes

Group all terms with $dy/dx$ on one side of the equation, then take out the common factor $dy/dx$.

Since $dy/dx$ is undefined here (the numerator is nonzero), its gradient is infinite, so the tangent at $P$ is a vertical line.

Answer key

$\frac{dy}{dx}=\frac{2x(2x+y)}{(y-x)(y+x)}$; (i) $(-1,1)$; (ii) $x=-1$; $OQPR$ is a square.

Native diagrams