Q.Find the points on the curve $9y^2 = x^3$, where the normal to the curve makes equal intercepts with both the axes. Also find the equation of the normals.

Concept: For a curve, the normal at a point makes equal intercepts on both axes when its slope is $\pm 1$ (since intercept form $x/a + y/a = 1$ gives slope $-1$).

Step 1 – Slope of tangent.

Differentiate $9y^2 = x^3$ implicitly:

$18y \frac{dy}{dx} = 3x^2 \quad\Rightarrow\quad \frac{dy}{dx} = \frac{x^2}{6y}$.

Step 2 – Slope of normal.

Slope of normal $= -\frac{1}{dy/dx} = -\frac{6y}{x^2}$.

For equal intercepts, the normal’s slope must be $\pm 1$ (line $x+y = c$ has slope $-1$; line $x-y = c$ has slope $1$).

So $-\frac{6y}{x^2} = \pm 1$.

Step 3 – Solve for points.

Case 1: $-\frac{6y}{x^2} = 1 \Rightarrow y = -\frac{x^2}{6}$. Substitute into $9y^2 = x^3$:

$9\left(\frac{x^4}{36}\right) = x^3 \Rightarrow \frac{x^4}{4} = x^3 \Rightarrow x^3(x-4)=0$.

$x=0$ gives $y=0$ (normal undefined here, slope infinite — discard). $x=4$ gives $y = -\frac{16}{6} = -\frac{8}{3}$. Point: $(4, -\frac{8}{3})$.

Case 2: $-\frac{6y}{x^2} = -1 \Rightarrow y = \frac{x^2}{6}$. Substitute: …

Solution

Differentiate $9y^2 = x^3$ implicitly:

Slope of the normal at $(x,y)$:

A line making equal intercepts on the axes has slope $\pm 1$, so $|m_N| = 1$, i.e. $6|y| = x^2$.

Find the points. From $6|y| = x^2$ we get $y^2 = \dfrac{x^4}{36}$. Substituting $y^2 = \dfrac{x^3}{9}$ from the curve:

Ignoring $x = 0$ (which gives the point $(0,0)$ where the normal is not defined in this sense), take $x = 4$. Then $9y^2 = 64$, so $y = \pm\dfrac{8}{3}$.

The required points are $\left(4,\ \tfrac{8}{3}\right)$ and $\left(4,\ -\tfrac{8}{3}\right)$.

Equations of the normals. …

Method: Finding Curve Points Where the Tangent/Normal Satisfies a Geometric Condition

This is the general approach for "find the point(s) on the curve where the tangent/normal is [parallel to a line / perpendicular to a line / makes equal intercepts / passes through a given external point]" — translate the geometric condition into an algebraic equation on the slope, then solve it together with the curve's own equation.

Steps

Step 1: Differentiate the curve to get $\frac{dy}{dx}$

If the curve is given implicitly (an equation in both $x$ and $y$), differentiate both sides with respect to $x$, treating $y$ as a function of $x$, and solve for $\frac{dy}{dx}$ in terms of $x$ and $y$.

Step 2: Write the slope of the line in question (tangent or normal)

The tangent's slope is $\frac{dy}{dx}$ itself. The normal's slope is the negative reciprocal:

Step 3: Translate the geometric condition into an equation …

Common Mistakes

Mistake 1: Considering only one sign of the slope

"Equal intercepts" is satisfied by BOTH $m=+1$ and $m=-1$ (the line could cross into different quadrant combinations) — solving only the $m=-1$ case and missing the $m=+1$ case (or vice versa) throws away half the valid points.

Mistake 2: Keeping the degenerate solution at the origin …