Q.Find the equation of the normal to the curve $x^2 = 4y$ which passes through the point $(-1, 4)$.

The normal at a point $(x_1, y_1)$ on $x^2 = 4y$ has slope $-\dfrac{2}{x_1}$, with $y_1 = \dfrac{x_1^2}{4}$.

Step 1: Impose that the normal passes through $(-1,4)$:

Step 2: Factor: $(x_1-2)(x_1^2+2x_1-4)=0$, giving $x_1 = 2,\ -1+\sqrt5,\ -1-\sqrt5$ — all three real, so all three give valid normals.

Step 3: Substituting each root back into $y-y_1=-\frac{2}{x_1}(x-x_1)$ and simplifying:

• $x_1=2$: $x+y=3$ …

Setting up the normal at a general point

The curve is $x^2 = 4y$. Differentiating implicitly with respect to $x$:

So at a point $(x_1, y_1)$ on the parabola, the tangent slope is $\dfrac{x_1}{2}$, and since the normal is perpendicular to the tangent, its slope is

Since $(x_1, y_1)$ lies on $x^2 = 4y$, we also have $y_1 = \dfrac{x_1^2}{4}$.

The equation of the normal at $(x_1, y_1)$ is therefore

Imposing the condition that it passes through $(-1, 4)$

First note $(-1,4)$ does not itself lie on the curve, since $(-1)^2 = 1 \neq 4(4) = 16$ — so we are looking for the point(s) of contact $(x_1, y_1)$ whose normal happens to pass through this external point.

Substituting $x = -1,\ y = 4$ into the normal's equation:

Multiplying both sides by $4x_1$:

Solving the cubic

Testing $x_1 = 2$: $2^3 - 8(2) + 8 = 8 - 16 + 8 = 0$ ✓. So $(x_1 - 2)$ is a factor:

Solving $x_1^2 + 2x_1 - 4 = 0$ by the quadratic formula:

So all three roots are $x_1 = 2,\ -1+\sqrt5,\ -1-\sqrt5$ — all real, so all three give genuine normal lines (this is why the question has three answers, not one).

Writing out all three normal lines

Root $x_1 = 2$: $y_1 = \dfrac{4}{4} = 1$, slope $= -\dfrac{2}{2} = -1$.

$y - 1 = -1(x - 2) \quad\Rightarrow\quad x + y = 3.$ …

Method: Finding a Normal Line to a Curve That Passes Through a Given External Point

Use this method whenever you're asked for the normal to a curve that passes through a point which is not stated to lie on the curve itself — you must first find which point(s) of contact on the curve produce a normal through the given point.

Steps

Step 1: Confirm the given point is not automatically the point of contact

Substitute the given point's coordinates into the curve's equation. If it doesn't satisfy the equation, the point of contact is unknown and must be found — this is the key difference from a "normal at a given point" problem.

Step 2: Set up the normal's equation at a general point of the curve

Let $(x_1, y_1)$ be the (unknown) point of contact on the curve. Differentiate the curve implicitly to get the tangent slope in terms of $x_1$, then take the negative reciprocal for the normal slope. Write the normal's equation in point-slope form using $(x_1, y_1)$, and use the curve's own equation to express $y_1$ in terms of $x_1$ (so only one unknown, $x_1$, remains).

Step 3: Force the normal through the given external point

Substitute the given point's coordinates into the normal equation from Step 2. This produces a single equation in $x_1$ alone — typically a polynomial, since the slope term introduces a variable in the denominator that clears when you multiply through.

Step 4: Solve the resulting equation for $x_1$ …

Common Mistakes

Mistake 1: Assuming the given point $(-1,4)$ lies on the parabola

$(-1,4)$ does not satisfy $x^2=4y$ (since $(-1)^2=1\ne 4(4)=16$), so it cannot be used directly as the point of contact $(x_1,y_1)$. The point of contact must be an unknown $(x_1, x_1^2/4)$ whose normal happens to pass through $(-1,4)$ — treating $(-1,4)$ itself as the point on the curve skips this step and gives a wrong slope.

Mistake 2: Stopping after finding one root of the cubic and not checking the others …