Built to get you above 85% in Methods.

Step 1

Let the tangent touch the parabola at the point \((p,\,p^2)\).

We don’t know the tangent yet, so we describe it by where it touches. Every point on \(y=x^2\) has the form \((p,p^2)\), so one unknown \(p\) covers every possible tangent.

Step 2

\(\dfrac{dy}{dx}=2x\), so at \(x=p\) the gradient of the tangent is \(2p\).

The derivative gives the gradient of the curve at any point, and the tangent shares that gradient.

Step 3

The same line passes through \(P(1,-3)\) and \((p,p^2)\), so its gradient is \(\dfrac{\text{rise}}{\text{run}}=\dfrac{p^2-(-3)}{p-1}\).

Gradient between two points: difference in \(y\) over difference in \(x\). Keep the \(-(-3)\) visible for now.

Step 4

\(p^2-(-3)=p^2+3\), so the gradient is \(\dfrac{p^2+3}{p-1}\).

Subtracting a negative is adding. A separate line, because this sign is where marks are lost.

Step 5

Both expressions describe the gradient of the same line, so \(\dfrac{p^2+3}{p-1}=2p\).

This is the key idea: one gradient, written two ways, gives an equation in \(p\).

Step 6

Multiply both sides by \((p-1)\): \(p^2+3=2p(p-1)\).

Clearing the fraction. Allowed because \(p\ne1\): a contact point at \(x=1\) would sit directly above \(P\), and a vertical line is not a tangent to this parabola.

Step 7

Expand the right-hand side: \(2p(p-1)=2p^2-2p\), so \(p^2+3=2p^2-2p\).

Multiply each term inside the bracket by \(2p\).

Step 8

Subtract \(p^2+3\) from both sides: \(0=2p^2-2p-p^2-3\).

Move everything to one side so the right-hand side is a quadratic equal to zero.

Step 9

Collect like terms: \(2p^2-p^2=p^2\), so \(p^2-2p-3=0\).

Standard form \(ap^2+bp+c=0\) with \(a=1,\ b=-2,\ c=-3\).

Step 10

Factorise: two numbers that multiply to \(-3\) and add to \(-2\) are \(-3\) and \(+1\), so \((p-3)(p+1)=0\).

Check by expanding: \(p^2+p-3p-3=p^2-2p-3\).

Step 11

So \(p-3=0\) or \(p+1=0\): \(p=3\) or \(p=-1\).

Two values of \(p\), so two points of contact, so two tangents.

Step 12

For \(p=3\): contact point \((3,\,3^2)=(3,9)\), gradient \(2(3)=6\).

Substitute back to find where this tangent touches and how steep it is.

Step 13

Point-gradient form: \(y-9=6(x-3)\). Expand: \(y-9=6x-18\). Add 9: \(y=6x-9\).

Each algebraic move on its own line so nothing is hidden.

Step 14

For \(p=-1\): contact point \((-1,\,(-1)^2)=(-1,1)\), gradient \(2(-1)=-2\).

Careful with \((-1)^2=+1\).

Step 15

\(y-1=-2(x-(-1))=-2(x+1)\). Expand: \(y-1=-2x-2\). Add 1: \(y=-2x-1\).

The double negative inside the bracket becomes \(x+1\).

Step 16

Check both lines pass through \(P(1,-3)\): \(6(1)-9=-3\) and \(-2(1)-1=-3\).

A one-line check that catches most algebra slips. Both correct.

Step 17

The picture: \(P\) sits below the parabola, and the two tangents leave it like a pair of sight-lines, each touching the curve at exactly one point.

This is the diagram in the hero above — the algebra you just did, drawn.

Step 18

Now replace \(-3\) by a general \(q\), so \(P\) is \((1,q)\). The gradient from \(P\) is \(\dfrac{p^2-q}{p-1}\), and equating to \(2p\) gives \(p^2-q=2p^2-2p\).

The same steps as before with \(q\) carried along.

Step 19

Rearrange: \(0=2p^2-2p-p^2+q\), so \(p^2-2p+q=0\).

A quadratic in \(p\) again, now with \(c=q\).

Step 20

How many solutions does a quadratic \(ap^2+bp+c=0\) have? The discriminant \(\Delta=b^2-4ac\) decides: \(\Delta\gt0\) gives two, \(\Delta=0\) gives one, \(\Delta\lt0\) gives none.

The discriminant is the part under the square root in the quadratic formula \(p=\dfrac{-b\pm\sqrt{\Delta}}{2a}\). A negative number has no real square root, so no real solutions.

Step 21

Here \(a=1,\ b=-2,\ c=q\): \(\Delta=(-2)^2-4(1)(q)=4-4q\).

Substitute carefully: \((-2)^2=+4\).

Step 22

\(\Delta\gt0\) when \(4-4q\gt0\), that is \(q\lt1\). \(\Delta=0\) when \(q=1\). \(\Delta\lt0\) when \(q\gt1\).

Solve each inequality: subtract 4, divide by \(-4\) and reverse the inequality sign.

Step 23

At \(x=1\) the parabola has height \(1^2=1\). So \(q\lt1\) means \(P\) is below the curve: two tangents. \(q=1\) means \(P\) is on the curve: one tangent. \(q\gt1\) means \(P\) is above the curve, inside the bowl: no tangent can reach it.

The algebra has become geometry. From inside the bowl you cannot see the curve edge-on, and the discriminant says exactly that. This final step, connecting the sign of \(\Delta\) to a picture, is what the top students do and what the harder questions on this site train.