Question

If the tangents at two points of a parabola are at right angles, then show that they intersect at a point on the directrix?

Answer 1

Let us take the standard equation form of a parabola which has the vertex at the origin and the directrix is the equation `x=-a` is:

`y^2=4ax`

Let us consider general coordinates `P(ap^2, 2ap)` and `Q(aq^2, 2aq)` that satisfy the required condition, and `p,q != 0`

The gradient of the tangent to a curve at any particular point is given by the derivative of the curve at that point.

Differentiating implicitly the parabola equation we get:

`2y dy/dx = 4a => dy/dx = (2a)/y`

So, the gradient of the tangent at `P(ap^2, 2ap)` is given by:

`m_P = (2a)/(2ap) = 1/p`

So the tangent passes through `P(ap^2, 2ap)` and has gradient `m_P=1/p`, so using the point/slope form `y-y_1=m(x-x_1)` the equation of that tangent is;

`y-2ap = 1/p(x - ap^2)`
`:. py - 2ap^2 = x - ap^2`
`:. py = x + ap^2`
`:. y = (x + ap^2 )/p` ..... [A]

Similarly for `Q(aq^2, 2aq)` we have:

`m_Q = 1/q`

And the tangent equation:

`qy = x + aq^2`
`:. y = (x + aq^2)/q` ..... [B]

The tangent equations at `P` and `Q` will therefore meet when we have a simultaneous solution of the equations [A] and [B]: ie if:

`(x + ap^2)/p = (x + aq^2)/q`
`:. q(x + ap^2) = p(x + aq^2)`
`:. qx + ap^2q = px + apq^2`

`:. px - qx = ap^2q - apq^2`
`:. (p-q)x = apq(p-q)`
`:. x = apq` ..... [C]

Now we also know that the tangents at `P` and `Q` are perpendicular, so the product of the tangents at `P` and `Q` are equal to `-1`, hence:

`m_P * m_Q = -1`
`:. 1/p * 1/q = -1`
`:. pq = -1`

Substituting this condition into Eq [C] we get:

`x = a*(-1)`
`x = -a`

Which, as we initially stated, is the equation of directrix

Hence the point of intersect lies on the directrix. QED