The curve C has equation xy=12. The tangents to the curve C at the distinct points P(p,12p) and Q(q,12q), where p and q are positive, intersect at T and the normals to C at these points intersect at N. Show that T is the point (2pqp+q,1p+q).
We have the points P(p,12p) and Q(q,12q) where, without loss of generality, 0<p<q as in the diagram. We can rearrange the equation of the curve to y=12x and then differentiate to obtain
dydx=−12x2.
At the point P, with coordinates (p,12p), the gradient is therefore −12p2. Thus the equation of the tangent through P is (y−12p)=−12p2(x−p), that is, y=1p−12p2x. Similarly, the equation of the tangent through Q is y=1q−12q2x.
The two tangents intersect when 1p−12p2x=1q−12q2x. i.e. when x(12q2−12p2)=1q−1p, We can simplify the left-hand bracket, using the difference of two squares, to give 12q2−12p2=12((1q)2−(1p)2)=12(1q−1p)(1q+1p)=12(1q−1p)(p+qpq). Then the equation involving the x-coordinate of T simplifies to 12x(1q−1p)(p+qpq)=(1q−1p) and so, dividing through by the bracketed expression on both sides (which is not zero since p≠q) and making x the subject, we see that x=2pqp+q. The y-coordinate of T is then y=1p−12p2×2pqp+q=1p(1−qp+q)=1p+q,as required.
In the case pq=12, find the coordinates of N.
It would be helpful to find the equations of the normals to the curve at P and Q.
We know that (gradient of the normal) × (gradient of the tangent) at any point on the curve is −1.
We can therefore use what we know about the gradients of the tangents to the curve at P and Q.
The gradients of the normals through P and Q are 2p2 and 2q2, respectively. The equation of the normal through P is therefore (y−12p)=2p2(x−p), which we can rearrange to give y=2p2x+12p−2p3. Similarly, the normal through Q has equation y=2q2x+12q−2q3.
The two normals intersect at N, whose x coordinate satisfies 2p2x+12p−2p3=2q2x+12q−2q3, that is, 2(p−q)(p+q)x=12(1q−1p)+2(p3−q3)=12(p−qpq)+2(p−q)(p2+pq+q2)and so x=1p+q(14pq+p2+pq+q2).(p≠q)
Using pq=12, this becomes x=1p+q((pq)2pq+p2+pq+q2)=1p+q(p2+2pq+q2)=1p+q(p+q)2=p+q.Substituting back into the equation of the normal through P we find y=2p2(p+q)+12p−2p3=2p2q+12p. Again, using pq=12 we find y=2p(12)+q=p+q. Thus N has coordinates (p+q,p+q) when pq=12.
Show (in this case) that T and N lie on the line y=x…
When pq=12, T has coordinates (1p+q,1p+q), so lies on y=x. And we have just shown that N has coordinates (p+q,p+q), which also lies in the line y=x.
Alternatively, when pq=12, then P=(p,12p) and Q=(q,12q)=(12p,p), which means by symmetry that T and N lie on y=x.
…and are such that the product of their distances from the origin is constant.
We need a way to find the distance of a point from the origin. We can use Pythagoras’s theorem to do this.
The point with coordinates (a,b) is at distance √a2+b2 from the origin.
The distance of T from the origin is |OT|=√2(p+q)2=√2p+q. Similarly, |ON|=√2(p+q)2=√2(p+q). Thus the product of their distances from the origin is |OT|×|ON|=√2p+q⋅√2(p+q)=2, which is a constant.