Question

Points P(2`p^2`,`1/p`) and Q(2`q^2`,`1/q`)lies on a curve.(pls see below) ?

Find the gradient of PQ and also the equation of the line PQ.
Deduce the equation of tangents at P and Q to the curve.Hence,find the coordinates of the intersection point of tangent at P and tangent at Q.

Answer 1

Please see below.

Explanation:

Equation of line joining `P(2p^2,1/p)` and `Q(2q^2,1/q)` is

`(y-1/p)/(1/q-1/p)=(x-2q^2)/(2p^2-2q^2)`

or `(y-1/p)=(x-2q^2)(2(p+q)(p-q))/((p-q)/(pq)`

or `(y-1/p)=2pq(x-2q^2)(p+q)`

or `y=2pq(p+q)x-4pq^3(p+q)+1/p`

which is in slope intercept form and hence gradient is `2pq(p+q)`

This has not been mentioned but parametric equation of curve appears to be `x=2t^2` and `y=1/t` and we can obtain equation of curve by eliminating `t`. As `t=1/y`, putting it in `x=2t^2`, we get

`x=2/y^2` or `xy^2=2`

Slope of tangent at `t` is given by `(dy)/(dx)` and differentiating `xy^2=2` implicitly, we have `y^2+2xy(dy)/(dx)=0`

or `(dy)/(dx)=-y/2x=-(1/t)/(2t^2)=-1/(2t^3)`

Equation of tangent at `P(2p^2,1/p)` is

`y-1/p=-1/(2p^3)(x-2p^2)` (A)

and equation of tangent at `Q(2q^2,1/q)` is

`y-1/q=-1/(2q^3)(x-2q^2)` (B)

Solving (A) and (B) for `x` and `y` will give us point of intersection of tangents at `P` and`Q`.