Show that the curves `x=y^2` and `x y=k` cut at right angles, if `8k^2=1` .

Explanation

The equation of the given curves are given as `x=y^2 ` and `xy=k`
Putting `x=y^2` and `xy =k` we get,
`y^3=k implies y=(k)^(1/3)`
`x=k^(2/3)` Thus, the point of intersection of the given curves as `(k^(2/3),k^(1/3))`
Differentiating `x=y^2` with respect to `x`, we have:
`1=2ydy/dx implies dy/dx=1/(2y)`
Therefore, the slope of the tangent to the curve is `x=y^2` at `(k^(2/3),k^(1/3))`
`dy/dx]_(k^(2/3),k^(1/3))=1/(2k)^(1/3)`
On differentiating `xy=k` with respect to x, we have:
`xdy/dx+y=0 implies dy/dx=-y/x`
Slope of a tangent to the curve `xy=k` at `(k^(2/3),k^(1/3))`
`dy/dx]_(k^(2/3),k^(1/3))=-y/x]_(k^(2/3),k^(1/3))=-k^(1/3)/k^(2/3)=-1/k^(1/3)`
We know that two curves intersect at right angles if the tangents to the curve at the point of intersection at `(k^(2/3),k^(1/3))` are perpendicular to each other. This implies that we should have the product of the tangents as-1.Thus , the given two curves cut at right angles if the product of the slopes of their respective tangents at `(k^(2/3),K^(1/3))` is `-1`.
`(1/(2k^(1/3)))(-1/k^(1/3))=-1`
`implies 2k^(2/3)=1`
`(2k^(2/3))^3=(1)^3`
`8k^2=1`
Hence the given two curves cut at right angles if `8k^2=1`