Question

Draw a diagram showing two perpendicular lines that intersect on the y-axis and are both tangent to the parabola \(y = {x^{\bf{2}}}\). Where do these lines intersect?

Short answer

The point of intersection of tangents to the parabola \(y = {x^2}\) is \(\left( {0, - \frac{1}{4}} \right)\).

Step-by-step solution

Find the condition for tangents to intersect at the y-axis

Two tangents to the parabola \(y = {x^2}\) intersect on the y-axis, the point of tangents should be equal distance from the point of intersection of tangents. The curve \(y = {x^2}\) is symmetric about the y-axis.

Let the point of tangents on \(y = {x^2}\) are \(\left( {a,{a^2}} \right)\) and \(\left( { - a,{a^2}} \right)\).

Find the equation of tangents

The slope of the tangent to the parabola can be calculated as follows,

\(\begin{aligned}\frac{{{\rm{d}}y}}{{{\rm{d}}x}} &= \frac{{\rm{d}}}{{{\rm{d}}x}}\left( {{x^2}} \right)\\ &= 2x\end{aligned}\)

So, the slope of the tangent at \(\left( {a,{a^2}} \right)\) is \(2a\) and the slope of the tangent at \(\left( { - a,{a^2}} \right)\) is \( - 2a\).

The equation of the tangent at \(\left( {a,{a^2}} \right)\) can be written as,

\(\begin{aligned}y - {a^2} &= 2a\left( {x - a} \right)\\y - {a^2} &= 2ax - 2{a^2}\\y &= 2ax - {a^2}\end{aligned}\)

The equation of the tangentat \(\left( { - a,{a^2}} \right)\) can be written as,

\(\begin{aligned}y - {a^2} &= - 2a\left( {x + a} \right)\\y - {a^2} &= - 2ax - 2{a^2}\\y &= - 2ax - {a^2}\end{aligned}\)

Draw the graph of two tangents

The intersection points of tangents \(y = 2ax - {a^2}\) and \(y = - 2ax - {a^2}\) is \(\left( {0, - {a^2}} \right)\). As the tangents are perpendicular to each other, then the product of their slope must be equal to \( - 1\).

\(\begin{aligned}\left( { - 2a} \right)\left( {2a} \right) &= - 1\\{a^2} &= \frac{1}{4}\end{aligned}\)

So, the point of intersection is \(\left( {0, - \frac{1}{4}} \right)\).

The figure below represents the curve \(y = {x^2}\) and its tangent.