Question

Find the equation(s) of the tangent line(s) to the parabola \(y=x^{2}\) through the given point. (a) \((0, a)\) (b) \((a, 0)\) Are there any restrictions on the constant \(a\) ?

Short answer

The equation of the tangent line for the point \((0, a)\) is \(y=0\), which is simply the x-axis. For the point \((a, 0)\), the equation of the tangent is \(y = 2a(x - a)\). There are no restrictions on \(a\) for the first case. However, for the second case, \(a\) cannot be 0.

Step-by-step solution

Compute the slope of the tangent line at any point on the parabola

The derivative \(y'(x)\) of the function \(y=x^{2}\) provides the slope of the tangent line at any point on the parabola. Hence, finding the derivative \(y'(x) = 2x\).

Find the point(s) of tangency for \((0, a)\)

The slope between any point on the parabola \((x, x^2)\) and the given point \((0, a)\) is \(\frac{x^2 - a}{x - 0} = x\). Correspondingly, by equating this slope to \(y'(x)\) gives \(x = 2x\), which suggests \(x = 0\) and the point(s) of tangency is/are \((0, 0)\).

Write down the equation of the tangent for \((0, a)\)

By utilizing the point-slope form \(y - y1 = m(x - x1)\), where \((x1, y1)\) is the point of tangency and \(m\) is the slope of the tangent, \(y = 0\) can be achieved.

Find the point(s) of tangency for \((a, 0)\)

The slope between any point on the parabola \((x, x^2)\) and the given point \((a, 0)\) is \(\frac{x^2 - 0}{x - a} = x\). By equating this slope to \(y'(x)\) gives \(x = 2x\), which implies \(x = a\) and the point of tangency is \((a, a^2)\).

Write down the equation of the tangent for \((a, 0)\)

By using the point-slope form results in \(y = 2a(x - a)\).

Determine the restrictions on \(a\)

Checking the equations derived above, one can see that there is no restriction on \(a\) for the first case \((0, a)\). For the second case \((a, 0)\), \(a\) can be any real number except that it cannot be 0 since in the slope formula, the denominator cannot be 0.