Question

Length of the tangent from $(-a,0)$ to $y^2 = 4ax$

Answer 1

$|2a|\sqrt{2}$

Answer 2

The equation of the parabola is $y^2 = 4ax$. The given point is $P(-a, 0)$. To find the length of the tangent from an external point $(x_1, y_1)$ to the parabola, we first find the points of tangency. The equation of the tangent to the parabola $y^2 = 4ax$ at a point $(x_0, y_0)$ on the parabola is $yy_0 = 2a(x + x_0)$. Since the tangent passes through the point $(-a, 0)$, we substitute these coordinates into the tangent equation: $(0)y_0 = 2a(-a + x_0)$ $0 = 2a(x_0 - a)$ Assuming $a \neq 0$, we have $x_0 - a = 0$, which gives $x_0 = a$.

Since the point $(x_0, y_0)$ lies on the parabola, it must satisfy the equation $y_0^2 = 4ax_0$. Substitute $x_0 = a$ into this equation: $y_0^2 = 4a(a) = 4a^2$ $y_0 = \pm \sqrt{4a^2} = \pm |2a|$.

So, the points of tangency are $Q_1(a, |2a|)$ and $Q_2(a, -|2a|)$. The external point is $P(-a, 0)$.

The length of the tangent from $P(-a, 0)$ to the point of tangency $Q(x_0, y_0)$ is the distance $PQ = \sqrt{(x_0 - (-a))^2 + (y_0 - 0)^2} = \sqrt{(x_0 + a)^2 + y_0^2}$. Using the point $Q_1(a, |2a|)$: Length $PQ_1 = \sqrt{(a + a)^2 + (|2a|)^2} = \sqrt{(2a)^2 + 4a^2} = \sqrt{4a^2 + 4a^2} = \sqrt{8a^2} = \sqrt{4a^2 \cdot 2} = \sqrt{4a^2} \sqrt{2} = |2a|\sqrt{2}$.

Using the point $Q_2(a, -|2a|)$: Length $PQ_2 = \sqrt{(a + a)^2 + (-|2a|)^2} = \sqrt{(2a)^2 + 4a^2} = \sqrt{4a^2 + 4a^2} = \sqrt{8a^2} = |2a|\sqrt{2}$.

Both tangent segments from $P$ to the parabola have the same length, which is $|2a|\sqrt{2}$. In the standard context of the parabola $y^2 = 4ax$, $a$ is usually taken as a positive parameter representing the focal length. In this case, $|2a| = 2a$. So, if $a > 0$, the length is $2a\sqrt{2}$. If $a < 0$, the length is $|2a|\sqrt{2} = -2a\sqrt{2}$. The expression $|2a|\sqrt{2}$ covers both cases.