Answer
$(2,4),(-2,4)$
Work Step by Step
Equation of parabola
$$ y=x^2 \hspace{3cm} (1)$$
Taking derivatives
$$\frac{dy}{dx}=\frac{d(x)^2}{dx}=2x$$
Let $m_1$ denote the slope of tangent through the point $(x_1,y_1)$ of the parabola .Then
$$\frac{dy}{dx}=2x_1=m_1 \hspace{3cm} (2) $$
The equation of tangent through the point $(x_1,y_1)$ with slope $m_1=2x_1$:
$$y-y_1=2x_1(x-x_1) \hspace{3cm} (3)$$
Since tangent passes through $(0,-4)$ .Putting $(x,y)=(0,-4)$ in equation (3)
$$-4-y_1=2x_1(0-x_1)$$
$$-4-y_1=-2x_1^2$$
$$-(4+y_1)=-2x_1^2$$
$$ 4+y_1=2x_1^2 \hspace{3cm} (4)$$
Putting $(x,y)=(x_1,y_1)$ in equation (1)
$$y_1=x_1^2 \hspace{3cm} (5) $$
Putting equation (5) in equation (4)
$$4+y_1=2y_1$$
$$4=2y_1-y_1$$
$$4=y_1$$
$$y_1=4 \hspace{3cm} (6)$$
Putting equation (6) in equation (5)
$$x_1^2=4$$
$x_1=2$ or $x_1=-2$
Hence passing through $(0,-4)$ ,there exist two tangents through points $(2,4)$ and $(-2,4) $ of parabola.
