91Ó°ÊÓ

A hyperbola is a type of conic section, which is a curve obtained from the intersection of a cone with a plane. In a standard coordinate system, the equation of a hyperbola takes the form \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \). In this formula, \(a^2\) and \(b^2\) are constants that dictate the shape and orientation of the hyperbola.

For the given problem, the hyperbola is described by the equation \(9x^2 - y^2 = 36\). This equation can be rewritten in its standard form by dividing every term by 36, resulting in \(\frac{x^2}{4} - \frac{y^2}{36} = 1\).

• \(a^2 = 4\) means \(a = 2\), indicating the distance from the center to each vertex along the \(x\)-axis.
• \(b^2 = 36\) gives \(b = 6\), the distance from the center to each co-vertex along the \(y\)-axis.

Understanding this form helps us deduce other properties of the hyperbola, such as its center at the origin \((0,0)\), and its axes being the \(x\) and \(y\)-axes.

Finding the points of tangency on a hyperbola is about calculating precisely where a tangent line touches the curve. For the hyperbola \(9x^2 - y^2 = 36\), this involves identifying specific points such that a tangent at that point intersects a certain location elsewhere - in our exercise, at the \(y\)-axis point \((0, 6)\).

To find these points, we start with the general equation for the tangent to a hyperbola at a point \((x_1, y_1)\), which for our hyperbola translates to \( \frac{xx_1}{4} - \frac{yy_1}{36} = 1 \). By setting \(x = 0\) (indicating the \(y\)-axis intersection), we solve for \(y_1\) when \(y = 6\).

• Solving the equation \(-\frac{6y_1}{9} = 4\), we find \(y_1 = -6\).
• Then substitute \(y_1 = -6\) back into the hyperbola's equation \(9x_1^2 - 36 = 36\) to get \(x_1 = \pm2\sqrt{2}\).

Thus, the points of tangency are \((2\sqrt{2}, -6)\) and \((-2\sqrt{2}, -6)\), where the tangent lines meet the \(y\)-axis at the desired intersection.

The concept of tangent line intersection involves finding where tangent lines to a curve meet a particular line or axis. In this exercise, the goal was to determine the points on the hyperbola where the tangent lines intersected the \(y\)-axis at \((0, 6)\).

To solve this, we used the equation for the tangent line derived from the hyperbola \(9x^2 - y^2 = 36\). The equation \( \frac{xx_1}{4} - \frac{yy_1}{36} = 1 \) was simplified, considering the line intersects at \((0, 6)\). By substituting \(x = 0\) and \(y = 6\), the equation simplifies, helping find the value of \(y_1\).

• This calculation resulted in the value \(y_1 = -6\).
• By substituting \(y_1\) back, we solve the hyperbola equation to find \(x_1 = \pm 2\sqrt{2}\).

The calculated points of tangency, \((2\sqrt{2}, -6)\) and \((-2\sqrt{2}, -6)\), are where the tangent lines intersect the hyperbola such that they meet the \(y\)-axis at \((0, 6)\). This process demonstrates how exact calculations can predict intersection points of tangents with an axis.