To determine the area of triangle \( \triangle AOB \), formed by the intersection of the tangent with the coordinate axes, we need to first find the equation of the tangent line to the ellipse \( \frac{x^2}{32} + \frac{y^2}{18} = 1 \) having a slope of \(-\frac{3}{4}\).
The general form of the tangent to the ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) with slope \(m\) is given by:
\(y = mx \pm \sqrt{a^2m^2 + b^2}\)
Here, \(a^2 = 32\) and \(b^2 = 18\). With \(m = -\frac{3}{4}\), the equation of the tangent becomes:
\(y = -\frac{3}{4}x \pm \sqrt{32 \left(-\frac{3}{4}\right)^2 + 18}\)
Simplifying:
\(y = -\frac{3}{4}x \pm \sqrt{32 \times \frac{9}{16} + 18}\)
\(y = -\frac{3}{4}x \pm \sqrt{18 + 18}\)
\(y = -\frac{3}{4}x \pm \sqrt{36}\)
\(y = -\frac{3}{4}x \pm 6\)
The equations of the tangents are:
\(y = -\frac{3}{4}x + 6\)
and
\(y = -\frac{3}{4}x - 6\)
Considering the equation \( y = -\frac{3}{4}x + 6 \), the intersection points with the axes are:
The vertices of triangle \( \triangle AOB \) are O(0, 0), A(0, 6), and B(8, 0).
The area of triangle OAB is given by:
\(Area = \frac{1}{2} \times \text{Base} \times \text{Height}\)
The base OA is along the y-axis with length 6, and the height OB is along the x-axis with length 8.
Thus, the area is:
\(Area = \frac{1}{2} \times 8 \times 6 = 24\text{ sq units}\)