Ellipse

Given, equation: $9x^2+4y^2=72$Differentiate w.r.t $x$, we get$18x+8y \cdot \frac{dy}{dx}=0$$\Rightarrow \frac{dy}{dx}=\frac{-18x}{8y}=\frac{-9x}{4y}$Slope of tangent at point $(2,3)$ is$m_t=\frac{-9(2)}{4(3)}=\frac{-18}{12}=\frac{-3}{2}$So, equation of tangent using point-slope form is$(y-3)=-\frac{3}{2}(x-2)$$\Rightarrow 2y-6=-3x+6$$\Rightarrow 3x+2y=12$Now, slope of normal, $m_n=\frac{-1}{m_t}=\frac{2}{3}$Now, slope of normal, $m_n=\frac{-1}{m_t}=\frac{2}{3}$So, equation of normal is$(y-3)=\frac{2}{3}(x-2)$$\Rightarrow 3y-9=2x-4$$\Rightarrow 2x-3y=-5$Now, for the $x$-intercept of the tangent line, put $y=0$, then$3x+2(0)=12$$\Rightarrow x=4$$\text{ so } (4,0)$For the $x$-intercept of the normal line, put $y=0$, then$2x-3(0)=-5$$\Rightarrow x=-\frac{5}{2}, \text{ so, } \left(-\frac{5}{2},0\right)$Now, base of triangle $=$ Distance $\text{b / w x}$-intercepts of the tangent and normal lines$b=4-\left(-\frac{5}{2}\right)=\frac{13}{2}$And, height of triangles is the $y$-coordinate of the point $(2,3)$ is $h=3$So, area $=\frac{1}{2} b h=\frac{1}{2}\left(\frac{13}{2}\right)\times 3=\frac{39}{4}$$\therefore \text{ Area }=\frac{39}{4} \text{ sq. units }$