The area (in sq. units) of the triangle forme | vedclass

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(B) Given the ellipse equation: $9x^2 + 4y^2 = 72$. Dividing by $72$,we get $\frac{x^2}{8} + \frac{y^2}{18} = 1$.At point $(2, 3)$,the equation of the

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$\frac{39}{4}$

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(B) Given the ellipse equation: $9x^2 + 4y^2 = 72$. Dividing by $72$,we get $\frac{x^2}{8} + \frac{y^2}{18} = 1$.At point $(2, 3)$,the equation of the tangent is $\frac{9(2)x}{72} + \frac{4(3)y}{72} = 1$,which simplifies to $\frac{x}{4} + \frac{y}{6} = 1$. The $X$-intercept is $x = 4$.The slope of the tangent is $m_t = -\frac{6}{4} = -\frac{3}{2}$.The slope of the normal is $m_n = -\frac{1}{m_t} = \frac{2}{3}$.The equation of the normal at $(2, 3)$ is $y - 3 = \frac{2}{3}(x - 2)$,which simplifies to $3y - 9 = 2x - 4$,or $2x - 3y = -5$.The $X$-intercept of the normal is found by setting $y = 0$: $2x = -5$,so $x = -\frac{5}{2}$.The base of the triangle on the $X$-axis is the distance between the intercepts: $|4 - (-\frac{5}{2})| = |\frac{8+5}{2}| = \frac{13}{2}$.The height of the triangle is the $Y$-coordinate of the point $(2, 3)$,which is $3$.Area $= \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes \frac{13}{2} 	imes 3 = \frac{39}{4}$.

The area (in sq. units) of the triangle formed by the tangent and normal to the ellipse $9x^2 + 4y^2 = 72$ at the point $(2, 3)$ with the $X$-axis is

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