The equation of the normal at the point (2, 3 | vedclass

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(B) The equation of the normal at point $(x_1, y_1)$ to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is given by $\frac{a^2x}{x_1} - \frac{b^2y

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$3y - 8x + 7 = 0$

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(B) The equation of the normal at point $(x_1, y_1)$ to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is given by $\frac{a^2x}{x_1} - \frac{b^2y}{y_1} = a^2 - b^2$.Given the ellipse $9x^2 + 16y^2 = 180$,we divide by $180$ to get $\frac{x^2}{20} + \frac{y^2}{11.25} = 1$.Here,$a^2 = 20$ and $b^2 = \frac{180}{16} = 11.25$.At point $(2, 3)$,the equation of the normal is $\frac{20x}{2} - \frac{11.25y}{3} = 20 - 11.25$.$10x - 3.75y = 8.75$.Multiplying by $4$ to clear decimals: $40x - 15y = 35$.Dividing by $5$: $8x - 3y = 7$,which simplifies to $3y - 8x + 7 = 0$.

The equation of the normal at the point $(2, 3)$ on the ellipse $9x^2 + 16y^2 = 180$ is:

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