The equations of the tangents drawn at the en | vedclass

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(C) The given equation is $9x^2 + 5y^2 - 30y = 0$. Completing the square for $y$: $9x^2 + 5(y^2 - 6y) = 0$. $9x^2 + 5(y^2 - 6y + 9) = 45$. Dividing by

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$y = 0, y = 6$

Vedclass · Answer

(C) The given equation is $9x^2 + 5y^2 - 30y = 0$. Completing the square for $y$: $9x^2 + 5(y^2 - 6y) = 0$. $9x^2 + 5(y^2 - 6y + 9) = 45$. Dividing by $45$,we get $\frac{x^2}{5} + \frac{(y - 3)^2}{9} = 1$. Since the denominator of $(y - 3)^2$ is greater than the denominator of $x^2$ $(9 > 5)$,the major axis is parallel to the $y$-axis. The vertices are found by setting $x = 0$ in the equation $\frac{x^2}{5} + \frac{(y - 3)^2}{9} = 1$,which gives $\frac{(y - 3)^2}{9} = 1$,so $(y - 3)^2 = 9$. Thus,$y - 3 = \pm 3$,which means $y = 6$ or $y = 0$. The tangents at the ends of the major axis are horizontal lines passing through these vertices,which are $y = 0$ and $y = 6$.

The equations of the tangents drawn at the ends of the major axis of the ellipse $9x^2 + 5y^2 - 30y = 0$ are:

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