If the line x - 2y = 12 is tangent to the ell | vedclass

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(B) The equation of the tangent to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ at point $(x_1, y_1)$ is $\frac{xx_1}{a^2} + \frac{yy_1}{b^2} =

Vedclass · Accepted Answer

$9$

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(B) The equation of the tangent to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ at point $(x_1, y_1)$ is $\frac{xx_1}{a^2} + \frac{yy_1}{b^2} = 1$.Given the point $(3, -4.5)$,the tangent equation is $\frac{3x}{a^2} - \frac{4.5y}{b^2} = 1$.Comparing this with the given tangent line $x - 2y = 12$,we rewrite the given line as $\frac{x}{12} - \frac{2y}{12} = 1$,which simplifies to $\frac{x}{12} - \frac{y}{6} = 1$.Equating the coefficients: $\frac{3}{a^2} = \frac{1}{12}$ $\Rightarrow a^2 = 36$ $\Rightarrow a = 6$.And $\frac{4.5}{b^2} = \frac{1}{6} \Rightarrow b^2 = 4.5 	imes 6 = 27$.The length of the latus rectum is $\frac{2b^2}{a} = \frac{2 	imes 27}{6} = \frac{54}{6} = 9$.

If the line $x - 2y = 12$ is tangent to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ at the point $(3, -4.5)$,then the length of the latus rectum of the ellipse is:

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