If frac{(3x - 4y - z)^2}{100} - frac{(4x + 3y | vedclass

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Question

(C) The given equation is $\frac{(3x - 4y - z)^2}{100} - \frac{(4x + 3y - 1)^2}{225} = 1$. Assuming the constant $z$ in the first term is a typo for $

Vedclass · Accepted Answer

$9$

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(C) The given equation is $\frac{(3x - 4y - z)^2}{100} - \frac{(4x + 3y - 1)^2}{225} = 1$. Assuming the constant $z$ in the first term is a typo for $1$,the equation becomes $\frac{(3x - 4y - 1)^2}{100} - \frac{(4x + 3y - 1)^2}{225} = 1$. Dividing the numerators by the square of the distance between the lines (which is $5^2 = 25$),we rewrite the equation as: $\frac{(\frac{3x - 4y - 1}{5})^2}{4} - \frac{(\frac{4x + 3y - 1}{5})^2}{9} = 1$. This is in the standard form $\frac{X^2}{a^2} - \frac{Y^2}{b^2} = 1$,where $a^2 = 4$ and $b^2 = 9$. Thus,$a = 2$ and $b = 3$. The length of the latus rectum is given by $\frac{2b^2}{a} = \frac{2 	imes 9}{2} = 9$.

If $\frac{(3x - 4y - z)^2}{100} - \frac{(4x + 3y - 1)^2}{225} = 1$,then the length of the latus rectum of the hyperbola is:

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