<s>[INST] Finding the length of latus rectum of an ellipse If the lines $2x+3y=10$ and $2x-3y=10$ are tangents at the extremities of its same latus rectum to an ellipse whose center is origin,then the length of the latus rectum is 
$(A)\frac{110}{27}\hspace{1cm}(B)\frac{98}{27}\hspace{1cm}(C)\frac{100}{27}\hspace{1cm}(D)\frac{120}{27}$
I found point of intersection of two tangents at $(5,0)$ but dont know how to solve further.Any guidance will be helpful. 
 [/INST] Using standard ellipse notation and relations for 
$ a, b, c, p $.
Tangent equation of ellipse
$$ \frac{x x_1}{a^2} + \frac{y y_1}{b^2} =1 \tag{1}$$
Given tangent equation
$$ \frac{x}{5} + \frac{y}{10/3} =1 \tag{2}$$
Comparing $ x, y $ coefficients,
$$ \frac{x_1}{a^2}= \frac{1}{5} \tag{3}$$
$$ \frac{y_1}{b^2}= \frac{3}{10} = \frac{p}{b^2} = \frac{1}{a} \tag{4}$$
$$ a =\frac{10}{3}\tag{5} $$
$$ x_1= \frac{a^2}{5} = \frac{100}{45} =c \tag{6}$$
$$b^2 = a^2 -c^2 = \frac{400}{81}, b =\frac{20}{9} \tag{7}$$
Semi-latus rectum or latus rectum?
$$ p =\frac{b^2}{a}= \frac{40}{27}  \tag{8} $$
I checked it graphically also, none of the given options tally. But there ought to be a more elegant way than this.
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