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.
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.
WLOG we choose the equation of the ellipse to be $$\dfrac{x^2}{a^2}+\dfrac{y^2}{a^2(1-e^2)}=1$$
The abscissa of the extremities of its one latus rectum to an ellipse $\pm ae$
$\implies y=\pm a(1-e^2)$
As the equation of the tangent at $(x_1,y_1)$ is $$\dfrac{xx_1}{a^2}+\dfrac{yy_1}{a^2(1-e^2)}=1$$
So, the equation of the tangents at $(ae,\pm a(1-e^2))$ will be
$$\dfrac{x e}a\pm\dfrac y a=1$$
These should be same as $$2x\pm3y=10$$
$$\implies\dfrac2{e/a}=\dfrac3{1/a}=\dfrac{10}1$$
$$\dfrac3{1/a}=\dfrac{10}1\iff3a=10$$ etc.
