Question

The equation of the latus-rectum of the hyperbola
$(10x-5{)}^2+(10y-2{)}^2=9(3x+4y-7{)}^2$ is :

• A. $y-\frac{1}{5}=-\frac{3}{4}(x-\frac{1}{2})$
• B. $y+\frac{1}{5}=\frac{3}{4}(x-\frac{1}{2})$
• C. $y-\frac{1}{5}=\frac{3}{4}(x-\frac{1}{2})$
• D. $y-\frac{1}{5}=-\frac{3}{4}(x+\frac{1}{2})$

Answer 1

The correct option is A $y-\frac{1}{5}=-\frac{3}{4}(x-\frac{1}{2})$

The given equation is ${(10x-5)}^2+{(10y-2)}^2=9{(3x+4y-7)}^2$.........$\left(1\right)$

This can be written as

${\Rightarrow (x-\frac{1}{2})}^2+{(y-\frac{1}{5})}^2=\frac{9}{4}{\left(\frac{3x+4y-7}{5}\right)}^2$

$\Rightarrow \sqrt{{(x-\frac{1}{2})}^2+{(y-\frac{1}{5})}^2}=\frac{3}{2}\left(\frac{3x+4y-7}{5}\right)$.......$\left(2\right)$

LHS is the distance from the point $\left(\frac{1}{2},\frac{1}{5}\right)$ and RHS is $\frac{3}{2}$ times distance from $3x+4y-7=0$

so equation 2 is focus-directrix definition of Hyperbola with $e=\frac{3}{2}$, focus at $\left(\frac{1}{2},\frac{1}{5}\right)$ and $3x+4y-7=0$ is a directrix.

The latus rectum through this focus is parallel to Directrix. So it's equation is then

$3(x-\frac{1}{2})+4(y-\frac{1}{5})=0$