Question

Show that the equation $(10x-5{)}^2+(10y-5{)}^2=(3x+4y-1{)}^2$ represents an ellipse. Find the length of its latus rectum.

• A. $\frac{5}{2}$
• B. $\frac{1}{2}$
• C. $\frac{3}{2}$
• D. $-\frac{1}{2}$

Answer 1

The correct option is C $\frac{3}{2}$

$(10x-5{)}^2+(10y-5{)}^2=(3x+4y-1{)}^2$

$25(2x-1{)}^2+25(2y-1{)}^2=(3x+4y-1{)}^2$

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

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

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

By observing equation $\left(1\right)$, we infer that

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

Centre of ellipse: $(\frac{1}{2},\frac{1}{2})$

$(e^2=1-\frac{b^2}{a^2})$

Equation of directrix $=(3x+4y-1=0)$

Also $d$ (Centre, directrix) $=\frac{3\left(\frac{1}{2}\right)+4\left(\frac{1}{2}\right)-1}{5}$

$=\frac{\frac{5}{2}+1}{5}=\frac{1}{2}$

$\therefore a+a_e=\frac{1}{2};a+\frac{a}{2}=\frac{1}{2}$

$\therefore \frac{3a}{2}=\frac{1}{2};a\frac{1}{3}$

$1-\frac{b^2}{a^2}=\frac{1}{4};\frac{b^2}{a^2}=\frac{3}{4};b=\frac{1}{\sqrt{12}}$

Length $=\frac{2b^2}{a}=2\left(\frac{1}{2}\right)\times \frac{1}{1/3}=\frac{1}{2}$