Question

Latus rectum of ellipse $4x^2+9y^2-8x-36y+4=0$ is

• A. $8/3$
• B. $4/3$
• C. $\frac{\sqrt{5}}{3}$
• D. $16/3$

Answer 1

The correct option is A $8/3$

R.E.F. Image.

Ellipse : $4x^2+9y^2-8x-36y+4=0$

$4x^2-8x+9y^2-36y+4=0$

$\Rightarrow 4(x^2-2x)+9(y^2-4y)+4=0$

$\Rightarrow 4(x^2-2x+2-1)+9(y^2-4y+4-4)+4=0$

$\Rightarrow 4(x^2-2x+1)+9(y^2-4y+4)+4-4-36=0$

$\Rightarrow 4(x-1{)}^2+9(y-2{)}^2=36$ (divide both side with 36.)

$\Rightarrow \frac{(x-1{)}^2}{9}+\frac{(y-2{)}^2}{4}=1$

now compare with standard equation where h,k are the centre of ellipse.

$\frac{(x-h{)}^2}{a^2}+\frac{(y-k{)}^2}{b^2}=1$

$a^2=9\Rightarrow a=3$

$b^2=4\Rightarrow b=2$

$b^2=a^2(1-e^2)$

$4=9(1-e^2)$

$1-e^2=\frac{4}{9}$

$e^2=1-\frac{4}{9}=\frac{5}{9}$

$e=\frac{\sqrt{5}}{3}$

co-ordinate point of A $(al,\frac{b^2}{a})$

$A=(\sqrt{5},\frac{4}{3})$

co-ordinate point of B (al,$\frac{-b^2}{a})\Rightarrow ({\sqrt{5}}^{x_2}\frac{-4}{3})$

distance of AB = latus return

$=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$

$\Rightarrow \sqrt{(\sqrt{5}-\sqrt{5}{)}^2+(\frac{-4}{3}-\frac{-4}{3}{)}^2}$

$\Rightarrow \sqrt{(\frac{8}{3}{)}^2}=8/3$