Question

From origin two perpendicular lines are drawn to intersect $\frac{x^2}{4}+\frac{Y^2}{1}=1$ at A and B then $\frac{1}{(OA{)}^2}+\frac{1}{(OB{)}^2}$ is equal to

• A. $1$
• B. $\frac{1}{4}$
• C. $\frac{5}{4}$
• D. $\frac{3}{4}$

Answer 1

The correct option is C $\frac{5}{4}$

Let the slope of two lines be $m$ & $\frac{-1}{m}$

$\therefore$ equation of lines as they pass from origin is

$y=mx$ & $y=\frac{-1}{m}x$

$\therefore$ putting $y=mx$ in equation $\frac{x^2}{4}+\frac{y^2}{1}=1$

$\frac{x^2}{4}+m^2{x}^2=1(m^2+\frac{1}{4})x^2=1$

$\therefore x^2=\frac{1}{m^2+\frac{1}{4}}$

Let $y=mx$ intersect the given equation at $A(x,y)$

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

Putting $y=\frac{-1}{m}x$ in the given equation

$\frac{x^2}{4}+\frac{x^2}{m^2}=1\therefore x^2=\frac{4m^2}{m^2+4}$

Let $y=\frac{x}{m}$ intersect the given equation equation at $B(x^9,y^{11})$

$\therefore (x^4{)}^2=\frac{4m^2}{m^2+4}(y^{11})=\frac{4}{m^2+4}$

$\therefore \frac{1}{(OA{)}^2}+\frac{1}{(OB{)}^2}=\frac{1}{(x^1{)}^2+(y^1{)}^2}+\frac{1}{(x^{11}{)}^2+(y^{11}{)}^2}$

$=\frac{m^2+\frac{1}{4}}{m^2+1}+\frac{m^2+4}{4(m^2+1)}=\frac{5}{4}$