Question

If the latus-rectum through one focus of a hyperbola subtends a right angle at the farther vertex,then write the eccentricity of the hyperbola.

Answer 1

$\text{It is given that}\Delta ABCisrightangletriangle.$

$\therefore A{B}^2=B{C}^2+A{C}^2$

$\left[Usingphyyhagorostheorem\right]$

$\Rightarrow (ae-ae{)}^2+(\frac{b^2}{a}+\frac{b^2}{a})=(ae+a^2)+\left(\frac{-b^2}{a}\right)+(ae+a{)}^2+{\left(\frac{b^2}{a}\right)}^2$

$\Rightarrow \frac{4b^2}{a^2}=2a^2(e+1{)}^2+\frac{2b^4}{a^2}$

$\Rightarrow \frac{4b^2-2b^4}{a^2}=2a^2(e+1{)}^2$

$\Rightarrow \frac{2b^4}{a^4}=2a^2(e+1{)}^2$

$\Rightarrow \frac{b^4}{a^4}=(e+1{)}^2...(i)$

$Now,$

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

$\Rightarrow b^4=a^4(e^2-1{)}^2[Usingbothsides]$

$\Rightarrow \frac{b^4}{a^4}=(e^2-1{)}^2$

$\Rightarrow (e+1{)}^2=(e^2-1\left)\right[Usingequation\left(I\right)]$

$\Rightarrow (e+1{)}^2=(e^2-1\left)\right(e^2-1)$

$\Rightarrow (e+1{)}^2=(e-1\left)\right(e+1\left)\right(e^2-1)$

$[\because a^2-b^2=(a-b\left)\right(a+b\left)\right]$

$\Rightarrow e+1=(e-1)(e^2-1)$

$\Rightarrow e+1=(e-1)(e-1)(e+1)$

$[\because a^2-b^2=(a-b\left)\right(a+b\left)\right]$

$\Rightarrow 1=(e-1{)}^2$

$\Rightarrow 1=e^2+1-2e$

$\Rightarrow e-2e=0$

$\Rightarrow e(e-2)=0$

$\Rightarrow e-2=0[\because e\ne 0]$

$\Rightarrow e=2$

$Hence,e=2$