Normal is drawn at one of the extremities of the latus rectum of the hyperbola `(x^2)/(a^2)-(y^2)/(b^2)=1` which meets the axes at points `Aa n dB` . Then find the area of triangle `O A B(O` being the origin).

Answer

`"Area"=(1)/(2)a^(2)e^(5)`

Explanation

Normal at point `P(x_(1),y_(1))` is `(a^(2)x)/(x_(1))+(b^(2)y)/(y_(1))=a^(2)+b^(2).`
It meets the axes at
`A(((a^(2)_b^(2))x_(1))/(a^(2)),0)and B(0,((a^(2)+b^(2))y_(1))/(b^(2)))`
`"Area of "DeltaOAB=(1)/(2)[((a^(2)+b^(2))x_(1))/(a^(2))][((a^(2)+b^(2))y_(1))/(b^(2))]`
`=(1)/(2)[((a^(2)+b^(2))x_(1)y_(1))/(a^(2)b^(2))]`
Now, normal is drawn at the extremity of latus rectum.
Hence, `(x_(1),y_(1))-=(ae,(b^(2))/(a))`
`therefore" Area"=(1)/(2)[((a^(2)+b^(2))^(2)b^(2)e)/(a^(2)b^(2))]`
`=(1)/(2)[(a^(4)(1+(b^(2))/(a^(2)))^(2)e)/(a^(2))]`
`=(1)/(2)a^(2)e^(5)`