Ifthe normal at P to the rectangular hyperbola `x^2-y^2=4` meets the axes in G and g and C is the centre of the hyperbola, then

Options

A. `PG=PC`
B. `Pg=PC`
C. `PG=Pg`
D. `Gg=2PC`

Answer

A, B, C, D

Explanation

Normal at point `P(2 sec theta, 2 tan theta)` is
`(2x)/(sec theta)+(2y)/(tan theta)=8`
It meets the axes at points `G(4 sec theta, 0) and g(0, 4 tan theta)`. Then,
`PG=sqrt(4 sec^(2)theta+4 tan^(2)theta)`
`Pg=sqrt(4 sec^(2)theta+4 tan^(2) theta)`
`PC=sqrt(4 sec^(2)theta+4 tan^(2)theta)`
`Gg=sqrt(16 sec^(2)theta+16 tan^(2)theta)`
`=2sqrt(4 sec^(2)theta+2 tan^(2)theta)=2PC`