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If the normal at P(asectheta,btantheta) ...

If the normal at `P(asectheta,btantheta)` to the hyperbola `x^2/a^2-y^2/b^2=1` meets the transverse axis in G then minimum length of PG is

A

`(b^(2))/(a)`

B

`|(a)/(b)(a+b)|`

C

`|(a)/(b)(a-b)|`

D

`|(a)/(b)(a-b)|`

Text Solution

Verified by Experts

The correct Answer is:
A
Equation of the normal is `ax cos theta + by cot theta = a^(2) +b^(2)` The normal at P meets the coordinate axes at `G((a^(2)+b^(2))/(a)sec theta,0)`
and `g (0,(a^(2)+b^(2))/(b)tan theta)`
`:. PG^(2) = ((a^(2)+b^(2))/(a)sec theta -a sec theta)^(2)+(b tan theta -0)^(2)`
`PG^(2) =(b^(2))/(a^(2)) (b^(2) sec^(2) theta + a^(2) tan^(2) theta)`
When `tan theta =0`
`PG =(b^(2))/(a)`
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