Tangents are drawn to `x^2+y^2=16` from the point `P(0, h)dot` These tangents meet the `x-a xi s` at `Aa n dB` . If the area of triangle `P A B` is minimum, then `h=12sqrt(2)` (b) `h=6sqrt(2)` `h=8sqrt(2)` (d) `h=4sqrt(2)`

Options

A. h=`12sqrt(2)`
B. h=`6sqrt(2)`
C. h=`8sqrt(2)`
D. h=`4sqrt(2)`

Answer

4

Explanation

Let `angle COA =theta Then OA =OC sec theta =4 sec theta`
Also `angleOPC =theta .Then OP =OC cosec theta =4 cosec theta`
Now `triangle_(PAB)=OA OP =(32)/(sin 2 theta)`
For `triangle_(PAB)` to be minimum `sin 2 theta =1 or theta =(pi)/(4)`
`therefore P=0,4 sqrt(2)`