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From point P(8,27) tangents PQ and PR ar...

From point `P(8,27)` tangents PQ and PR are drawn to the ellipse `(x^(2))/(4)+(y^(2))/(9)=1.` Then, angle subtended by `QR` at origin is

A

`tan^(-1).(sqrt(6))/(65)`

B

`tan^(-1).(4sqrt(6))/(65)`

C

`tan^(-1).(8sqrt(2))/(65)`

D

`tan^(-1).(48sqrt(6))/(455)`

Text Solution

Verified by Experts

The correct Answer is:
D
The equation of Qr is T=o (chord of contact)
`(8x)/(4)+(27y)/(9)=1`
or 2x+3y=1
Now, the equation or the pair of lines passing through the origin and point Q and R is given by
`((x^(2))/(4)+(y^(2))/(9))=(2x+3y)^(2)` [ Making the equation of ellispe homogeneous using (1)]
or `9x^(2)+4y^(2)=36(4x^(2)+12xy+9y^(2))`
or `135x^(2)+432xy+320y^(2)=0`
Therefore, the required angle is
`tan^(-1).(2sqrt(216^(2)-135xx320))/(455)=tan^(-1).(8sqrt(2916-2700))/(455)`
`=tan^(-1).(8sqrt(216))/(455)=tan^(-1).(48sqrt(6))/(455)`
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