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Question

The area of triangle formed by the tangents from point (3,2) to hyperbola x29y2=9 and the chord of contact w.r.t. point (3,2) is :

A
8
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B
4
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C
6
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D
16
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Solution

The correct option is D 8
Hyperbola is x29y2=9 or x29y21=1
Equation of tangent is y=mx±a2m2b2 ..... (i)
It passes through (3,2)
2=3m±9m21
or 4+9m212m=9m21
m1=512 and m2=
Equation of tangent (i) for m1=512
y=512x±9(512)21
or y=512x±34
On taking (-)ve sign, point P(3,2) does not satisfy the equation of tangent therefore rejecting (-)ve sign.
Hence equation of tangent is y=5x12+34 .....(ii)
Now, equation of tangent (i) for m2= is x±3=0
rejecting (+) sign (since taking (+) sign point P(3,2) does not satisfy this equation.)
Hence, equation of tangent is x3=0 ....(iii)
Now equation of chord of contact w.r.t. point P(3,2) is T=0
or 3x18y=9
or x6y=3 .....(iv)
Solving (ii) and (iv); x=5,y=43
Solving (iii) and (iv); x=3,y=0
Now vertices of triangle are (3,2),(3,0),(5,43)
Area=12∣ ∣ ∣ ∣3213015431∣ ∣ ∣ ∣
=12|4164|
=8sq.units

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