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Question

The area of triangle (in sq.units) formed by the tangents from point (3,2) to hyperbola x29y2=9 and the chord of contact with respect to the point (3,2)  is


Solution



Hyperbola is x29y21=1
Equation of tangent is : y=mx±a2m2b2
It passes through (3,2) 2=3m ± 9m21
 9m2+412m=9m21
 m=512 and it can aso be observed that equation of tangent at vetex is x=3,
So, slope of other tangent line is  m=
Equation of tangents are :
x3=0(1) and y=512x+34(2)
Now, equation of chord of contact is T=0 :
 3x18y=9 x6y=3(3)
Solving (1), (2) and (3), we get point of intersection of three lines as :(3,2),(3,0),(5,43)
Area=12∣ ∣3(0(4)3)+3(432)+(5)(20)∣ ∣
=12|44610|
=162
=8 sq. units

Alternative method:
Hyperbola is x29y21=1x29y29=0(1)
Equation of chord of contact is T=0
 3x18y=9 x6y=3(2)
Using (1),(2), we can find the points of intersection of the chord of contact with the hyperbola.
Thus (3+6y)29y29=0
Thus points are (3,0),(5,43)
Area=12∣ ∣3(0(4)3)+3(432)+(5)(20)∣ ∣
=12|44610|
=162
=8 sq. units
 

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