  Question

# The area of triangle (in sq.units) formed by the tangents from point (3,2) to hyperbola x2−9y2=9 and the chord of contact with respect to the point (3,2)  is

Solution

## Hyperbola is x29−y21=1 Equation of tangent is : y=mx±√a2m2−b2 It passes through (3,2)⇒ 2=3m ± √9m2−1 ⇒ 9m2+4−12m=9m2−1 ⇒ 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 : x−3=0⋯(1) and y=512x+34⋯(2) Now, equation of chord of contact is T=0 : ⇒ 3x−18y=9⇒ x−6y=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(−43−2)+(−5)(2−0)∣∣ ∣∣ =12|4−4−6−10| =162 =8 sq. units Alternative method: Hyperbola is x29−y21=1⇒x2−9y2−9=0⋯(1) Equation of chord of contact is T=0 ⇒ 3x−18y=9⇒ x−6y=3⋯(2) Using (1),(2), we can find the points of intersection of the chord of contact with the hyperbola. Thus (3+6y)2−9y2−9=0 Thus points are (3,0),(−5,−43) Area=12∣∣ ∣∣3(0−(−4)3)+3(−43−2)+(−5)(2−0)∣∣ ∣∣ =12|4−4−6−10| =162 =8 sq. units  Suggest corrections   