Question

Passage: Tangents are drawn to the hyperbola x² − 9y² = 9 from (3, 2). 

Answer the following questions.

(i) The area of the triangle formed by the tangents and the chord contact of (3, 2) is 

(A)  6 

(B)  8 

(C)  10 

(D)  12

(ii) The area of the triangle formed by the tangent to the hyperbola at (3, 0) and the two asymptotes is

(A)   3

(B)  6 

(C)   9 

(D)  2

(iii)  The midpoint of the intercept of the tangent at (3, 0) between the asymptotes is

(A)  (1, 0) 

(B)  (2, 0) 

(C)  (3, 1) 

(D)  (3, 0)

Answer 1

Correct option (i) (b)(ii) (a)(iii)(d)

Explanation :

(i)  y = mx + √(9m² - 1) is a tangent to the hyperbola

x²/9 - y²/1 = 1

This passes through the point (3, 2). This implies

(3m - 2)² = 9m² - 1

⇒ -12m = -5

⇒ m = 5/12

Therefore, one tangent is 5x − 12y + 9 = 0. Also the tangent at the vertex (3, 0) passes through (3, 2). Hence, the other tangent through (3, 2) is x = 3. The chord of contact is

3x/9 - 2y/1 = 1

⇒ x - 6y = 3

Therefore, the sides of the triangle are 

5x - 12y + 9 = 0

x = 3 and x - 6y = 3

Solving these equations, the vertices of the triangle are (3, 2), (3, 0) and (-5,-4/3). Hence, the area of the triangle is

(ii) The area of the triangle formed by two asymptotes and a tangent to the hyperbola

x²/a² - y²/b² = 1

is always constant which is equal to ab .

(iii) We know that the portion of the tangent is intercepted between the asymptote is bisected at the point of contact . In fact the asymptotes are x = ± 3y and the tangent at (3, 0) is x = 3. Hence, the tangent at (3,0) intersects the asymptotes at points (3, 1) and (3, − 1) so that the midpoint of the segment is (3, 0).