The foci of a hyperbola are ( ± 2,0) and its eccentricity is 3/2. A tangent, perpendicular to the line 2 x+3 y=6, is drawn at a point in the first quadrant on the hyperbola. If the intercepts made by the tangent on the x - and y-axes are a and b respectively, then |6 a|+|5 b| is equal to __________

Solution

Maths sir · 5 steps

Fix the hyperbola from ae and e, then use c² = a²m² - b² for the tangent of slope 3/2.

1. Build the hyperbola. The foci give ae = 2. With e = 3/2, this means a = 4/3, so a² = 16/9. Then b² = a²(e² - 1) = 16/9 × 5/4 = 20/9.
2. Slope of tangent. The line 2x + 3y = 6 has slope -2/3. A line perpendicular to it has slope m = 3/2.

   c² = a²m² - b² = 16/9 × 9/4 - 20/9 = 16/9 ⇒ c = ±4/3

3. Pick the sign. The line y = mx + c touches the hyperbola at (-a²m/c, -b²/c). This point is in the first quadrant only when c < 0. So c = -4/3, and the point of contact is (2, 5/3).
4. Find the intercepts. The tangent is y = 3/2x - 4/3. Putting y = 0 gives a = 8/9, and putting x = 0 gives b = -4/3.
5. So. |6a| + |5b| = 16/3 + 20/3 = 36/3 = 12.

The common slip: 18. Taking (± 2, 0) as the vertices, so that a = 2, gives intercepts 4/3 and -2 and a total of 18. The foci are at (± ae, 0), not at (± a, 0).

The idea to keep. For a tangent of known slope to a hyperbola, use c² = a²m² - b². The quadrant of the point of contact decides the sign of c.

Answer from the NTA final answer key. Solution by PrepWiser.