Let the equation of two diameters of a circle x²+y²-2 x+2 f y+1=0 be 2 p x-y=1 and 2 x+p y=4 p. Then the slope m ∈ (0, ∞) of the tangent to the hyperbola 3 x²-y²=3 passing through the centre of the circle is equal to _______________.

Solution

Maths sir · 5 steps

Diameters meet at the centre: find it, then use c² = a²m² - b² for the hyperbola tangent.

1. Centre on both lines. The circle has centre (1, -f), and every diameter passes through it. Putting (1,-f) in 2px - y = 1 gives 2p + f = 1, so f = 1 - 2p. Putting it in 2x + py = 4p gives 2 - pf = 4p.
2. Solve for p. Substituting f = 1 - 2p gives 2 - p + 2p² = 4p, that is 2p² - 5p + 2 = 0. So p = 2 or p = 1/2, which give f = -3 or f = 0.
3. Reject f = 0. The radius is given by r² = 1 + f² - 1 = f². With f = 0 the radius is zero, which is not a real circle. So f = -3 and the centre is (1, 3).
4. Tangency condition. The hyperbola is x²/1 - y²/3 = 1, so a² = 1 and b² = 3. A tangent y = mx + c through (1,3) has c = 3 - m.

   c² = a²m² - b² ⇒ (3 - m)² = m² - 3

5. So. Expanding gives 9 - 6m = -3, so m = 2 and the tangent is y = 2x + 1 (check: c² = 1 = 4 - 3). The other tangent from (1,3) is the vertical line x = 1, which has no finite slope. The required slope is 2.

The common slip: 1. Using the ellipse condition c² = a²m² + b² instead of c² = a²m² - b² gives (3-m)² = m² + 3, so m = 1, not 2.

The idea to keep. Two diameters of a circle intersect at its centre. For a tangent y = mx + c to x²/a² - y²/b² = 1, use c² = a²m² - b², with a minus sign, not a plus.

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