Let A be a point on the x-axis. Common tangents are drawn from A to the curves x²+y²=8 and y²=16 x. If one of these tangents touches the two curves at Q and R, then (Q R)² is equal to :

Solution

Maths sir · 4 steps

Write the parabola's tangent in slope form, then make its distance from the centre equal the radius.

1. Tangent to parabola. For y² = 16x we have a = 4. Every tangent can be written as y = mx + 4/m, touching the parabola at (4/m², 8/m).
2. Also touch circle. The circle x² + y² = 8 has centre (0, 0) and radius 2√2. The line is tangent to it when its distance from the origin equals the radius.

   (|4/m|)/√(1+m²) = 2√2 ⇒ m⁴ + m² - 2 = 0 ⇒ m² = 1

3. Find Q and R. Take m = 1, so the tangent is y = x + 4. It meets the x-axis at A(-4, 0). It touches the parabola at R(4, 8). It touches the circle at the foot of the perpendicular from the origin, Q(-2, 2).
4. So. (QR)² = (4 - (-2))² + (8 - 2)² = 36 + 36 = 72. The answer is (d). Taking m = -1 gives the mirror image in the x-axis and the same value.

The idea to keep. For a common tangent, start with the curve whose tangent has the neater slope form. Then put the condition for the other curve into it. For a circle, that condition is distance from the centre equals the radius.

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