The distance of the point (6,-2√2) from the common tangent y=mx+c,m > 0, of the curves x=2y² and x=1+y² is :

Solution

Maths sir · 5 steps

Write the slope-form tangent of each parabola and make their intercepts equal. That fixes m in one line.

1. First parabola. x = 2y² is y² = 1/2x, so 4a = 1/2 and a = 1/8. Its tangent of slope m is y = mx + a/m = mx + 1/8m.
2. Second parabola. x = 1 + y² is y² = x - 1, a parabola with a = 1/4 and vertex (1, 0). Replace x by x - 1 in the standard tangent: y = m(x-1) + 1/4m, which is y = mx - m + 1/4m.
3. Match the intercepts. Both lines have slope m, so they are the same line when their intercepts agree.

   1/8m = -m + 1/4m ⇒ m = 1/8m ⇒ m² = 1/8

4. The tangent. With m > 0, m = 1/2√2 and c = 1/8m = √2/4. Multiplying y = x/2√2 + √2/4 by 4√2 gives x - 2√2 y + 1 = 0.
5. So. The distance from (6, -2√2) is (|6 - 2√2(-2√2) + 1|)/√(1 + 8) = (|6 + 8 + 1|)/3 = 5. The answer is (b).

The idea to keep. For a common tangent of two parabolas, write each one's tangent in slope form with the same m, then equate the intercepts. For a shifted parabola, shift x in the tangent too.

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