ParabolaeasyFree

Minimum of SQ+PQ for Point Inside x²+4y=0 | JEE Parabola

JEE Maths question with a full step-by-step solution.

Question
Consider the parabola x2+4y=0x^2 + 4y = 0. Let P(a,b)P(a, b) be a fixed point inside the parabola and let SS be the focus. Then the minimum value of SQ+PQSQ + PQ, as the point QQ moves on the parabola, is:
A∣1−a∣|1 - a|
B∣ab∣+1|ab| + 1
Ca2+b2\sqrt{a^2 + b^2}
D1−b1 - bcorrect
Solution
Step 1: For x2=−4yx^2 = -4y the focus is S(0,−1)S(0, -1) and the directrix y=1y = 1. If NN is the foot of perpendicular from QQ to the directrix, then SQ=QNSQ = QN, so SQ+PQ=QN+PQSQ + PQ = QN + PQ. Step 2: This is least when P,Q,NP, Q, N are collinear (perpendicular to the directrix), giving the distance from PP to the directrix:
min⁡(SQ+PQ)=1−b.\min(SQ + PQ) = 1 - b.
 Correct answer: (4)
Still stuck on this question?Ask your doubt on WhatsApp
Similar questions
Parabola · easy
The Cartesian equation of the curve whose parametric equations are x=t2+2t+3x = t^2 + 2t + 3 and y=t+1y = t + 1 is:
Parabola · easy
If x+a=0x + a = 0 is the directrix of the parabola y2=2y+ax+2y^2 = 2y + ax + 2, then the value of aa is:
Parabola · medium
The length of the latus rectum of the parabola x=t2+t+1x = t^2 + t + 1, y=t2+2t+3y = t^2 + 2t + 3 is:
Parabola · medium
If any point P(x,y)P(x, y) which satisfies the relation (5x−1)2+(5y−2)2=λ(3x−4y−1)2(5x - 1)^2 + (5y - 2)^2 = \lambda(3x - 4y - 1)^2 represents a parabola, then:

Solve more, learn faster

Sign up free to solve more JEE Maths questions and explore doMath — timed drills, mastery sprints, bookmarks, and chapter-wise progress tracking.

Sign up freeExplore doMath