Question

Previous year Question of jee mains

mathamatics shift :1
24 feb 2021
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Answer 1

Answer:

Option (B)

Step-by-step explanation:

Given :- y² = 4ax is parabola in which the mid-point of line segment joining focus of parabola and one point is moving on the parabola .

Let's assume the mid-point is M (h,k)

Then Focus for y² = 4ax will be S(a,0)

The coordinates of moving point P(at²,2at)

Now, from mid-point formula

• h = at²+a/2 , k = 2at+0/2

• → t² = 2h -a/a and t = k/a

• → k²/a² = 2h - a /a

• → K² = (2h-a/a )a²

• → k² = (2h - a)*a

• → locus of (h,k) is y² = a(2x - a )

→making equation of parabola in standard form .

• y² = 2a ( x - a /2 )

• y² = 2a ( x- a/2)

As we know when equation of parabola is y² = 4ax then equation of directrix becomes x + a = 0

• Similarly, equation of directrix of y² = 2a (x-a/2) is x - a /2 = -a/2

• x = 0 Answer