This question was previously asked in

JEE Mains Previous Paper 1 (Held On: 11 Jan 2019 Shift 2)

Option 4 : A parabola

JEE Mains Previous Paper 1 (Held On: 12 Apr 2019 Shift 2)

7876

90 Questions
360 Marks
180 Mins

Let the equation of circle be:

⇒ x^{2 }+ y^{2 }+ 2gx + 2fy + c = 0 ----(1)

The length of intercepts made by the circle x^{2 }+ y^{2 }+ 2nx + 2my + q = 0 with x axis is \(2\sqrt {{n^2} - q} \)

For this problem, the length of intercepts made by the circle with x axis from question is:

\(\Rightarrow 4a = 2\sqrt {{g^2} - c} \)

∴ c = g^{2 }- 4a^{2 ----}(2)

Also, as the circle is passing throughP(0, 2b).

⇒ 0 + 4b^{2 }+ 0 + 4bf + c = 0

∴ 4b^{2 }+ 4bf + c = 0 ----(3)

Eliminating ‘c’ from equation (2) and equation (3),

⇒ g^{2 }- 4a^{2 }= -4b^{2 }- 4bf

⇒ 4b^{2 }+ 4bf + g^{2 }- 4a^{2 }= 0

So, the locus of (-g, -f) is:

⇒ 4b^{2 }- 4by + x^{2 }- 4a^{2 }= 0

∴ x^{2 }= 4by + 4a^{2 }- 4b^{2}