NIMCET Parabola Previous Year Question Paper and Solution with PDF

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NIMCET Previous Year Questions (PYQs)

NIMCET Parabola PYQ

Qus : 1

A point P in the first quadrant, lies on

$y^2 = 4ax$ , a > 0, and keeps a distance of 5a units from its focus. Which of the following points lies on the locus of P?

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Solution

Locus of Point on Parabola

Given: Point on parabola $y^2 = 4a x$ is at distance $5a$ from focus $(a, 0)$ .

Distance Equation:

$(x - a)^2 + y^2 = 25a^2 \Rightarrow (x - a)^2 + 4a x = 25a^2 \Rightarrow x^2 + 2a x - 24a^2 = 0$

Solving gives: $x = 4a$ , $y = 4a$

✅ Final Answer: $\boxed{(4a,\ 4a)}$

Qus : 2

A circle touches the x–axis and also touches the circle with centre (0, 3) and radius 2. The locus of the centre of the circle is

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Solution

Qus : 3

The two parabolas

$y^2 = 4a(x + c)$ and

$y^2 = 4bx, a > b > 0$ cannot have a common normal unless

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Solution

Qus : 4

Coordinate of the focus of the parabola

$4y^2+12x-20y+67=0$ is

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Solution

Qus : 5

An equilateral triangle is inscribed in the parabola

$y^2 = x$ . One vertex of the triangle is at the vertex of the parabola. The centroid of triangle is

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Solution

Qus : 6

An equilateral triangle is inscribed in the parabola

$y^{2} = 4ax$ , such that one of the vertices of the triangle coincides with the vertex of the parabola. The length of the side of the triangle is:

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Solution

Qus : 7

The locus of the mid points of all chords of the parabola

$y^{2}=4x$ which are drawn through its vertex, is

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Solution

Qus : 8

$x = 1$ is the directrix of the parabola

$y^{2} = kx - 8$ , then k is:

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Solution

Qus : 9

A normal to the curve

$x^{2} = 4y$ passes through the point (1, 2). The distance of the origin from the normal is

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Solution

Qus : 10

The equation of the tangent at any point of curve

$x=a cos2t, y=2\sqrt{2} a sint$ with

$m$ as its slope is

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Solution

Qus : 11

The locus of the mid-point of all chords of the parabola

$y^2 = 4x$ which are drawn through its vertex is

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Solution

Locus of Midpoint of Chords

Given Parabola: $y^2 = 4x$

Condition: Chords pass through the vertex $(0, 0)$

Let the other end of the chord be $(x_1, y_1)$ , so the midpoint is:

$M = \left( \frac{x_1}{2}, \frac{y_1}{2} \right) = (h, k)$

Since the point lies on the parabola: $y_1^2 = 4x_1$

⇒ $(2k)^2 = 4(2h)$

⇒ $4k^2 = 8h$

⇒ $\boxed{k^2 = 2h}$

✅ Locus of midpoints: $y^2 = 2x$

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NIMCET Previous Year Questions (PYQs)

NIMCET Parabola PYQ

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Locus of Point on Parabola

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Locus of Midpoint of Chords

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