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NIMCET Previous Year Questions (PYQs)
NIMCET Parabola PYQ
NIMCET PYQ
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NIMCET Previous Year PYQ NIMCET NIMCET 2023 PYQ
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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NIMCET Previous Year PYQ NIMCET NIMCET 2023 PYQ
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: \( \boxed{(4a,\ 4a)} \)
NIMCET PYQ
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NIMCET Previous Year PYQ NIMCET NIMCET 2023 PYQ
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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NIMCET Previous Year PYQ NIMCET NIMCET 2023 PYQ
Solution
NIMCET PYQ
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NIMCET Previous Year PYQ NIMCET NIMCET 2012 PYQ
Focus of the parabola
$x^2 + y^2 - 2xy - 4(x + y - 1) = 0$ is:
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NIMCET Previous Year PYQ NIMCET NIMCET 2012 PYQ
Solution
Given equation: $x^2 + y^2 - 2xy - 4x - 4y + 4 = 0$
Group terms:
$(x - y)^2 - 4(x + y) + 4 = 0$
Let:
$u = x - y,\;\; v = x + y$
Then equation becomes:
$u^2 - 4v + 4 = 0$
$u^2 = 4(v - 1)$
This is the standard parabola:
$u^2 = 4p(v - 1)$
Comparing gives:
$4p = 4 \Rightarrow p = 1$
Vertex in $(u,v)$:
$(0,1)$
Focus in $(u,v)$:
$(0, 1 + p) = (0,2)$
Convert to $(x,y)$:
$x - y = 0$
$x + y = 2$
Solving:
$x = 1,\; y = 1$
Therefore, the focus is:
$\boxed{(1,1)}$
NIMCET PYQ
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NIMCET Previous Year PYQ NIMCET NIMCET 2011 PYQ
Vertex of parabola $ y^{2} - 8y + 19 = 0 $
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NIMCET Previous Year PYQ NIMCET NIMCET 2011 PYQ
Solution
$ y^{2} - 8y + 16 = -3 $ $ (y - 4)^{2} = -3 $ Axis horizontal, vertex is $(h,k) = (1,4)$
NIMCET PYQ
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NIMCET Previous Year PYQ NIMCET NIMCET 2024 PYQ
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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NIMCET Previous Year PYQ NIMCET NIMCET 2024 PYQ
Solution
NIMCET PYQ
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NIMCET Previous Year PYQ NIMCET NIMCET 2022 PYQ
Coordinate of the focus of the parabola $4y^2+12x-20y+67=0$ is
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NIMCET Previous Year PYQ NIMCET NIMCET 2022 PYQ
Solution
NIMCET PYQ
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NIMCET Previous Year PYQ NIMCET NIMCET 2010 PYQ
If the tangents at the extremities of a focal chord of the parabola $x^2 = 4ay$ meet at a point where the abscissas are $x_1$ and $x_2$, then
$x_1 x_2 =$
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NIMCET Previous Year PYQ NIMCET NIMCET 2010 PYQ
Solution
Parametric form of parabola: x = 2at, y = at² Focal chord endpoints: t and –1/t Slope of tangent at t: 1/t Equation of tangent meets the tangent at –1/t Product of x-intercepts = a²
NIMCET PYQ
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NIMCET Previous Year PYQ NIMCET NIMCET 2025 PYQ
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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NIMCET Previous Year PYQ NIMCET NIMCET 2025 PYQ
Solution
Given parabola $y^{2} = x. $ $ A(0,0) $ is one vertex of the equilateral triangle.Let the other two vertices be
$ B(t^{2}, t) $ and $ C(t^{2}, -t) $,
since they lie on the parabola and are symmetric.
Distance:
$ BC = \sqrt{(t^{2}-t^{2})^{2} + (t - (-t))^{2}} = 2t. $
And
$ AB = \sqrt{(t^{2}-0)^{2} + (t-0)^{2}}
= \sqrt{t^{4}+t^{2}}
= t\sqrt{t^{2}+1}. $
For equilateral triangle:
$ AB = BC $
$ t\sqrt{t^{2}+1} = 2t $
$ \sqrt{t^{2}+1} = 2 $
$ t^{2}+1 = 4 $
$ t^{2} = 3. $
So the points become:
$ B = (3,\sqrt{3}), \quad C = (3,-\sqrt{3}). $
Centroid:
$ G = \left( \frac{0+3+3}{3},; \frac{0+\sqrt3-\sqrt3}{3} \right)
= (2,0). $
NIMCET PYQ
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NIMCET Previous Year PYQ NIMCET NIMCET 2014 PYQ
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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NIMCET Previous Year PYQ NIMCET NIMCET 2014 PYQ
Solution
NIMCET PYQ
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NIMCET Previous Year PYQ NIMCET NIMCET 2015 PYQ
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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NIMCET Previous Year PYQ NIMCET NIMCET 2015 PYQ
Solution
NIMCET PYQ
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NIMCET Previous Year PYQ NIMCET NIMCET 2014 PYQ
If $x = 1$ is the directrix of the parabola $y^{2} = kx - 8$, then k is:
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NIMCET Previous Year PYQ NIMCET NIMCET 2014 PYQ
Solution
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NIMCET Previous Year PYQ NIMCET NIMCET 2014 PYQ
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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NIMCET Previous Year PYQ NIMCET NIMCET 2014 PYQ
Solution
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NIMCET Previous Year PYQ NIMCET NIMCET 2016 PYQ
The vertex is the midpoint of focus and foot of perpendicular from focus to directrix
Foot of perpendicular from $S(-1, 1)$ to $4x + 3y - 24 = 0$:
$\dfrac{x + 1}{4} = \dfrac{y - 1}{3} = -\dfrac{4(-1) + 3(1) - 24}{4^2 + 3^2}$
$= -\dfrac{-4 + 3 - 24}{16 + 9}$
$= -\dfrac{-25}{25}$
$= 1$
$\therefore x + 1 = 4 \Rightarrow x = 3$
$\therefore y - 1 = 3 \Rightarrow y = 4$
$\therefore$ Foot of perpendicular $Z = (3, 4)$
Since vertex $V$ is midpoint of $SZ$:
$V = \left(\dfrac{-1 + 3}{2},\ \dfrac{1 + 4}{2}\right)$
$V = \left(\dfrac{2}{2},\ \dfrac{5}{2}\right)$
$\therefore \boxed{V = \left(1,\ \dfrac{5}{2}\right)}$
The vertex of the parabola whose focus is (-1,1) and directrix is 4x + 3y - 24 = 0 is
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NIMCET Previous Year PYQ NIMCET NIMCET 2016 PYQ
Solution
Given: Focus $S(-1, 1)$ and directrix $4x + 3y - 24 = 0$The vertex is the midpoint of focus and foot of perpendicular from focus to directrix
Foot of perpendicular from $S(-1, 1)$ to $4x + 3y - 24 = 0$:
$\dfrac{x + 1}{4} = \dfrac{y - 1}{3} = -\dfrac{4(-1) + 3(1) - 24}{4^2 + 3^2}$
$= -\dfrac{-4 + 3 - 24}{16 + 9}$
$= -\dfrac{-25}{25}$
$= 1$
$\therefore x + 1 = 4 \Rightarrow x = 3$
$\therefore y - 1 = 3 \Rightarrow y = 4$
$\therefore$ Foot of perpendicular $Z = (3, 4)$
Since vertex $V$ is midpoint of $SZ$:
$V = \left(\dfrac{-1 + 3}{2},\ \dfrac{1 + 4}{2}\right)$
$V = \left(\dfrac{2}{2},\ \dfrac{5}{2}\right)$
$\therefore \boxed{V = \left(1,\ \dfrac{5}{2}\right)}$
NIMCET PYQ
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NIMCET Previous Year PYQ NIMCET NIMCET 2023 PYQ
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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NIMCET Previous Year PYQ NIMCET NIMCET 2023 PYQ
Solution
NIMCET PYQ
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NIMCET Previous Year PYQ NIMCET NIMCET 2023 PYQ
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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NIMCET Previous Year PYQ NIMCET NIMCET 2023 PYQ
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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