NIMCET 2027 Parabola Previous Year Questions (PYQs) – Page 2 of 2
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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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4
Solution
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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3
4
Solution
If $x = 1$ is the directrix of the parabola $y^{2} = kx - 8$, then k is:
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4
Solution
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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4
Solution
The vertex of the parabola whose focus is (-1,1) and directrix is 4x + 3y - 24 = 0 is
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4
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 $SZ$:
$V = \left(\dfrac{-1 + 3}{2},\ \dfrac{1 + 4}{2}\right)$
$V = \left(\dfrac{2}{2},\ \dfrac{5}{2}\right)$
$\therefore \boxed{V = \left(1,\ \dfrac{5}{2}\right)}$
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 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
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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4
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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