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CUET PG MCA Previous Year Questions (PYQs)
CUET PG MCA Parabola PYQ
CUET PG MCA PYQ
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CUET PG MCA Previous Year PYQ CUET PG MCA CUET 2025 PYQ
If $x^2 =-16y$ is an equation of parabala then:
(A) directrix is y = 4
(B) directrix is x = 4
(C) co-ordinates of focus are (0,- 4)
(D) co-ordinates of focus are (-4,-0)
(E) length of latusrectum =16
Choose the correct answer from the options given below:
1. (A) and (E) only
2. (B), (C) and (E) only
3. (A), (C) and (E) only
4. (B), (D) and (E) only
Go to Discussion
CUET PG MCA Previous Year PYQ CUET PG MCA CUET 2025 PYQ
Solution
Given \(x^{2}=-16y\).
Compare with the standard form \(x^{2}=-4ay\) ⇒ \(4a=16\Rightarrow a=4\).
Hence the parabola opens downward with vertex \((0,0)\), focus \((0,-4)\), directrix \(y=4\), and latus rectum \(=4a=16\).
Checking options:
(A) \(y=4\) ✅,
(B) \(x=4\) ❌,
(C) \((0,-4)\) ✅,
(D) \((-4,0)\) ❌,
(E) \(16\) ✅.
Therefore, the correct choice is \(\boxed{3\text{ — (A), (C), and (E) only}}\).
CUET PG MCA PYQ
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CUET PG MCA Previous Year PYQ CUET PG MCA CUET 2024 PYQ
An equilateral triangle is inscribed in a parabola $y^2=8x$ whose one vertix is at the vertex of the parabola then the length of the side of the triangle is:
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CUET PG MCA Previous Year PYQ CUET PG MCA CUET 2024 PYQ
Solution
Shortcut (Formula):
For a parabola \(y^2=4ax\), an equilateral triangle inscribed with one vertex at the parabola’s vertex has side
\[ s = 8a\sqrt{3}. \]
Here \(y^2=8x \Rightarrow 4a=8 \Rightarrow a=2\). Hence
\[ s = 8\cdot 2\sqrt{3}=16\sqrt{3}. \]
Final Answer: \(16\sqrt{3}\)
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