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AMU MCA Previous Year Questions (PYQs)

AMU MCA Conic Section PYQ

AMU MCA PYQ
Let $G$ be a group having elements $a$ and $b$ such that $O(a)=4$, $O(b)=2$ and $a^3 b = ba$. Then $O(ab)$ is






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AMU MCA Previous Year PYQ AMU MCA AMU MCA 2020 PYQ

Solution

Given:

$a^4 = e$

$b^2 = e$

Also $a^3 b = ba$

Multiply by $a$:

$a^4 b = baa$

$b = baa$

$ab = a(baa) = (ab)^{-1}$

Hence $(ab)^2 = e$

Therefore order of $ab = 2$

AMU MCA PYQ
The plane $ax+by+cz=0$ cuts the cone $yz+zx+xy=0$ in perpendicular lines if





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AMU MCA Previous Year PYQ AMU MCA AMU MCA 2022 PYQ

Solution

AMU MCA PYQ
The tangents at the extremities of a focal chord of a parabola intersect on the directrix at an angle





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AMU MCA Previous Year PYQ AMU MCA AMU MCA 2022 PYQ

Solution

AMU MCA PYQ
The equation $yz + zx + xy = 0$ represents:





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AMU MCA Previous Year PYQ AMU MCA AMU MCA 2025 PYQ

Solution

$yz + zx + xy = 0$ can be factored into linear terms, representing a pair of planes.
AMU MCA PYQ
Condition that the plane $lx + my + nz = p$ should touch the ellipsoid $\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1$ is:





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AMU MCA Previous Year PYQ AMU MCA AMU MCA 2025 PYQ

Solution

For a plane to be tangent to the ellipsoid, the condition is $a^2l^2 + b^2m^2 + c^2n^2 = p^2$.
AMU MCA PYQ
Number of arbitrary constants in the equation of a cone is:





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AMU MCA Previous Year PYQ AMU MCA AMU MCA 2025 PYQ

Solution

General second-degree homogeneous equation representing a cone contains 5 arbitrary constants.
AMU MCA PYQ
The equation of a circular cylinder, whose guiding curve is $ x^2+y^2+z^2=9,\quad x-y+z=3 $ will be





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AMU MCA Previous Year PYQ AMU MCA AMU MCA 2021 PYQ

Solution

Eliminate the plane equation from sphere to get quadratic surface representing cylinder. Correct reduction gives: $ x^2+y^2+z^2+xy+yz-zx-9=0 $
AMU MCA PYQ
The equation of the cone reciprocal to $x^2 + 2y^2 + 3z^2 = 0$ is






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AMU MCA Previous Year PYQ AMU MCA AMU MCA 2020 PYQ

Solution

Reciprocal cone coefficients are reciprocals of given coefficients.

So equation becomes

$\frac{x^2}{1} + \frac{y^2}{1/2} + \frac{z^2}{1/3} = 0$

Multiplying gives

$6x^2 + 3y^2 + 2z^2 = 0$

AMU MCA PYQ
The equation of axis of the conic $\sqrt{ax} + \sqrt{by} = 1$ is






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AMU MCA Previous Year PYQ AMU MCA AMU MCA 2020 PYQ

Solution


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