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NIMCET Previous Year Questions (PYQs)
NIMCET Hyperbola PYQ
NIMCET PYQ
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NIMCET Previous Year PYQNIMCET NIMCET 2022 PYQ
If the foci of the ellipse $\frac{x^2}{25}+\frac{y^2}{b^2}=1$ and the hyperbola $\frac{x^2}{144}-\frac{y^2}{81}=\frac{1}{25}$ are coincide, then the value of $b^2$
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NIMCET Previous Year PYQNIMCET NIMCET 2022 PYQ
Solution
NIMCET PYQ
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NIMCET Previous Year PYQNIMCET NIMCET 2024 PYQ
At how many points the following curves intersect $\frac{{y}^2}{9}-\frac{{x}^2}{16}=1$ and $\frac{{x}^2}{4}+\frac{{(y-4)}^2}{16}=1$
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NIMCET Previous Year PYQNIMCET NIMCET 2024 PYQ
Solution
NIMCET PYQ
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NIMCET Previous Year PYQNIMCET NIMCET 2022 PYQ
If ${\Bigg{(}\frac{x}{a}\Bigg{)}}^2+{\Bigg{(}\frac{y}{b}\Bigg{)}}^2=1$, $(a{\gt}b)$ and ${x}^2-{y}^2={c}^2$ cut at right angles, then
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NIMCET Previous Year PYQNIMCET NIMCET 2022 PYQ
Solution
If $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and $\frac{x^2}{c^2}+\frac{y^2}{d^2}=1$ are orthogonal.Then
$a^2-b^2=c^2-d^2$
Similarly
If $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and $x^2-y^2=c^2$ are orthogonal.
It means
$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and $\frac{x^2}{c^2}+\frac{y^2}{-c^2}=1$ are orthogonal
Then
$a^2-b^2=c^2-(-c^2)$
$a^2-b^2=2c^2$
NIMCET PYQ
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NIMCET Previous Year PYQNIMCET NIMCET 2018 PYQ
Equation of the common tangents with a positive slope to the circle
and 
is
and isGo to Discussion
NIMCET Previous Year PYQNIMCET NIMCET 2018 PYQ
Solution
NIMCET PYQ
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NIMCET Previous Year PYQNIMCET NIMCET 2017 PYQ
The equation of the hyperbola with centre at the region, length of the transverse axis is 6 and
one focus (0, 4) is
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NIMCET Previous Year PYQNIMCET NIMCET 2017 PYQ
Solution
NIMCET PYQ
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NIMCET Previous Year PYQNIMCET NIMCET 2023 PYQ
Find foci of the equation $x^2 + 2x – 4y^2 + 8y – 7 = 0$
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NIMCET Previous Year PYQNIMCET NIMCET 2023 PYQ
Solution
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