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NIMCET Previous Year Questions (PYQs)

NIMCET Hyperbola PYQ

NIMCET 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
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

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NIMCET 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
If the line $a^2 x + ay +1=0$, for some real number $a$, is normal to the curve $xy=1$ then





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NIMCET Previous Year PYQNIMCET NIMCET 2024 PYQ

Solution

Problem:

The line \[ a^2x + ay + 1 = 0 \] is normal to the curve \[ xy = 1 \]. Find possible values of \( a \in \mathbb{R} \).

Step 1: Slope of Line

Rewrite: \[ y = -a x - \frac{1}{a} \] → slope = \( -a \)

Step 2: Curve Derivative

\[ xy = 1 \Rightarrow \frac{dy}{dx} = -\frac{y}{x} \] Slope of normal = \( \frac{x}{y} \)

Match Slopes

\[ -a = \frac{x}{y} \Rightarrow x = -a y \]

Plug into Curve

\[ xy = 1 \Rightarrow (-a y)(y) = 1 \Rightarrow y^2 = -\frac{1}{a} \]

For real \( y \), we need \( a < 0 \)

✅ Final Answer:

\[ \boxed{a < 0} \]

NIMCET PYQ
Equation of the common tangents with a positive slope to the circle and  is





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NIMCET Previous Year PYQNIMCET NIMCET 2018 PYQ

Solution

NIMCET 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
The locus of the intersection of the two lines $\sqrt{3} x-y=4k\sqrt{3}$ and $k(\sqrt{3}x+y)=4\sqrt{3}$, for different values of k, is a hyperbola. The eccentricity of the hyperbola is:





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NIMCET Previous Year PYQNIMCET NIMCET 2014 PYQ

Solution

NIMCET PYQ
If the foci of the ellipse $b^{2}x^{2}+16y^{2}=16b^{2}$ and the hyperbola $81x^{2}-144y^{2}=\frac{81 \times 144}{25}$ coincide, then the value of $b$, is





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NIMCET Previous Year PYQNIMCET NIMCET 2014 PYQ

Solution

NIMCET PYQ
If $3x + 4y + k = 0$ is a tangent to the hyperbola ,$9x^{2}-16y^{2}=144$ then the value of $K$ is





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NIMCET Previous Year PYQNIMCET NIMCET 2015 PYQ

Solution

NIMCET PYQ
If PQ is a double ordinate of the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ such that OPQ is an equilateral triangle, where O is the centre of the hyperbola, then which of the following is true?





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NIMCET Previous Year PYQNIMCET NIMCET 2014 PYQ

Solution

NIMCET 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

Finding Foci of a Conic

Given Equation: \( x^2 + 2x - 4y^2 + 8y - 7 = 0 \)

Step 1: Complete the square

⇒ \( (x + 1)^2 - 4(y - 1)^2 = 4 \)

Rewriting: \( \frac{(x + 1)^2}{4} - \frac{(y - 1)^2}{1} = 1 \)

This is a horizontal hyperbola with:

  • Center: \( (-1, 1) \)
  • \( a^2 = 4 \), \( b^2 = 1 \)
  • \( c = \sqrt{a^2 + b^2} = \sqrt{5} \)

✅ Foci: \( (-1 \pm \sqrt{5},\ 1) \)


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