NIMCET Hyperbola Previous Year Question Paper and Solution with PDF

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NIMCET Previous Year Questions (PYQs)

NIMCET Hyperbola PYQ

Qus : 1NIMCET PYQ 2022

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

Qus : 2NIMCET PYQ 2024

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

Qus : 3NIMCET PYQ 2022

${\Bigg{(}\frac{x}{a}\Bigg{)}}^2+{\Bigg{(}\frac{y}{b}\Bigg{)}}^2=1$ ,

$(a{\gt}b)$ and

${x}^2-{y}^2={c}^2$ cut at right angles, then

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NIMCET Previous Year PYQNIMCET NIMCET 2022 PYQ

Solution

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and

$\frac{x^2}{c^2}+\frac{y^2}{d^2}=1$ are orthogonal.

Then

$a^2-b^2=c^2-d^2$

Similarly

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ 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

$a^2-b^2=c^2-(-c^2)$

$a^2-b^2=2c^2$

Qus : 4NIMCET PYQ 2024

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:

$\boxed{a < 0}$

Qus : 5NIMCET PYQ 2018

Equation of the common tangents with a positive slope to the circle

and

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NIMCET Previous Year PYQNIMCET NIMCET 2018 PYQ

Solution

Qus : 6NIMCET PYQ 2017

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

Qus : 7NIMCET PYQ 2014

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

Qus : 8NIMCET PYQ 2014

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

Qus : 9NIMCET PYQ 2015

$3x + 4y + k = 0$ 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

Qus : 10NIMCET PYQ 2014

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

Qus : 11NIMCET PYQ 2023

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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NIMCET Previous Year Questions (PYQs)

NIMCET Hyperbola PYQ

Solution

Solution

Solution

Solution

Problem:

Step 1: Slope of Line

Step 2: Curve Derivative

Match Slopes

Plug into Curve

✅ Final Answer:

Solution

Solution

Solution

Solution

Solution

Solution

Solution

Finding Foci of a Conic

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