MCA NIMCET Hyperbola Previous Year Question Paper and Solution with PDF

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MCA NIMCET Hyperbola PYQ

Qus : 1 MCA NIMCET PYQ

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2012 PYQ

Solution

Qus : 2 MCA 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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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2022 PYQ

Solution

Qus : 3 MCA 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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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2024 PYQ

Solution

Qus : 4 MCA 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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MCA NIMCET Previous Year PYQ MCA NIMCET 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

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

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

Qus : 5 MCA 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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MCA NIMCET Previous Year PYQ MCA NIMCET 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 : 6 MCA NIMCET PYQ

The angle between the asymptotes of the hyperbola $27x^2 - 9y^2 = 24$ is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2010 PYQ

Solution

Given hyperbola: 27x² – 9y² = 24 Divide both sides by 24: x²/(24/27) – y²/(24/9) = 1 ⇒ x²/(8/9) – y²/(8/3) = 1 Here: a² = 8/9, b² = 8/3 For hyperbola x²/a² – y²/b² = 1, angle between asymptotes = 2 tan⁻¹(b/a) Compute b/a: b/a = √( (8/3) / (8/9) ) = √3 Angle = 2 tan⁻¹(√3) = 2 × 60° = 120°

Qus : 7 MCA NIMCET PYQ

Equation of the common tangents with a positive slope to the circle $x^2+y^2-8x=0$ and$\dfrac{x^2}{9}-\dfrac{y^2}{4}=1$ is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2018 PYQ

Solution

Qus : 8 MCA NIMCET PYQ

Let $F_1, F_2$ be foci of hyperbola $\frac{{x}^2}{{a}^2}-\frac{{y}^2}{{b}^2}=1$, a>0, b>0, and let O be the origin. Let M be an arbitrary point on curve C and above X-axis and H be a point on $MF_1$ such that $MF_2\perp{{F}}_1{{F}}_2$, $MF_1\perp{{O}}{{H}}$, $|OH|=\lambda |OF_2|$ with $\lambda \in(2/5, 3/5)$, then the range of the eccentricity $e$ is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2025 PYQ

Solution

Qus : 9 MCA 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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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2017 PYQ

Solution

Qus : 10 MCA 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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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2014 PYQ

Solution

Qus : 11 MCA 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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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2014 PYQ

Solution

Qus : 12 MCA 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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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2015 PYQ

Solution

Qus : 13 MCA 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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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2014 PYQ

Solution

Qus : 14 MCA NIMCET PYQ

Find the area of the triangle formed in the right half-plane by the lines $x-y=0$ and $x+y=0$, and a tangent to the hyperbola $x^2-y^2=a^2$, where $a$ is a non-zero number.

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2026 PYQ

Solution

The lines $x-y=0$ and $x+y=0$ are the asymptotes of the hyperbola

$x^2-y^2=a^2$.

For a hyperbola, the area of the triangle formed by any tangent and its two asymptotes is constant.

Area $=a^2$

Qus : 15 MCA NIMCET PYQ

Find foci of the equation $x^2 + 2x – 4y^2 + 8y – 7 = 0$

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MCA NIMCET Previous Year PYQ MCA NIMCET 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) $

Qus : 16 MCA NIMCET PYQ

The foci of the ellipse $\dfrac{x^2}{16} + \dfrac{y^2}{b^2} = 1$ and the hyperbola $\dfrac{x^2}{144} - \dfrac{y^2}{81} = \dfrac{1}{25}$ coincide. Then the value of $b^2$ is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Given ellipse: $\dfrac{x^2}{16} + \dfrac{y^2}{b^2} = 1$ ...(i)
Given hyperbola: $\dfrac{x^2}{144} - \dfrac{y^2}{81} = \dfrac{1}{25}$ ...(ii)
**For Hyperbola:**
Rewriting (ii) in standard form:
$\dfrac{x^2}{\dfrac{144}{25}} - \dfrac{y^2}{\dfrac{81}{25}} = 1$
Here, $a^2 = \dfrac{144}{25}$ and $b^2 = \dfrac{81}{25}$
For hyperbola, $c^2 = a^2 + b^2$
$\Rightarrow c^2 = \dfrac{144}{25} + \dfrac{81}{25} = \dfrac{225}{25} = 9$
$\therefore c = 3$
So, foci of hyperbola $= (\pm 3,\ 0)$
**For Ellipse:**
Here, $a^2 = 16$
For ellipse, $c^2 = a^2 - b^2$
$\Rightarrow c^2 = 16 - b^2$
Since foci of ellipse and hyperbola coincide:
$\Rightarrow 16 - b^2 = 9$
$\Rightarrow b^2 = 16 - 9$
$\therefore \boxed{b^2 = 7}$

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MCA NIMCET Previous Year Questions (PYQs)

MCA NIMCET Hyperbola PYQ

Solution

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

Solution

Solution

Solution

Finding Foci of a Conic

Solution

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