NIMCET Hyperbola Previous Year Questions (PYQs) – Page 1 of 2
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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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Solution
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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Solution
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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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
$a^2-b^2=c^2-(-c^2)$
$a^2-b^2=2c^2$
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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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:
\[ \boxed{a < 0} \]
The angle between the asymptotes of the hyperbola $27x^2 - 9y^2 = 24$ is
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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°
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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Solution
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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Solution
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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Solution
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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Solution
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