datavorous/entrance-exam-dataset huggingface dataset:
A hot body placed in air cools down to a lower temperature. The rate of decrease of temperature is proportional to the temperature difference from the surrounding. The body loses $60 \%$ and $80 \%$ of maximum heat it can loose in time $t_1$ and $t_2$ respectively. The ratio $t_2 / t_1$ will be

Options:
A $\frac{\ln (10)}{\ln (2)}$
B $\frac{\ln (8)}{\ln (6)}$
C $\frac{\ln (1)}{\ln (3)}$
D $\frac{\ln (5)}{\ln \left(\frac{5}{2}\right)}$

The correct answer is:
$\frac{\ln (5)}{\ln \left(\frac{5}{2}\right)}$

Solution:
According to Newton's law of cooling,
$$
-\frac{d T}{d t} \alpha \Delta \mathrm{T} \text { or } \frac{d T}{d t}=-K \Delta \mathrm{T}
$$
Solved the above Eq., we get
$$
\begin{aligned}
T(t) & =T_0+\left(T_1-T_0\right) e^{-k t} \Rightarrow e^{-k t}=\frac{T(t)-T_0}{T_1-T_0} \\
\text { or } \quad t & =\frac{1}{k} \ln \frac{T_1-T_0}{T(t)-T_0}
\end{aligned}
$$
Where, $T_1=$ temperature of body at $t=0$
$T_0=$ temperature of surrounding.
Let $T_1=0^{\circ} \mathrm{C}$
So, $t=\frac{1}{k} \ln \frac{-T_0}{T(t)-T_0}=\frac{1}{k} \ln \frac{-Q_0}{Q(t)-Q_0}[\because Q(t)=m c T(t)]$
According to the question, the body loses $60 \%$ and $80 \%$ of maximum heat in time $t_1$ and $t_2$
$$
\text { then } \quad \begin{aligned}
t_1 & =\frac{1}{k} \ln \left(\frac{-Q_0}{\frac{60}{100} Q_0-Q_0}\right) \\
t_2 & =\frac{1}{k} \ln \left(\frac{-Q_0}{\frac{80}{100} Q_0-Q_0}\right)
\end{aligned}
$$
Divide Eqs. (ii) by (i), we get
$$
\frac{t_2}{t_1}=\frac{\frac{1}{k} \ln \left(\frac{-Q_0}{\frac{80}{100} Q_0-Q_0}\right)}{\frac{1}{k} \ln \left(\frac{-Q_0}{100-Q_0}\right)}
$$
$$
\Rightarrow \frac{t_2}{t_1}=\frac{\ln \left(\frac{-Q_0}{\left.\frac{80 Q_0-100 Q_0}{100}\right)}\right.}{\ln \left(\frac{-Q_0}{\left.\frac{60 Q_0-100 Q_0}{100}\right)}\right.}=\frac{\ln \left(\frac{-Q_0 \times 100}{-20 Q_0}\right)}{\ln \left(\frac{Q_0 \times 5}{2 Q_0}\right)}
$$
Hence, the correct option is (4).

Topic: Physics, Thermal Properties of Matter, JEE Main
A house of height \( 100 \mathrm{~m} \) subtends a right angle at the window of an opposite house. If the height of the window be \( 64 \mathrm{~m} \), then the distance between the two houses is

Options:
A \( 48 \mathrm{~m} \)
B \( 36 \mathrm{~m} \)
C \( 54 \mathrm{~m} \)
D \( 72 \mathrm{~m} \)

The correct answer is:
\( 48 \mathrm{~m} \)

Solution:
In ∆DAB, tan θ=64d
⇒ d=64 cot θ 1
In ∆CDE, tan 90°-θ=100-64d
⇒ d=36 tan θ …2
From Equations 1 and 2, we get
d2=36×64 ⇒ d=48 m

Topic: Mathematics, Heights and Distances, JEE Main
A huge circular arc of length 4.4 ly subtends an angle 4s at the centre of the circle. How long it would take for a body to complete 4 revolution if its speed is 8 AU per second?
Given : 1 ly=9.46×1015 m

1 AU=1.5×1011 m

Options:
A 3.5×106 s
B 4.5×1010 s
C 4.1×108 s
D 7.2×108

The correct answer is:
4.5×1010 s

Solution:
ℓ=4.4ly=4.4×9.46×1015

length of Arc. =ℓ=Rθ

4.4×9.46×1015=Rθ

θ=4 S=4×4.843×10-6=1.94×10-5 rad

4.4×9.46×1015=R×1.94×10-5

R=2.1455×1021 meter

Speed =8Au=8×1.5×1011 ms-1=12×1011 ms-1

4 revolution means distance =4×2πR metre

time = distance speed =4×2πR12×1011; time =8×3.14×2.1455×102112×1011⇒4.5×1010sec

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2021 (27 Aug Shift 1)
A human female experiences two major changes, menarche and menopause during her life. Mention the significance of both the events.

Options:

None

Solution:
The initiation of menstruation at puberty at the age between 9-15 years in human beings is called menarche. While around 50 years of age menstrual cycles ceases that i termed as menopause. Menarche is the stage that signifies the maturation and readiness of the female reproductive system for bearing child.
It signifies the capability of the ovaries to produce mature oocyte (female gamete) that can now be fertilised by the sperm and also that the uterus is capable of supporting the foetal growth and development.
Menopause signifies the end of child bearing age. At this age supply of healthy eggs is very low, the levels of the hormones secreted by the ovaries decline and the menstruation stops. The uterus no longer remains conductive for foetal growth.

Topic: Biology, Human Reproduction
A human RBC is placed in $1.5 \%$ salt solution. It will

Options:
A swell up
B shrink
C remain unaffected
D burst.

The correct answer is:
shrink

Solution:
For human $\mathrm{RBC}, 0.9 \%$ salt solution is isotonic. Hence a medium of $1.5 \%$ salt solution would be hypertonic i.e. more concentrated. Hence, the RBC will shrink due to exosmosis.

Topic: Biology, Transport in Plants, AIIMS, AIIMS 2001
A: Hybridization of carbon is sp2 in all its crystalline allotropes.

R: There are alternate double-single bonds in each allotrope of carbon.

Options:
A Both Assertion & Reason are true and the reason is the correct explanation of the assertion.
B Both Assertion & Reason are true but the reason is not the correct explanation of the assertion.
C Assertion is true statement but Reason is false.
D Both Assertion and Reason are false statements.

The correct answer is:
Both Assertion and Reason are false statements.

Solution:
(i) In diamond, all C-atom are sp3 hybridized.
(ii) In graphite hybridization is s p 2

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main
A hydrated solid X on heating initially gives a monohydrated compound Y. Y upon heating above 373 K leads to an anhydrous white powder Z. X and Z, respectively, are;

Options:
A washing soda and soda ash.
B baking soda and soda ash.
C washing soda and dead burnt plaster.
D baking soda and dead burnt plaster.

The correct answer is:
washing soda and soda ash.

Solution:
'X' is washing soda Na2CO3.10H2O which on heating gives a monohydrate Na2CO3.H2O. The monohydrate on further heating gives an anhydrous residue of Na2CO3
∴X and Z are washing soda and soda ash.

Topic: Chemistry, s Block Elements, JEE Main, JEE Main 2019 (10 Apr Shift 2)
A hydraulic automobile lift is designed to lift vehicles of mass 5000 kg. The area of cross section of the cylinder carrying load is 250 cm2. The maximum pressure the smaller piston would have to bear is [Assume g=10 m s-2]

Options:
A 20×106 Pa
B 2×105 Pa
C 200×106 Pa
D 2×106 Pa

The correct answer is:
2×106 Pa

Solution:
The given data is

m=5000 kgg=10 m s-2A=250×10-4 m2

Using the formula of pressure,

P=FA=mgA
=5000×10250×10-4
=2×106 N m-2=2×106 Pa

Topic: Physics, Mechanical Properties of Fluids, JEE Main, JEE Main 2023 (08 Apr Shift 2)
A hydraulic press containing water has two arms with diameters as mentioned in the figure. A force of $10 \mathrm{~N}$ is applied on the surface of water in the thinner arm. The force required to be applied on the surface of water in the thicker arm to maintain equilibrium of water is _____N.

Options:

The correct answer is:
1000

Solution:
$\begin{aligned} & \frac{F_1}{A_1}=\frac{F_2}{A_2} \\ & \frac{F_1}{\pi(7)^2}=\frac{10}{\pi \times(0.7)^2} \\ & F_1=1000 \mathrm{~N}\end{aligned}$

Topic: Physics, Mechanical Properties of Fluids, JEE Main, JEE Main 2024 (05 Apr Shift 2)
A hydrocarbon A CnH2n-4 on ozonolysis gives
CH32CHCH2CHO,OHCCH2CH2CHO and CH3COCH3 .
The value of n is

Options:

The correct answer is:
16

Solution:
Based on the formula of hydrocarbon CnH2n-4 the degree of unsaturation is 3. So, there is two case:
(I) 3 double bonds
(II) 1 double bond and 1 triple bond
Second case can be easily rejected because products of ozonolysis are aldehyde or ketone rather than carboxylic acid.
So the alkene would be:
(CH3)2CHCH2CH=CHCH2CH2CH=CHCH2CH2CH=CCH32
Which on ozonolysis will give products as:
CH32CHCH2CH=CHCH2CH2CH=CHCH2CH2CH=CCH32
→Ozonolysis (CH3)2CHCH2CHO+2OHCCH2CH2CHO+CH3COCH3
The formula of alkene is C16H28
So the value of ‘x’ is 16.

Topic: Chemistry, Hydrocarbons, JEE Main
A hydrocarbon contains $86 \%$ carbon, $488 \mathrm{ml}$ of the hydrocarbon weight $1.68 \mathrm{~g}$ at STP. Then the hydrocarbon is an

Options:
A Alkane
B Alkene
C Alkyne
D Arene

The correct answer is:
Alkene

Solution:
\(\begin{array}{lllll}
\text{Element} & \text{Wt.} & \text{Mole} & \text{Ratio} & \text{Empirical formula} \\
C=86 \% & 12 & 7.1 & 1 & C H_2 \\
H=14 \% & 1 & 14 & 2 & \text { Belongs to alkene } \mathrm{C}_n \mathrm{H}_{2 n}
\end{array}\)

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
A hydrogen atom in ground state is given an energy of $10.2 \mathrm{eV}$. How many spectral lines will be emitted due to transition of electrons?

Options:
A 6
B 3
C 10
D 1

The correct answer is:
1

Solution:
Hydrogen will be in first excited state therefore it will emit one spectral line corresponding to transition b/w energy level 2 to 1

Topic: Physics, Atomic Physics, JEE Main, JEE Main 2024 (09 Apr Shift 2)
A hydrogen atom in its ground state absorbs 10.2eV of energy. The angular momentum of electron of the hydrogen atom will increase by the value of (Given, Planck's constant =6.6×10-34Js).

Options:
A 2.10×10-34Js
B 1.05×10-34Js
C 3.15×10-34Js
D 4.2×10-34Js

The correct answer is:
1.05×10-34Js

Solution:
Energy absorbed is given by,

ΔE=13.6×1n12-1n22

10.2=13.6×1n12-1n22

Therefore, we get, n1=1 & n2=2

The change in angular momentum

ΔL=n2h2π-n1h2π=h2π

=1.05×10-34 J s

Topic: Physics, Atomic Physics, JEE Main, JEE Main 2022 (27 Jun Shift 1)
A hydrogen atom initially in the ground level absorbs a photon which excites it to the n=4 level. Determine the wavelength of the photon.

Options:
A 8.6Ã 10 â7 m
B 9.8Ã 10 â8 m
C 7.4Ã 10 â8 m
D 8.2Ã 10 â8 m

The correct answer is:
9.8Ã 10 â8 m

Solution:
1 Î» =R( 1 n 1 2 â 1 n 2 2 )

1 Î» =1.09Ã 10 7 ( 1 1 2 â 1 4 2 )

=1.09Ã 10 7 ( 16â1 16 )

=1.09Ã 10 7 Ã 15 16

Î»= 16 1.09Ã15Ã 10 7 =9.8Ã 10 â8 âm

Topic: Physics, Atomic Physics, JEE Main
A hydrogen atom, initially in the ground state is excited by absorbing a photon of wavelength $980 Ã…$. The radius of the atom in the excited state, in terms of Bohr radius $a_{0}$, will be:

Options:
A $25 \mathrm{a}_{0}$
B $9 a_{0}$
C $16 \mathrm{a}_{0}$
D None of the above

The correct answer is:
$16 \mathrm{a}_{0}$

Solution:
Energy of photon $=\frac{\mathrm{hc}}{\lambda}=\frac{12500}{980}=12.75 \mathrm{eV}$

Energy of electron in $\mathrm{n}^{\text {th }}$ orbit is given by

$\mathrm{En}=\frac{-13.6}{\mathrm{n}^{2}} \Rightarrow \mathrm{E}_{\mathrm{n}}-\mathrm{E}_{1}=-13.6\left[\frac{1}{\mathrm{n}^{2}} \frac{-1}{1^{2}}\right]$

$\Rightarrow 12.75=13.6\left[\frac{1}{1^{2}} \frac{-1}{\mathrm{n}^{2}}\right] \Rightarrow \mathrm{n}=4$

$\therefore$ Electron will excite to $\mathrm{n}=4$ We know that ' $R^{\prime} \propto n^{2}$

$\therefore$ Radius of atom will be $16 \mathrm{a}_{0}$

Topic: Physics, Atomic Physics, JEE Main, JEE Main 2019 (11 Jan Shift 1)
A hydrogen atom is in ground state. Then to get six lines in emission spectrum, wavelength of incident radiation should be

Options:
A $800 Å$
B $825 Å$
C $975 Å$
D $1025 Å$

The correct answer is:
$975 Å$

Solution:
Number of possible spectral lines emitted when an electron jumps back to ground
$$
\text { state from } \mathrm{n}^{\text {th }} \text { orbit }=\frac{\mathrm{n}(\mathrm{n}-1)}{2}
$$
Here, $\frac{\mathrm{n}(\mathrm{n}-1)}{2}=6 \Rightarrow \mathrm{n}=4$
Wavelength $\lambda$ from transition from $\mathrm{n}=1$ to $\mathrm{n}=4$ is given by,
$$
\frac{1}{\lambda}=R\left(\frac{1}{1}-\frac{1}{4^{2}}\right) \Rightarrow \lambda=\frac{16}{15 R}=975 Å
$$

Topic: Physics, Atomic Physics, JEE Main
A hydrogen atom is in the ground state. Then to get six lines in the emission spectrum, the wavelength of incident radiation should be ( Rydberg constant R =1.1Ã107 m-1)

Options:
A 800 A∘
B 825 A∘
C 975 Aâ
D 1025 A∘

The correct answer is:
975 Aâ

Solution:
Number of possible spectral lines emitted when an electron jumps back to the ground state from nthorbit=n(n−1)2
Here, n(n−1)2=6⇒n=4
Wavelength λ from the transition from n=1 to n=4 is given by,
1λ=R(11−142)⇒λ=1615R=975 A∘

Topic: Physics, Atomic Physics, JEE Main
A hydrogen atom moving at speed v collides with another hydrogen atom kept at rest. Find the minimum value of v for which one of the atoms may get ionised. The mass of a hydrogen atom =1.67Ã10-27 kg. Assume completely inelastic collision.

Options:
A 7.2×104 m s-1
B 2.3×104 m s-1
C 2.3×105 m s-1
D 7.2×105 m s-1

The correct answer is:
7.2×104 m s-1

Solution:
Inelastic collision will take place, if a part of incident kinetic energy is utilised in exciting the atom. Here, one atom is to be ionised, i.e. âE=13.6 eV
Assuming completely inelastic collision.
For an inelastic collision
v1=v2=v
By momentum conservation,
mv=2mV
â V=v2
12mv2=12â2mV2+âE
=mv22+âEâµV=v/2
14mv2=âE
v=4âEm
vmin=4Ã13.6Ã1.6Ã10-191.67Ã10-27
=7.2Ã104 m/s

Topic: Physics, Atomic Physics, JEE Main
A hydrogen sample is prepared in a particular excited state $A$ of quantum number, $n_A=3$. The ground state energy of hydrogen atom is $-|E|$. The photons of energy $\frac{|E|}{12}$ are absorbed in the sample which results in the excitation of some electrons to excited state $B$ of quantum number $n_B$, whose value is

Options:
A 6
B 4
C 5
D 7

The correct answer is:
6

Solution:
Given, energy of hydrogen atom in ground state $=-|E|$ and $n_A=3$
$\therefore$ Energy of electrons in excited state, $n_A$
$(E)_{n_A}=\frac{-|E|}{n_A^2}$
Energy of electrons in excited state, $n_B$
$=(E)_{n B}=\frac{-(E)}{n_B^2}$
When sample absorbs the photon of energy, $\frac{|E|}{12}$, then its electrons reaches from energy state $n_A$ to energy state $n_B$.
$\begin{aligned}
& \text { Hence, }(E)_{n_B}-(E)_{n_A}=\frac{|E|}{12} \\
& \frac{-|E|}{n_B^2}-\left\{\frac{-|E|}{n_A^2}\right\}=\frac{|E|}{12} \\
& \frac{-1}{n_B^2}+\frac{1}{n_A^2}=\frac{1}{12} \Rightarrow-\frac{1}{n_B^2}=\frac{1}{12}-\frac{1}{n_A^2} \\
& =\frac{1}{12}-\frac{1}{9} \Rightarrow-\frac{1}{n_B^2}=-\frac{1}{36} \\
& \Rightarrow n_B^2=36 \\
& \therefore n_B=6
\end{aligned}$

Topic: Physics, Atomic Physics, JEE Main
A hydrogen-like atom (atomic number Z) is in a higher excited state of quantum number n. This excited atom can make a transition to the first excited state by successively emitting two photons of energies 10.20 eV and 17.00 eV respectively. Alternatively, the atom from the same excited state can make a transition to the second excited state by successively emitting two photons of energy 4.25 eV and 5.95 eV respectively. Determine the value of Z. [Ionization energy of hydrogen atom =13.6 eV]

Options:

The correct answer is:
3

Solution:
10.2+17=13.6×Z2122-1n2; 4.25+5.95=13.6×Z2132-1n2
Solving the above two equation we get, Z=3 and n=6.

Topic: Physics, Atomic Physics, JEE Main
A hydrogen-like atom (described by the Bohr model) is observed to emit six wavelengths, originating from all possible transitions between a group of levels. These levels have energies between -0.85 eV and -0.544 eV (including both these values). Find the atomic number of the atom. (Take hc=1240 eV nm, ground state energy of hydrogen atom = -13.6 eV)

Options:

The correct answer is:
3

Solution:
Total 6 lines are emitted. Therefore,
n(n-1)2=6
or n = 4
So, transition is taking place between mth energy state and (m + 3)th energy state.
Em = - 0. 85 eV
Or
-13.6(z2m2)=-0.85
Or
zm= 0. 25 ... (i)
Similarly, Em + 3 = - 0. 544 eV
or -13.6z2(m+3)2 = - 0. 544
or z(m+3)=0.2â¦(ii) ...(ii)
Solving Eqs. (i) and (ii) for z and m, we get
m = 12 and z = 3

Topic: Physics, Atomic Physics, JEE Main
A hydrogen-like atom of atomic number Z is in an excited state of quantum number 2n. It can emit a maximum energy photon of 204 eV. If it makes a transition to the quantum state n, a photon of energy 40.8 eV is emitted. Calculate the atomic number Z. Ground state energy of hydrogen atom is -13.6 eV

Options:

The correct answer is:
4

Solution:
Let ground state energy (in eV) be E1
Then, from the given condition
E2n- E1= 204 eV
Or E 1 4 n 2 - E 1 = 204 eV
Or E 1 1 4 n 2 - 1 = 204 eV ... (i)
And E2n - En = 40. 8 eV
Or E14n2-E1n2=40.8eV
or E 1 = - 3 4 n 2 = 40.8 eV .... (ii)
From Eqs. (i) and (ii),
1 - 1 4 n 3 4 n 2 = 5
Or 1 = 1 4 n 2 + 1 5 4 n 2
Or 4 n 2 = 1
or n = 2
From Eq. (ii),
E 1 = - 4 3 n 2 40.8 eV
=-43(2)2(40.8)eV
or E1 =- 217. 6 eV
E1= - (13. 6) Z2
Therefore Z 2 = E 1 - 1 3 Â· 6 = - 2 1 7 - 6 - 1 3 Â· 6 = 1 6
Therefore Z = 4

Topic: Physics, Atomic Physics, JEE Main
A hydrogen-like neutral species in some excited state A, on absorbing a photon of energy 3.066 eV get excited to a new state B. When the electron from state B returns back, photons of a maximum ten different wavelengths can be observed in which some photons are of energy smaller than 3.066 eV, some are of equal energy and only four photons are having energy greater than 3.066 eV. The ionization energy of this atom is

Options:
A 14.6 eV
B 3.066 eV
C 6.132 eV
D 9.2 eV

The correct answer is:
14.6 eV

Solution:
nn-12=10 ⇒n=5
For the transition, n = 5 to n = 2, photon should be of equal energy
⇒ kZ2122-152=3.066 eV
⇒ kZ2=14.6 eV

Topic: Physics, Atomic Physics, JEE Main
A hyperbola has foci 4,2,2,2 and it passes through P2,4. The eccentricity of the hyperbola is

Options:
A tan⁡3π10
B tan⁡5π12
C tan⁡π3
D tan⁡3π8

The correct answer is:
tan⁡3π8

Solution:
Let 4,2=S1 and 2,2=S2 and

the eccentricity of the hyperbola is e, then

S1S2=2ae and PS1-PS2=2a (where, 2a is the length of the transverse axis)
⇒e=S1S2PS1-PS2=222-2
⇒e=12-1=2+1=tan⁡3π8

Topic: Mathematics, Hyperbola, JEE Main
A hyperbola has its centre at the origin, passes through the point 4, 2 and has transverse axis of length 4 along the x-axis. Then the eccentricity of the hyperbola is:

Options:
A 3
B 32
C 23
D 2

The correct answer is:
23

Solution:
Equation of hyperbola having tranverse axis along x-axis is x2a2-y2b2=1

From the given information, equation of hyperbola can be taken as
x24-y2b2=1, since a=2
Since it passes through 4, 2,
Hence b2=43
Now e=1+b2a2= 1+13=23

Topic: Mathematics, Hyperbola, JEE Main, JEE Main 2019 (09 Jan Shift 2)
A hyperbola having its centre at the origin is passing through the point $(5,2)$ and has transverse axis of length 8 along the $\mathrm{X}$-axis. Then the eccentricity of its conjugate hyperbola is

Options:
A $\frac{\sqrt{13}}{3}$
B $\sqrt{\frac{13}{3}}$
C $\frac{\sqrt{13}}{2}$
D $\sqrt{\frac{13}{2}}$

The correct answer is:
$\frac{\sqrt{13}}{2}$

Solution:
Let equation of hyperbola is $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
Given, $2 \mathrm{a}=8 \Rightarrow \mathrm{a}=4$
It passes through $(5,2)$
Hence, on solving we get, $\mathrm{b}=\frac{8}{3}$
Now, eccentricity of its conjugate hyperbola is
$$
=\sqrt{\frac{\mathrm{a}^2+\mathrm{b}^2}{\mathrm{~b}^2}}=\frac{\sqrt{13}}{2}
$$

Topic: Mathematics, Hyperbola, JEE Main
A hyperbola having the transverse axis of length \( 1 \) unit is confocal with the ellipse \( 3 x^{2}+4 y^{2}=12 . \) The square of length of conjugate axis of hyperbola is

Options:

The correct answer is:
3

Solution:
3x2+4y2=12⇒x24+y23=1

Electricity of ellipse =1−34=12

Hence, the focus of ellipse is 2×12,0=1,0

As the hyperbola is confocal with the ellipse, the focus of the hyperbola is 1,0.

Now for hyperbola ae=1 and 2a=1⇒a=12

ae=12e=1⇒e=2

e=1+b2a2⇒2=1+b214⇒4=1+4b2

b2=34⇒b=32

Length of conjugate axis =2×32=3

∴ 2b2=32=3

Topic: Mathematics, Hyperbola, JEE Main
A hyperbola having the transverse axis of length, 2 has the same foci as that of the ellipse, 3x2+4y2=12 then this hyperbola does not pass through which of the following points?

Options:
A 12,0
B -32,1
C 1,-12
D 32,12

The correct answer is:
32,12

Solution:
Equation of ellipse : x24+y23=1

⇒e=1-34=12

Foci of the ellipse: ±2×12,0 i.e.,±1,0

For hyperbola: Length of the transverse axis=2a=2⇒a=12

Foci of hyperbola ±1,0⇒ae=1

b2=a2e2-a2=1-12=12

Equation of hyperbola :x21/2-x21/2=1

⇒x2-y2=12⇒2x2-2y2=1

Only the point, 32,12 does not satisfy the above equation.

Topic: Mathematics, Hyperbola, JEE Main, JEE Main 2020 (03 Sep Shift 1)
A hyperbola, having the transverse axis of length $2 \sin \theta$, is confocal with the ellipse $3 x^2+4 y^2=12$. Then, its equation is

Options:
A
$x^2 \operatorname{cosec}^2 \theta-y^2 \sec ^2 \theta=1$

B
$x^2 \sec ^2 \theta-y^2 \operatorname{cosec}^2 \theta=1$

C
$x^2 \sin ^2 \theta-y^2 \cos ^2 \theta=1$

D
$x^2 \cos ^2 \theta-y^2 \sin ^2 \theta=1$

The correct answer is:
$x^2 \operatorname{cosec}^2 \theta-y^2 \sec ^2 \theta=1$

Solution:
The given ellipse is
$$
\begin{aligned}
& \frac{x^2}{4}+\frac{y^2}{3}=1 \Rightarrow \alpha=2, \beta=\sqrt{3} \\
& \therefore \quad 3=4\left(1-e^2\right) \Rightarrow e=\frac{1}{2} \\
&
\end{aligned}
$$
$$
\therefore \quad a e=1
$$
Hence the eccentricity $e_1$, of the hyperbola is given by
$$
\begin{array}{rlrl}
& & 1 & =e_1 \sin \theta \\
\Rightarrow & & e_1 & =\operatorname{cosec} \theta \\
\Rightarrow & b^2 & =\sin ^2 \theta\left(\operatorname{cosec}^2 \theta-1\right)=\cos ^2 \theta
\end{array}
$$
Hence, the hyperbola is
$$
\frac{x^2}{\sin ^2 \theta}-\frac{y^2}{\cos ^2 \theta}=1 \text { or } x^2 \operatorname{cosec}^2 \theta-y^2 \sec ^2 \theta=1
$$

Topic: Mathematics, Ellipse, JEE Main
A hyperbola having the transverse axis of length 2 units has the same focii as that of ellipse 3x2+4y2=12, then its equation is

Options:
A 2x2-2y2=1
B 2x2-2y2=3
C x2-y2=-2
D x2-y2=2

The correct answer is:
2x2-2y2=1

Solution:
For the hyperbola x2a2-y2b2=1, length of semi transverse axis a=12,

For the ellipse x24+y23=1, the foci is ±1,0
i.e. ±1=±a2+b2
⇒b2=1-a2=1-12=12

Hence, the equation of the required hyperbola is x212-y212=1

Topic: Mathematics, Hyperbola, JEE Main
A hyperbola passes through the foci of the ellipse x225+y216=1 and its transverse and conjugate axes coincide with major and minor axes of the ellipse, respectively. If the product of their eccentricities is one, then the equation of the hyperbola is:

Options:
A x29-y216=1
B x2-y2=9
C x29-y225=1
D x29-y24=1

The correct answer is:
x29-y216=1

Solution:
For ellipse e1=1-b2a2=35

for hyperbola e2=53

Let hyperbola be

x2a2-y2b2=1

∵ it passes through 3,0⇒9a2=1

⇒a2=9

⇒b2=a2e2-1

=9259-1=16

∴ Hyperbola is

x29-y216=1

option 1

Topic: Mathematics, Hyperbola, JEE Main, JEE Main 2021 (25 Feb Shift 2)
A hyperbola passes through the point P2,3 and has foci at ± 2,0. Then the tangent to this hyperbola at P also passes through the point

Options:
A 32,23
B 22,33
C 3,2
D -2,-3

The correct answer is:
22,33

Solution:
From the standard equation of hyperbola

⇒±ae=± 2

As we know, b2=a2e2-1.

⇒b2=4-a2

∴ Equation of hyperbola is x2a2-y24-a2=1

∵ It passes through 2,3

⇒2a2-34-a2=1

⇒a2=t

⇒24-t-3t-t4-t=0

⇒8-2t-3t-4t+t2=0

⇒t2-8t-t+8=0

⇒ t= 8, 1

So, a2=8, 1.

⇒a=22, 1

But for a=22, b becomes imaginary. So this case is rejected.

∴ Hyperbola is x21-y23=1.

The equation of the tangent at P is 2x-y3=1.

Now checking each points.

Topic: Mathematics, Hyperbola, JEE Main, JEE Main 2017 (02 Apr)
A hyperbola passes through the points $(3,2)$ and $(-17,12)$ and has its centre at origin and transverse axis is along $x$-axis. The length of its transverse axis is

Options:
A $2$
B $4$
C $6$
D None of these

The correct answer is:
$2$

Solution:
Let the equation of hyperbola is $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
But it passes through $(3,2) \Rightarrow \frac{9}{a^2}-\frac{4}{b^2}=1$ ....(i)
Also its passes through $(-17,12)$
$\Rightarrow \quad \frac{(-17)^2}{a^2}-\frac{(12)^2}{b^2}=1$ ....(ii)
Solving these, we get $a=1$ and $b=\sqrt{2}$
Hence length of transverse axis $=2 a=2$.

Topic: Mathematics, Hyperbola, JEE Main