datavorous/entrance-exam-dataset huggingface dataset:
Explain why sexual reproduction in angiosperms is said to take place through double fertilisation and triple fusion. Also draw a labelled diagram of embryo sac to explain the phenomena.

Options:

None

Solution:
An angiospermic plants reproduces sexually by the formation of male and female gametes. The male gamete is a pollen which contains two male nuclei and the female gamete is an egg cell produced in ovule (female gametophyte).
The pollen grains germinate on the stigma of a flower and the results in growth of pollen through the tissues of stigma and style and reach the egg apparatus. The two male gametes are discharged within the embryo sac. One of the male gamete fuses with the egg cell to form a diploid zygote.
This fusion is known as fertilisation or syngamy. The .second male gamete fuses with the diploid secondary nucleus and forms the triploid Primary Endosperm Nucleus (PEN). This fusion is known as triple fusion.
Because of the involvement of two fusion, this event in angiosperms is termed as double fertilisation. The zygote then develops into embryo and PEN develops into endosperm which provides nourishment to the developing embryo.

Topic: Biology, Plant Kingdom
Exponential growth in human population is represented in

Options:
A Lag phase
B Log phase
C Plateau stage
D First stage

The correct answer is:
Log phase

Solution:
The log phase or exponential phase is the geometrical increase in population size owing to the abundance of food and other favourable conditions.

Topic: Biology, Reproductive Health, NEET
Exponential growth occurs when there is

Options:
A a great environmental resistance
B no environmental resistance
C no biotic potential
D a fixed carrying capacity

The correct answer is:
no environmental resistance

Solution:
Exponential growth in a population occurs when the resources are abundant in the environment and organism is not faced by any resistance from the environment like calamities.

Topic: Biology, Organisms and Populations, NEET
Extrusion of second polar body from egg nucleus occurs

Options:
A after entry of sperm but before fertilization.
B after fertilization.
C before entry of sperm into ovum.
D simultaneously with first cleavage.

The correct answer is:
after entry of sperm but before fertilization.

Solution:
The sperm entry stimulates the secondary oocyte to undergo a meiotic II division, which produces the ovum and the second polar body. This event occurs before fertilization.

Topic: Biology, Human Reproduction, NEET, NEET 2019 (All India)
Extrusion of second polar body from egg nucleus occurs:

Options:
A After the entry of sperm, before the completion of fertilization
B After the completion of fertilization
C Before the entry of sperm
D Has no relation to sperm entry.

The correct answer is:
After the entry of sperm, before the completion of fertilization

Solution:
The entry of sperm stimulates the secondary oocyte to resume and complete the suspended meiosis II. This produces a haploid mature ovum and a second polar body. The second polar body immediately degenerates and sperm tail as well.

Topic: Biology, Human Reproduction, NEET
$f:\{1,2,3\} \rightarrow\{4,5\}$ is not a function if it is defined by which one of the following?

Options:
A $\{(2,4),(3,5),(1,5)\}$
B $\{(1,4),(2,4),(3,4)\}$
C $\{(1,4),(2,5),(3,4)\}$
D $\{(1,4),(1,5),(2,4),(2,5),(3,4),(3,5)\}$

The correct answer is:
$\{(1,4),(1,5),(2,4),(2,5),(3,4),(3,5)\}$

Solution:
If the relation is defined by option (d), then each 1,2 and 3 has two images. $\mathrm{So}$, it is not a function.

Topic: Mathematics, Sets and Relations, NDA, NDA 2008 (Phase 2)
f $A$ is a finite set having $n$ elements, then the number of relations which can be defined in $A$ is

Options:
A $2^{n}$
B $n^{2}$
C $2^{n}$
D $n^{n}$

The correct answer is:
$2^{n}$

Solution:
If A is a finite set having $n$ elements, then the number of relations which can be defined in A set is
$2^{n \times n}=2^{n^{2}}$

Topic: Mathematics, Sets and Relations, NDA, NDA 2013 (Phase 2)
$f(\mathrm{x})=\left\{\begin{array}{ll}3 x^{2}+12 x-1 & -1 \leq x \leq 2 \\ 37-x, & 2 < x \leq 3\end{array}\right.$
Which of the following statements are correct?
1- $\mathrm{f}(\mathrm{x})$ is continuous at $\mathrm{x}=2$.
2- $\mathrm{f}(\mathrm{x})$ attains greatest value at $\mathrm{x}=2$.
3- $\mathrm{f}(\mathrm{x})$ is differentiable at $\mathrm{x}=2$. Select the correct answer using the code given below:

Options:
A 1 and 2 only
B 2 and 3 only
C 1 and 3 only
D 1,2 and 3

The correct answer is:
1 and 2 only

Solution:
For continuity at $x=2$. RHL
$\lim _{x \rightarrow 2^{+}} f(x)=\lim _{x \rightarrow 2^{+}} 37-x=37-2=3 \sigma$
LHL
$\lim _{x \rightarrow 2^{-}} f(x)=\lim _{x \rightarrow 2^{-}} 3 x^{2}+12 x-1$
$=3(2)^{2}+12 \times 2-1=12+24-1=35 .$
$f(2)=3 \times 4+12 \times 2-1=12+24-1=35$
So. $\mathrm{RHL}=\mathrm{LHL}$
$\Rightarrow f(x)$ is continuous at $x=2$. For differentiability at $x=2$.
$\mathrm{LHD}=\lim _{x \rightarrow 2^{-}} \frac{f(x)-f(2)}{x-2}=\lim _{h \rightarrow 0} \frac{f(2-h)-f(2)}{2-h-2}$
$=\lim _{h \rightarrow 0} \frac{3(2-h)^{2}+12(2-h)-(12+24-1)}{-h}$
$=\lim _{h \rightarrow 0} \frac{3 h^{2}-24 h}{-h}=\lim _{h \rightarrow 0} 24-3 h=24$
$\mathrm{RHD}=\lim _{x \rightarrow 2^{+}} \frac{f(x)-f(2)}{x-2}=\lim _{h \rightarrow 0} \frac{f(2+h)-f(2)}{2+h-2}$
$=\lim _{h \rightarrow 0} \frac{37-2-h-35}{h}=-1$
$\mathrm{LHD} \neq \mathrm{RHD}$
$\Rightarrow f(x)$ is not differentiable at $x=2$.
To check 2 .
For $x=2$. $f(x)=3 x^{2}+12 x-1$
$=3\left(x^{2}+4 x-\frac{1}{3}\right)=3\left((x+2)^{2}-\frac{13}{3}\right)$
On putting $x=2$
$f(x)=3\left(16-\frac{13}{3}\right)=35$
On putting $x=1$
$f(1)=3\left((1+2)^{2}-\frac{13}{3}\right)=3\left(9-\frac{13}{3}\right)=14$
So $f(x)$ attains greatest value at $x=2$.

Topic: Mathematics, Continuity and Differentiability, NDA, NDA 2016 (Phase 2)
$f(\mathrm{x})=\left\{\begin{array}{ll}3 x^{2}+12 x-1 & -1 \leq x \leq 2 \\ 37-x, & 2 < x \leq 3\end{array}\right.$
Which of the following statements is/are correct?
1- $\quad \mathrm{f}(\mathrm{x})$ is increasing in the interval $[-1,2]$.
2- $\mathrm{f}(\mathrm{x})$ is decreasing in the interval $(2,3]$. Select the correct answer using the code given below:

Options:
A 1 only
B 2 only
C Both 1 and 2
D Neither 1 nor 2

The correct answer is:
Both 1 and 2

Solution:
For $-1 \leq x \leq 2$
$f(x)=3 x^{2}+12 x-1$
$f^{\prime}(x)=6 x+12$
If we take any point in the interval $[-1,2]$ then
$f^{\prime}(1)=6 \times 1+12=18>0$
$\Rightarrow f(x)$ is increasing in the interval $[-1,2]$.
For $2 < x \leq 3$
$f(x)=37-x$
$f^{\prime}(x)=-1 < 0$
$\Rightarrow f(x)$ is decreasing in the interval $(2,3]$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2016 (Phase 2)
$f(x)=\cos x$ is monotonic decreasing under which one of the following conditions? $\quad$

Options:
A $0 < x < \frac{\pi}{2}$ only
B $\frac{\pi}{2} < x < \pi$ only
C $\quad 0 < x < \pi$
D $\quad 0^{2} < x < 2 \pi$

The correct answer is:
$\quad 0 < x < \pi$

Solution:
Given, $f(x)=\cos x$,
$\Rightarrow f^{\prime}(x)=-\sin x$
Now $f^{\prime}(x)=0$
$\Rightarrow-\sin x=0$
$\Rightarrow \sin x=\sin (0)$
$\Rightarrow x=n \pi$
Clearly, $f^{\prime}(x) < 0$, when $0 < x < _{x}$
Hence $f(x)$ is decreasing when $0 < x < \pi$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2008 (Phase 2)
$\mathrm{f}(\mathrm{xy})=\mathrm{f}(\mathrm{x})+\mathrm{f}(\mathrm{y})$ is true for all

Options:
A Polynomial functions $\mathrm{f}$
B Trigonometric functions $\mathrm{f}$
C Exponential functions $\mathrm{f}$
D Logarithmic functions $\mathrm{f}$

The correct answer is:
Logarithmic functions $\mathrm{f}$

Solution:
Let $f(x)=\log x$
$\therefore f(y)=\log y$
$\& f(x y)=\log (x y)=\log x+\log y$
$\Rightarrow \mathrm{f}(\mathrm{xy})=\mathrm{f}(\mathrm{x})+\mathrm{f}(\mathrm{y})$

Topic: Mathematics, Functions, NDA, NDA 2015 (Phase 2)
Fertilization membrane in ovum prevents

Options:
A Meiosis-II
B Polyspermy
C Meiosis-I
D Phase of Multiplication

The correct answer is:
Polyspermy

Solution:
Process of Fertilisation :
I. Entry of a sperm induces changes in the zona pellucida layer of the ovum and blocks the entry of other sperms preventing polyspermy.
II. Acrosomal secretions (including enzymes) help the sperm to enter into the cytoplasm of the ovum through the zona pellucida.
III. Secondary oocyte undergoes completion of meiosis -II, resulting in formation of a ovum (n) and a Second polar body (n).
IV. Nuclei of both gametes fuse to form a zygote.

Topic: Biology, Human Reproduction, NEET
Figure (a) shows a spring of force constant $k$ clamped rigidly at one end and a mass $m$ attached to its free end. A force $\boldsymbol{F}$ applied at the free end stretches the spring. Figure (b) shows the same spring with both ends free and attached to a mass $m$ at either end. Each end of the spring in Fig. (b) is stretched by the same force $F$.
(a) What is the maximum extension of the spring in the two cases?
(b) If the mass in Fig. (a) and the two masses in Fig.
(b) are released, what is the peried of oscillation in each case?

Options:

None

Solution:
(a) Let $y$ be the maximum extension produced in the spring in Fig. (a)
Then $F=k y$ (in magnitude) $\therefore y=\frac{F}{k}$
In Fig. (b), the force on one mass acts as the force of reaction due to the force on the other mass.
Therefore, each mass behaves as if it is fixed with respect to the other.
Therefore, $F=k y \Rightarrow y=\frac{F}{k}$
(b) In Fig. (a), $F=-k y$
$$
\begin{aligned}
&\Rightarrow m a=-k y \Rightarrow a=-\frac{k}{m} y \\
&\therefore \omega^2=\frac{k}{m} \text { i.e., } \omega=\sqrt{\frac{k}{m}}\left[\because a=\omega^2 y\right]
\end{aligned}
$$
Therefore, period $T=\frac{2 \pi}{\omega}=2 \pi \sqrt{\frac{m}{k}}$
In Fig. (b), we may consider that the centre of the system is $\mathrm{O}$ and there are two springs each of length $\frac{l}{2}$ attached to the two masses, each $m$, so that $K^{\prime}$ is the spring factor of each of the springs.
Then, $K^{\prime}=2 k$
$$
\therefore T=2 \pi \sqrt{\frac{m}{K^{\prime}}}=2 \pi \sqrt{\frac{m}{2 k}}
$$

Topic: Physics, Oscillations
Figure shows a potentiometer circuit for comparison of two resistances. The balance point with a standard resistor $\mathrm{R}=10.0 \Omega$ is found to be $58.3 \mathrm{~cm}$, while that with the unknown resistance $X$ is $68.5 \mathrm{~cm}$. Determine the value of $X$. What might you do if you failed to find a balance point with the given cell of emf $E$ ?

Options:

None

Solution:
In this case, $\frac{R}{X}=\frac{l_1}{l_2}$
or $\quad X=R \frac{l_2}{l_1}=\frac{10 \times 68.5}{58.3}=11.75 \Omega$
In case of failure, a high resistance is put in series with the cell $\mathrm{E}$. This will reduce current through $\mathrm{R}$ and $\mathrm{X}$ and potential difference across them will be reduced to values lower than the potential difference across wire $\mathrm{AB}$.

Topic: Physics, Current Electricity
Figure shows a standard two slit arrangement witn slits $\mathrm{S}_1, \mathrm{~S}_2, \mathrm{P}_1, \mathrm{P}_2$ are the two minima points on either side of $\mathrm{P}$ (figure).
At $\mathrm{P}_2$ on the screen, there is a hole and behind $\mathrm{P}_2$ is a second 2-slit arrangement with slits $\mathrm{S}_3, \mathrm{~S}_4$ and a second screen behind them.

Options:
A
There would be no interference pattern on the second screen but it would be lighted

B
The second screen would be totally dark

C
There would be a single bright point on the second screen

D
There would be a regular two slit pattern on the second screen

The correct answer is:
There would be a regular two slit pattern on the second screen

Solution:
Consider the given figure there is a hole at point $\mathbf{P}_2$. By Huygen's principle, wave will propagates from the sources $S_1$ and $S_2$. Each point on the screen will acts as sources of secondary wavelets. Wavefront starting from $\mathrm{P}_2$ reaches at $\mathrm{S}_3$ and $\mathrm{S}_4$ which will again act as two monochromatic or coherent sources.
Hence, there will be always a regular two slit pattern on the second screen.

Topic: Physics, Wave Optics
Fill in the blanks:

(i) The ____a____ lymphoid organs provide the sites for interaction of lymphocytes with antigen, which then proliferate to become _____b_____.

(ii) The thymus is a lobed organ located near the heart and beneath the ____c____.

(iii) Spleen has a large reservoir of ___d___.

Options:
A a-primary, b-antigen sensitive cells, c-collar bone, d-lymphocytes
B a-secondary, b-receptor cells, c-breastbone, d-erythrocytes
C a-primary, b-effector cells, c-breastbone, d-lymphocytes
D a-secondary, b-effector cells, c-breastbone, d-erythrocytes

The correct answer is:
a-secondary, b-effector cells, c-breastbone, d-erythrocytes

Solution:
Lymphoid organs can be primary or secondary. Primary organs are those organs that form and mature the lymphocytes. These are bone marrow and thymus. Bone marrow forms B and T lymphocytes while thymus helps in maturing T lymphocytes.Secondary organs provide the sites for interaction of formed lymphocytes and thus these form effector cells. These organs are spleen, lymph nodes, peyer's patches etc. Thymus is also known as 'throne of immunity' and is located in the chest region located near the heart and beneath the breastbone. Spleen is known as the graveyard of red blood cells and thus is a reservoir of it.

Topic: Biology, Human Health and Diseases, NEET
Find beat frequency if the motion of two particles is given by
y1=0.25sin⁡310t
y2=0.25sin⁡316t

Options:
A 3
B 3π
C 6π
D 6

The correct answer is:
3π

Solution:
The given equations of waves be written as
y1=0.25sin⁡310t …i
And y2=0.25sin⁡316t …ii
Comparing Eqs. (i) and (ii) with the standard wave equation, written as
y=asin⁡ωt …iii
We have, ω1=310
⟹ν1=3102π unit
And ω2=316
⟹ν2=3162π unit
Hence, beat frequency=ν2-ν1
=3162π-3102π=3π unit

Topic: Physics, Waves and Sound, NEET
Find beat frequency if the motion of two particles is given by
y1=0.25sin⁡310t
y2=0.25sin⁡316t

Options:
A 3
B 3π
C 6π
D 6

The correct answer is:
3π

Solution:
The given equations of waves be written as
y1=0.25sin⁡310t …i
And y2=0.25sin⁡316t …ii
Comparing Eqs. (i) and (ii) with the standard wave equation, written as
y=asin⁡ωt …iii
We have, ω1=310
⟹v1=3102π unit
And ω2=316
⟹v2=3162π unit
Hence, beat frequency=v2-v1
=3162π-3102π=3π unit

Topic: Physics, Waves and Sound, NEET
Find incorrect statement w.r.t. pollen grains.

Options:
A Generally spherical measuring about 25-50 micro meters
B Exineless areas in pollen wall are called germ pore
C The vegetative cell is bigger, has abundant food reserves and a large irregularly shaped nucleus.
D The generative cell has abundant cytoplasm and an irregular shaped nucleus

The correct answer is:
The generative cell has abundant cytoplasm and an irregular shaped nucleus

Solution:
The generative cell is small and floats in the cytoplasm of the vegetative cell. It is spindle-shaped with dense cytoplasm and a nucleus.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET
Find ' \(\lambda\) ' if \(\mathbf{a}, \mathbf{b}, \mathbf{c}\) are three non-coplanar vectors such that
\(\left[\begin{array}{ccc}
4 a+3 b-c & 4 a+3 b+2 c \quad a-4 b-c
\end{array}\right]\)
\(=\left(\lambda^2+\lambda+1\right)\left[\begin{array}{lll}a & b & c\end{array}\right]\)

Options:
A \(-7,8\)
B \(-7,-6\)
C \(7,-8\)
D \(-7,-8\)

The correct answer is:
\(7,-8\)

Solution:
\(\begin{array}{rlrl}
\left|\begin{array}{ccc}
4 & 3 & -1 \\
4 & 3 & 2 \\
1 & -4 & -1
\end{array}\right|[\bar{a} \bar{b} \bar{c}] & =\left(\lambda^2+\lambda+1\right)[\bar{a} \bar{b} \bar{c}] \\
4(5)-3(-6)-1(-19) & =\lambda^2+\lambda+1 \\
57 & =\lambda^2+\lambda+1 \\
\Rightarrow & & \lambda^2+\lambda-56 & =0 \\
\Rightarrow & & (\lambda-7)(\lambda+8) & =0 \\
\Rightarrow & & \lambda & =7 \text { (or) }-8
\end{array}\)
Hence, option (c) is correct.

Topic: Mathematics, Vector Algebra, AP EAMCET, AP EAMCET 2020 (18 Sep Shift 2)
Find out the correct relation for the dependance of change in acceleration due to gravity on the angle at the latitude, due to rotation of earth

Options:
A $d g \propto \cos \phi$
B $d g \propto \cos ^2 \phi$
C $d g \propto \cos ^{3 / 2} \phi$
D $d g \propto \frac{1}{\cos \phi}$

The correct answer is:
$d g \propto \cos ^2 \phi$

Solution:
Effective acceleration due to gravity due to rotation earth is given as
$\begin{aligned} & g^{\prime}=g-R \omega^2 \cos ^2 \phi \\ \therefore & \text { Change in acceleration, } d g=R \omega^2 \cos ^2 \phi \\ \therefore & d g \propto \cos ^2 \phi\end{aligned}$

Topic: Physics, Gravitation, AIIMS, AIIMS 2012
Find out the correct statement.

Options:
A Totally unrelated species could also compete for the same resources.
B Resources need not to be limiting for competition to occur.
C In interference competition, feed efficiency of one species might be reduced due to the interfering and inhibitory presence of the other species, even if resources are abundant.
D All are true

The correct answer is:
All are true

Solution:
In some shallow South American lakes, visiting flamingoes and resident fishes compete for the same resource: the food zooplanktons present in the lakes. Resources need not be limiting for competition to occur. In interference competition, the feeding efficiency of one species might be reduced due to the interfering and inhibitory presence of the other species.

Topic: Biology, Organisms and Populations, NEET
Find out the incorrect statements:
(1) Blue-green alga is a prokaryotic cell.
(2) WBC is ameboid in shape.
(3) Mycoplasma is the largest cell.
(4) Microbodies are cell organelles.
(5) Prokaryotes do not exhibit a wide variety of shapes.

Options:
A 2, 4 and 5
B 2, 3 and 5
C 3, 4 and 5
D 3 and 5 only

The correct answer is:
3 and 5 only

Solution:
Mycoplasma is the smallest cell, they are 0.3 microns in diameter. The organization of the prokaryotic cell is fundamentally similar even though prokaryotes exhibits a wide variety of shape and function. The four basic shapes of bacteria are bacilli (rod-like), coccus (spherical), vibrio (comma-shaped) and spirillum (spiral).

Topic: Biology, Biological Classification, NEET
Find out the MISMATCHED pair.

Options:
A Haemophilia - Sex linked recessive
B Cystic fibrosis - Autosomal dominant
C Down's syndrome - Trisomy 21
D Turner's syndrome - X-linked

The correct answer is:
Cystic fibrosis - Autosomal dominant

Solution:
The chromosomal disorder occurs when there is an addition or deletion of the chromosome in the individual. In the case of trisomy of the 21st chromosome causes Down's syndrome. Turner's syndrome is a type of chromosomal disorder occurring due to loss of one X chromosome resulting in the genotype as 44+X or 44+XO. Such females are characterized by stunted height and webbed-neck. They are sterile as ovaries are rudimentary and lack secondary sexual characteristics.
Haemophilia is an X-linked recessive blood disorder. In it, the individual has delayed clotting time for blood which can be fatal. X-lined disorders are inherent disorders due to mutation in the X-chromosome which causes the phenotype to be always expressed in male and in female only in homozygous condition.
Cystic fibrosis is a hereditary genetic autosomal recessive disorder of the respiratory system. It is caused due to mutation in the CFTR gene. This condition occurs in an individual when both the alleles for CFTR are affected.

Topic: Biology, Principles of Inheritance and Variation, NEET
Find out the value of equilibrium constant for the following reaction at 298 K - 2NH3(g) + CO2(g) ⇌ NH2CONH2(aq) + H2O(l) Standard Gibbs energy change, ∆rG at the given temperature is –13.6 kJ mol–1.

Options:
A 2.4 × 102
B 3.6 × 102
C 1.2 × 102
D 4.8 × 102

The correct answer is:
2.4 × 102

Solution:
Correct Option is : (A)
2.4 × 102

Topic: Chemistry, Thermodynamics (C), NEET
Find out the wrongly matched pair.

Options:
A Secondary metabolite â Anthocyanin
B Protein – Insulin
C Chitin – Polysaccharide
D Cellulose – Heteropolymer

The correct answer is:
Cellulose – Heteropolymer

Solution:
Cellulose is the most important structural component of the cell wall of plants. It is a linear polymer of β - D glucose units connected through β - 1 , 4-glycosidic linkage. The linear chains form microfibrillae or bundles of parallel chains held together by hydrogen bonds.

Topic: Chemistry, Biomolecules, NEET
Find the area of the region bounded by the curves $y=x^2+2, y=x, x=0$ and $x=3$.

Options:

None

Solution:
Equation of the parabola is
$$
y=x^2+2 \text { or } x^2=(y-2)
$$
Its vertrx is $(0,2)$ axis is $y$-axis.
Boundary lines are $\mathrm{y}=\mathrm{x}$, $x=0, x=3$. Graphs of the
curve and lines have been
shown in the figure.
Area of the region $\mathrm{PQRO}$
$=$ Area of the region $\mathrm{OAQR}$ - Area of region $\mathrm{OAP}$
$$
=\int_0^3\left(x^2+2\right) d x-\int_0^3 x d x=\left[\frac{x^3}{3}+2 x\right]_0^3-\left[\frac{x^2}{2}\right]_0^3
$$
$=\left[\left(\frac{27}{3}+6\right)-0\right]-\left(\frac{9}{2}-0\right)=\frac{21}{2}$ sq. units.

Topic: Mathematics, Area Under Curves
Find the coordinates of the foci and the vertices, the eccentricity and the length of the latus rectum of the hyperbola
$\frac{y^2}{9}-\frac{x^2}{27}=1$

Options:

None

Solution:
Comparing the equation $\frac{y^2}{9}-\frac{x^2}{27}=1$ with the standard equation.
we have, $a=3, b=3 \sqrt{3}$
and $c=\sqrt{a^2+b^2}=\sqrt{9+27}=\sqrt{36}=6$
Therefore, the coordinates of the foci are $(0, \pm 6)$ and that of vertices are $(0, \pm 3)$. Also, the eccentricity $e=\frac{c}{a}=\frac{5}{4}$ and the length of latus rectum $=\frac{2 b^2}{a}=\frac{2(3 \sqrt{3})^2}{3}=\frac{54}{3}=18$ units

Topic: Mathematics, Conic Sections
Find the coordinates of the foci and the vertices, the eccentricity and the length of the latus rectum of the hyperbola
$\frac{x^2}{16}-\frac{y^2}{9}=1$

Options:

None

Solution:
Comparing the equation $\frac{x^2}{16}-\frac{y^2}{9}=1$ with the standard equation $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$, we have, $a=4, b=3$ and $c=\sqrt{a^2+b^2}=\sqrt{16+9}=\sqrt{25}=5$ Therefore, the coordinates of the foci are $(\pm 5,0)$ and that of vertices are $(\pm 4,0)$.
Also, the eccentricity $e=\frac{c}{a}=\frac{5}{4}$ and the length of latus rectum $=\frac{2 b^2}{a}=\frac{2(3)^2}{4}=\frac{9}{2}$ units.

Topic: Mathematics, Conic Sections
Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola
$49 y^2-16 x^2=784$

Options:

None

Solution:
Dividing the equation by 784 on both sides, we have
$\frac{49 y^2}{a}-\frac{16 x^2}{784}=1 \Rightarrow \frac{y^2}{16}-\frac{x^2}{49}=1$
Comparing the equation with the standard equation $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$, we find that $a=4, b=7$ and $c=\sqrt{a^2+b^2}=\sqrt{16+49}=\sqrt{65}$
Therefore the coordinates of the foci are $(0, \pm \sqrt{65})$ and that of vertices are $(0, \pm 4)$. Also the eccentricity $e=\frac{c}{a}=\frac{\sqrt{65}}{4}$ and the length of latus rectum is $\frac{2 b^2}{a}=\frac{2(7)^2}{4}=\frac{49}{2}$ units.

Topic: Mathematics, Conic Sections
Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse.
$36 x^2+4 y^2=144$

Options:

None

Solution:
The given equation of the ellipse can be written if standard form as
$\frac{36 x^2}{144}+\frac{4 y^2}{144}=1 \text { or } \frac{x^2}{4}+\frac{y^2}{36}=1$
Since the denominator of $x^2 < $ denominator of $y^2$, the major axis is along the $y$-axis. Comparing the given equation with the standard equation
$\frac{x^2}{b^2}+\frac{y^2}{a^2}=1,$
we have $b=2$ and $a=6$.
Also, $c=\sqrt{a^2-b^2}=\sqrt{36-4}=\sqrt{32}=4 \sqrt{2}$
and $e=\frac{c}{a}=\frac{\sqrt{34}}{6}=\frac{4 \sqrt{2}}{6}=\frac{2 \sqrt{2}}{3}$
Hence, the foci, are $(0,4 \sqrt{2})$ and $(0,-4 \sqrt{2})$. Vertices, are $(0,6)$ and $(0,-6)$; the length of the major axis, $2 a$ is 12 units, the length of the minor axis, $2 b$ is 4 units and the eccentricity of the ellipse, $e$ is $\frac{2 \sqrt{2}}{3}$.
The length of latus rectum,
$\frac{2 b^2}{a}$ is $\frac{2\left(2^2\right)}{6}=\frac{4}{3}$ units.

Topic: Mathematics, Conic Sections