datavorous/entrance-exam-dataset huggingface dataset:
The equation of a sphere is $x^{2}+y^{2}+z^{2}-10 z=0$. If one end point of a diameter of the sphere is $(-3,-4,5)$, what is the other end point?

Options:
A $(-3,-4,-5)$
B $(3,4,5)$
C $(3,4,-5)$
D $(-3,4,-5) \quad$

The correct answer is:
$(3,4,5)$

Solution:
The equation of the given sphere is $\mathrm{x}^{2}+\mathrm{y}^{2}+\mathrm{z}^{2}-10 \mathrm{z}=0$
Its centre is $(0,0,5)$. Coordinates of one end point of a diameter of the sphere is given as $(-3,-4,5)$ Let Coordinates of another end point of this diameter $\left(\mathrm{x}_{1}, \mathrm{y}_{1}, \mathrm{z}_{1}\right)$
$\frac{-3+\mathrm{x}_{1}}{2}=0 \quad \Rightarrow \mathrm{x}_{1}=3$
$\frac{-4+y_{1}}{2}=0 \quad \Rightarrow y_{1}=4$
and $\frac{5+z_{1}}{2}=5 \Rightarrow z_{1}=5$
$\therefore \quad$ Required coordinates are $(3,4,5)$.

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2007 (Phase 2)
The equation of a straight line which cuts off an intercept of 5 units on negative direction of y-axis and makes an angle $120^{\circ}$ with positive direction of $\mathrm{x}$ -axis is

Options:
A $\mathrm{y}+\sqrt{3} \mathrm{x}+5=0$
B $y-\sqrt{3} x+5=0$
C $y+\sqrt{3} x-5=0$
D $y-\sqrt{3} x-5=0$

The correct answer is:
$\mathrm{y}+\sqrt{3} \mathrm{x}+5=0$

Solution:
Slope of the required line, $\mathrm{m}=\tan 120^{\circ}=-\sqrt{3}$ (since it is making angle with positive direction of $\mathrm{x}$ -axis). Given, line cuts off an intercept of 5 units on negative direction of y-axis. This means line is intersecting $\mathrm{y}$ axis at $(0,-5)$
$\mathrm{P}=(0,-5)$ and $\mathrm{m}=-\sqrt{3}$.
Equation of line is $\mathrm{y}-(-5)=-\sqrt{3}(\mathrm{x}-0)$
$\Rightarrow \mathrm{y}+5=-\sqrt{3} \mathrm{x}$
$\Rightarrow \mathrm{y}+\sqrt{3} \mathrm{x}+5=0$

Topic: Mathematics, Straight Lines, NDA, NDA 2017 (Phase 2)
The equation of a straight line which makes an angle $45^{\circ}$ with the $\mathrm{x}$ -axis with y-intercept 101 units is:

Options:
A $10 \mathrm{x}+101 \mathrm{y}=1$
B $101 \mathrm{x}+\mathrm{y}=1$
C $x+y-101=0$
D $x-y+101=0$

The correct answer is:
$x-y+101=0$

Solution:
The equation of the required line is, $\mathrm{y}=\mathrm{mx}+\mathrm{c}$$\ldots \ldots$ (i)
where $m=\tan 45^{\circ}=1$
and $\mathrm{c}=\mathrm{y}-$ intercept $=101$ units
from (i) $\mathrm{y}=\mathrm{x}+101 \Rightarrow \mathrm{x}-\mathrm{y}+101=0$

Topic: Mathematics, Straight Lines, NDA, NDA 2012 (Phase 2)
The equation of a transverse wave is given by $y=0.05 \sin \pi(2 t-0.02 x)$, where $x, y$ are in metre and $t$ is in second. The minimum distance of separation between two particles which are in phase and the wave velocity are respectively

Options:
A $50 \mathrm{~m}, 50 \mathrm{~ms}^{-1}$
B $100 \mathrm{~m}, 100 \mathrm{~ms}^{-1}$
C $50 \mathrm{~m}, 100 \mathrm{~ms}^{-1}$
D $100 \mathrm{~m}, 50 \mathrm{~ms}^{-1}$

The correct answer is:
$100 \mathrm{~m}, 100 \mathrm{~ms}^{-1}$

Solution:
Given, $y=0.05 \sin \pi(2 t-0.02 x)$
or $\quad y=0.05 \sin (2 \pi t-0.02 \pi x)$
On comparing this equation with standard equation
$y=a \sin (\omega t-k x)$
where $\omega=\frac{2 \pi}{T}$ and $K=\frac{2 \pi}{\lambda}$, we have
$\frac{2 \pi}{\lambda}=0.02 \pi \Rightarrow \lambda=100 \mathrm{~m}$
Also, $\quad \omega=\frac{2 \pi}{T}=2 \pi$
or $\quad \frac{1}{T}=1 \Rightarrow v=1$
So, $v=r \lambda=1 \times 100=100 \mathrm{~m} / \mathrm{s}$

Topic: Physics, Waves and Sound, KCET, KCET 2013
The equation of a transverse wave is given by y=10sinπ0.01x-2t where x and y are in cm and t is in seconds. Its frequency is

Options:
A 10 s-1
B 2 s-1
C 1 s-1
D 0.01 s-1

The correct answer is:
1 s-1

Solution:
Comparing with the standard equation, y=Asinkx-ωt, we have
ω=2π rad/s

Frequency n=ω2π=2π2π=1 s-1
⇒n=1 s-1

Topic: Physics, Waves and Sound, NEET
The equation of a wave is $y=5 \sin \left(\frac{t}{0.04}-\frac{x}{4}\right)$; where, $x$ is in cm and $t$ is in second. The maximum velocity of the wave will be

Options:
A $1 \mathrm{~ms}^{-1}$
B $2 \mathrm{~ms}^{-1}$
C $1.5 \mathrm{~ms}^{-1}$
D $1.25 \mathrm{~ms}^{-1}$

The correct answer is:
$1.25 \mathrm{~ms}^{-1}$

Solution:
Equation of wave
$y=5 \sin \left(\frac{t}{0.04}-\frac{x}{4}\right)$
The standard equation of a wave in the given form is
$y=a \sin \left(\omega t-\frac{2 \pi x}{\lambda}\right)$
Comparing the given equation with the standard equation, we get
$a=5 \text { and } \omega=\frac{1}{0.04}=25$
Therefore, maximum velocity of particles of the medium,
$v_{\max }=a \omega=5 \times 25=125 \mathrm{cms}^{-1}=1.25 \mathrm{~ms}^{-1}$

Topic: Physics, Waves and Sound, JIPMER, JIPMER 2010
The equation of an ellipse in its standard form, given the distance between its foci is 2 units and the length of its latusrectum is $\frac{15}{2}$ units, is

Options:
A $15 x^2+4 y^2=15$
B $4 x^2+15 y^2=60$
C $15 x^2+16 y^2=240$
D $16 x^2+15 y^2=40$

The correct answer is:
$15 x^2+16 y^2=240$

Solution:
Standard form of ellipse, $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
a > b
$\because$ Distance between foci $=2 a e$
2 = 2ae
$\Rightarrow \quad a e=1$...(ii)
$\begin{aligned} & \text { and length of latusrectum }=\frac{2 b^2}{a} \\ & \qquad \begin{aligned} \frac{15}{2} & =\frac{2 b^2}{a} \Rightarrow b^2=\frac{15 a}{4} \\ \because \quad b^2 & =a^2\left(-e^2+1\right)=-a^2 e^2+a^2 \\ \frac{15 a}{4} & =-1+a^2\end{aligned}\end{aligned}$
$\begin{aligned} & \Rightarrow \quad 4 a^2-15 a-4=0 \\ & 4 a^2-16 a+a-4=0 \\ & 4 a(a-4)+(a-4)=0 \\ &(4 a+1)(a-4)=0 \\ & \Rightarrow \quad a=-1 / 4 \text { and } 4\end{aligned}$
When, a = + 4
$$
\Rightarrow \quad b^2=\frac{15 a}{4}=\frac{15 \times 4}{4} \Rightarrow b=\sqrt{15}
$$
$\therefore$ Equation of ellipse, $\frac{x^2}{16}+\frac{y^2}{15}=1$
$$
\Rightarrow \quad 15 x^2+16 y^2=240
$$

Topic: Mathematics, Ellipse, AP EAMCET, AP EAMCET 2021 (25 Aug Shift 2)
The equation of motion of a particle performing linear S.H.M is $x=5 \sin \left[4 t-\frac{\pi}{6}\right]$, where $x$ is its displacement in $\mathrm{cm}$. The velocity of the particle when its displacement is $3 \mathrm{~cm}$, is

Options:
A $8 \mathrm{~cm} / \mathrm{s}$
B $6 \mathrm{~cm} / \mathrm{s}$
C $16 \mathrm{~cm} / \mathrm{s}$
D $10 \mathrm{~cm} / \mathrm{s}$

The correct answer is:
$16 \mathrm{~cm} / \mathrm{s}$

Solution:
Comparing with standard equation:
$x=a \sin [\omega t+\phi]$
The amplitude of the wave is $a=5$ and the angular velocity of the wave is $\omega=4$ The velocity of the wave can be obtained by taking derivative as:
$\begin{aligned} & \frac{d x}{d t}=\omega a \cos [\omega t+\phi]=\omega \sqrt{a^2-x^2} \\ & \therefore v=\omega \sqrt{a^2-y^2}=4 \sqrt{(5)^2-(3)^2}=16 \mathrm{~cm} / \mathrm{s}\end{aligned}$

Topic: Physics, Oscillations, MHT CET, MHT CET 2022 (06 Aug Shift 2)
The equation of simple harmonic motion of a particle is a+4π2x=0 where a is the instantaneous linear acceleration at displacement x . The frequency of motion is

Options:
A 1 Hz
B 4π Hz
C 14 Hz
D 4 Hz

The correct answer is:
1 Hz

Solution:
Comparing a+4π2x=0 with standard equation of simple harmonic motion a+ω2x=0
ω2=4π2,ω=2π
2πn=2π,n=1 Hz

Topic: Physics, Oscillations, NEET
The equation of simple harmonic wave produced in the string under tension $0.4 \mathrm{~N}$ is given by $\mathrm{y}=4 \sin (3 \mathrm{x}+60 \mathrm{t}) \mathrm{m}$. The mass per unit length of the string is

Options:
A $10^{-3} \mathrm{~kg} \mathrm{~m}^{-1}$
B $10^{-5} \mathrm{~kg} \mathrm{~m}^{-1}$
C $10^{-3} \mathrm{~g} \mathrm{~cm}^{-1}$
D $10^{-5} \mathrm{~g} \mathrm{~cm}^{-1}$

The correct answer is:
$10^{-3} \mathrm{~kg} \mathrm{~m}^{-1}$

Solution:
The standard equation of a wave can be written as
$$
\mathrm{y}=\mathrm{A} \sin (\mathrm{kx}+\omega \mathrm{t})
$$
Speed of wave $\mathrm{V}=\frac{\omega}{\mathrm{k}}=\frac{60}{3}=20 \mathrm{~m} / \mathrm{s}$
Also, $V=\sqrt{\frac{T}{m}}$
$$
\therefore \mathrm{m}=\frac{\mathrm{T}}{\mathrm{V}^2}=\frac{0.4}{400}=10^{-3} \mathrm{~kg} \mathrm{~m}^{-1}
$$

Topic: Physics, Waves and Sound, MHT CET, MHT CET 2021 (21 Sep Shift 2)
The equation of sound wave is y=0.0015 sin(62.4x+316t ). The wavelength of this wave is

Options:
A 0.2 unit
B 0.1 unit
C 0.3 unit
D 0.4 unit

The correct answer is:
0.1 unit

Solution:
Sound wave equation is,
y=0.0015 sin(62.4x + 316t)
Standard equation of wave motion is,
y = A sin 2 π x λ + t T
Comparing the given equation with the standard equation, we get
2πλ=62.4
or, the wavelength of the wave, λ=2π62.4=0.1 unit

Topic: Physics, Waves and Sound, NEET
The equation of stationary wave along a stretched string is given by $y=5 \sin \frac{\pi x}{3} \cos 40 \pi t$, where, $x$ and $y$ are in cm and $t$ in second. The separation between two adjacent nodes is :

Options:
A 1.5 cm
B 3 cm
C 6 cm
D 4 cm

The correct answer is:
3 cm

Solution:
Given, equation
$y=5 \sin \frac{\pi x}{3} \cos 40 \pi t$
Comparing with standard equation $y=A \sin k x \cos \omega t$
$\begin{array}{rlrl}\therefore & k & =\frac{\pi}{3} \\ \Rightarrow & \frac{2 \pi}{\lambda} & =\frac{\pi}{3} \\ \lambda & =6 \mathrm{~cm}\end{array}$
Separation between two consecutive nodes $=\frac{\lambda}{2}=\frac{6}{2}=3 \mathrm{~cm}$

Topic: Physics, Waves and Sound, JIPMER, JIPMER 2006
The equation of stationary wave on a string clamped at both ends and vibrating in third harmonic is $Y=0.5 \sin (0.314 x) \cos (600 \pi t)$ where $x$ and $y$ are in $\mathrm{cm}, t$ in second. The length of the vibrating string is

Options:
A $20 \mathrm{~cm}$
B $10 \mathrm{~cm}$
C $40 \mathrm{~cm}$
D $30 \mathrm{~cm}$

The correct answer is:
$30 \mathrm{~cm}$

Solution:
The standard equation of a standing wave is
$y=2 A \sin (k x) \cos (\omega t)$
On comparing with standard equation,
$\omega=600 \pi \mathrm{s}^{-1}$ and $k=\frac{\pi}{10}$
The wavelength of the wave can be obtained by using the wave number: $k=\frac{2 \pi}{\lambda}=\frac{\pi}{10}$
The wavelength is $\lambda=20 \mathrm{~cm}$
Frequency of the third harmonics is three times the fundamental.
Therefore, length of the string is: $L=3\left(\frac{\lambda}{2}\right)=3(10)=30 \mathrm{~cm}$

Topic: Physics, Waves and Sound, MHT CET, MHT CET 2022 (10 Aug Shift 1)
The equation of the circle which cuts the circles
\(\begin{aligned}
& S_1 \equiv x^2+y^2-4=0 \\
& S_2 \equiv x^2+y^2-6 x-8 y+10=0 \\
& S_3 \equiv x^2+y^2+2 x-4 y-2=0
\end{aligned}\)
at the extremities of diameters of these circles is

Options:
A \(x^2+y^2-4 x-6 y-4=0\)
B \(x^2+y^2+4 x-4=0\)
C \(x^2+y^2=25\)
D \(x^2+y^2+x+y+1=0\)

The correct answer is:
\(x^2+y^2-4 x-6 y-4=0\)

Solution:
Let the equation of required circle is
\(S \equiv x^2+y^2+2 g x+2 f y+c=0\)
and equation of given circles
\(\begin{aligned}
& S_1 \equiv x^2+y^2-4=0 \\
& S_2 \equiv x^2+y^2-6 x-8 y+10=0 \\
& S_3=x^2+y^2+2 x-4 y-2=0
\end{aligned}\)
\(\because\) Circle \(S=0\) cuts the circles \(S_1=0, S_2=0\) and \(S_3=0\) at the extermities of the diameters, so common chord of \(S=0\) and \(S_1=0\) passes through the centre of the circle \(S_1=0\), so \(c=-4\)
Similarly \((2 g+6) x+(2 f+8) y-14=0\) passes through \((3,4)\), so \(6 g+8 f+36=0\)
\(\Rightarrow \quad 3 g+4 f+18=0\)...(i)
and \((2 g-2) x+(2 f+4) y-2=0\), passes through \((-1,2)\), so
\(-2 g+4 f+8=0\)...(ii)
From Eqs. (i) and (ii), we get
\((g, f)=(-2,-3)\)
So equation of required circle is
\(x^2+y^2-4 x-6 y-4=0\)

Topic: Mathematics, Circle, AP EAMCET, AP EAMCET 2019 (22 Apr Shift 1)
The equation of the circle which passes through the points $(1,0),(0,-6)$ and $(3,4)$ is

Options:
A $4 x^{2}+4 y^{2}+142 x+47 y+140=0$
B $4 x^{2}+4 y^{2}-142 x-47 y+138=0$
C $4 x^{2}+4 y^{2}-142 x+47 y+138=0$
D $4 x^{2}+4 y^{2}+150 x-49 y+138=0$

The correct answer is:
$4 x^{2}+4 y^{2}-142 x+47 y+138=0$

Solution:
Let $\mathrm{A}=(1,0), \mathrm{B}=(0,-6), \mathrm{C}=(3,4)$
Equation of $\mathrm{AB}$ is $\mathrm{L}: \frac{\mathrm{y}-0}{-6-0}=\frac{\mathrm{x}-1}{0-1}$
$\Rightarrow \frac{y}{-6}=\frac{x-1}{-1} \Rightarrow y=6 x-6$
$\Rightarrow 6 x-y-6=0 .$
Equation of circle (c) with $\mathrm{AB}$ as diameter is $(\mathrm{x}-1)(\mathrm{x}-$ $0)+(y-0)(y+6)=0$
$\Rightarrow x^{2}-x+y^{2}+6 y=0 .$
The system of circle passing through the intersection of the circle $\mathrm{C}$ and the line $\mathrm{L}$ is given by $\mathrm{C}+\mathrm{kL}=0$ $\Rightarrow x^{2}-x+y^{2}+6 y+k(6 x-y-6)=0$
This circle is passing through $(3,4)$. $\therefore(3)^{2}-3+(4)^{2}+6(4)+\mathrm{k}[6(3)-4-6]=0$
$\Rightarrow 9-3+16+24+\mathrm{k}(18-10)=0$
$\Rightarrow 46+8 \mathrm{k}=0 \Rightarrow 8 \mathrm{k}=-46 \Rightarrow \mathrm{k}=\frac{-46}{8}=\frac{-23}{4}$
$\therefore$ Equation of circle is
$x^{2}-x+y^{2}+6 y+\left(\frac{-23}{4}\right)(6 x-y-6)=0$
$\Rightarrow 4 x^{2}-4 x+4 y^{2}+24 y-138 x+23 y+138=0$
$\Rightarrow 4 x^{2}+4 y^{2}-142 x+47 y+138=0 .$

Topic: Mathematics, Circle, NDA, NDA 2017 (Phase 2)
The equation of the circle which touches the axes at a distance 5 from the origin is $y^{2}+x^{2}-2 \alpha x-2 \alpha y+\alpha^{2}=0$. What is the value of $u ?$

Options:
A 4
B 5
C 6
D 7

The correct answer is:
5

Solution:
Coordinates of the centre of given circle $=(\alpha, \alpha)$ and
radius $=\sqrt{(\alpha)^{2}+(\alpha)^{2}-\alpha^{2}}=\sqrt{\alpha^{2}}=\alpha$
$\therefore(\alpha-5)^{2}+(\alpha)^{2}=(\alpha)^{2}$
$\Rightarrow \alpha^{2}+25-10 \alpha=0$
$\Rightarrow(\alpha-5)^{2}=0$
$\Rightarrow \alpha=5$
Then, other root will always real.

Topic: Mathematics, Circle, NDA, NDA 2008 (Phase 2)
The equation of the curve passing through the point
$(-1,-2)$ which satisfies $\frac{\mathrm{dy}}{\mathrm{dx}}=-\mathrm{x}^{2}-\frac{1}{\mathrm{x}^{3}}$ is

Options:
A $17 x^{2} y-6 x^{2}+3 x^{5}-2=0$
B $6 x^{2} y+17 x^{2}+2 x^{5}-3=0$
C $6 x y-2 x^{2}+17 x^{5}+3=0$
D $17 x^{2} y+6 x y-3 x^{5}+5=0$

The correct answer is:
$6 x^{2} y+17 x^{2}+2 x^{5}-3=0$

Solution:
$\frac{d y}{d x}=-x^{2}-\frac{1}{x^{3}}$
$\Rightarrow \quad \int d y=\int\left(-x^{2}-\frac{1}{x^{3}}\right) d x$
$\Rightarrow \quad y=-\frac{x^{3}}{3}+\frac{1}{2 x^{2}}+c$
Putting $(-1,-2)$, we get $-2=\frac{1}{3}+\frac{1}{2}+c \Rightarrow \mathrm{C}=-\frac{17}{6}$
$\therefore \quad y=-\frac{x^{3}}{3}+\frac{1}{2 x^{2}}-\frac{17}{6}$
$\Rightarrow \quad 6 x^{2} y=-2 x^{5}+3-17 x^{2}$
$\Rightarrow 6 x^{2} y+17 x^{2}+2 x^{5}-3=0$

Topic: Mathematics, Differential Equations, NDA, NDA 2018 (Phase 2)
The equation of the ellipse in the standard form whose length of the latus rectum is 4 and whose distance between the foci is 42, is

Options:
A x22+y23=1
B 2x2+y2=8
C x2+2y2=16
D x2+5y2=25

The correct answer is:
x2+2y2=16

Solution:
Given 2ae=42, 2b2a=4

Now b2=2a ...1

ae=22

⇒a2e2=8

⇒a2-b2=8 ∵e2=1-b2a2

⇒a2-2a-8=0

⇒a=4, -2

a=-2 not possible b2≠-4

So a2=16, b2=8

Required ellipse x216+y28=1

Topic: Mathematics, Ellipse, TS EAMCET, TS EAMCET 2020 (09 Sep Shift 1)
The equation of the ellipse whose centre is at origin, major
axis is along $x$ -axis with eccentricity $\frac{3}{4}$ and latus rectum 4
units is

Options:
A $\frac{x^{2}}{1024}+\frac{7 y^{2}}{64}=1$
B $\frac{49 \mathrm{x}^{2}}{1024}+\frac{7 \mathrm{y}^{2}}{64}=1$
C $\frac{7 x^{2}}{1024}+\frac{49 y^{2}}{64}=1$
D $\frac{x^{2}}{1024}+\frac{y^{2}}{64}=1$

The correct answer is:
$\frac{49 \mathrm{x}^{2}}{1024}+\frac{7 \mathrm{y}^{2}}{64}=1$

Solution:
Given, $b^{2}=2 a, c^{2}=\left(\frac{3}{4}\right)^{2} a^{2}=\frac{9}{16} a^{2}$
We know, $a^{2}=b^{2}+c^{2}$
So
$\mathrm{a}^{2}=2 \mathrm{a}+\frac{9}{16} \mathrm{a}^{2} \Rightarrow 16 \mathrm{a}^{2}=32 \mathrm{a}+9 \mathrm{a}^{2} \Rightarrow 7 \mathrm{a}^{2}=32 \mathrm{a}$
$\Rightarrow \mathrm{a}=\frac{32}{7}$
$\therefore b^{2}=\frac{64}{7}$
Equation of ellipse is
$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 \Rightarrow \frac{x^{2}}{\left(\frac{32}{7}\right)^{2}}+\frac{y^{2}}{\frac{64}{7}}=1 \Rightarrow \frac{49 x^{2}}{1024}+\frac{7 y^{2}}{64}=1$

Topic: Mathematics, Ellipse, NDA, NDA 2017 (Phase 2)
The equation of the ellipse whose vertices are $(\pm 5,0)$ and foci at $(\pm 4,0)$ is

Options:
A $\frac{x^{2}}{25}+\frac{y^{2}}{9}=1$
B $\frac{x^{2}}{9}+\frac{y^{2}}{25}=1$
C $\frac{x^{2}}{16}+\frac{y^{2}}{25}=1$
D $\frac{x^{2}}{25}+\frac{y^{2}}{16}=1$

The correct answer is:
$\frac{x^{2}}{25}+\frac{y^{2}}{9}=1$

Solution:
Since vertices of an ellipse are $(\pm a, 0)$ and foci are $(\pm \mathrm{ae}, 0)$
$\therefore \quad a=5$ and $a e=4$
$\Rightarrow e=\frac{4}{5}$
$\Rightarrow \frac{16}{25}=1-\frac{\mathrm{b}^{2}}{25}$
$\Rightarrow \frac{\mathrm{b}^{2}}{25}=\frac{9}{25} \Rightarrow \mathrm{b}^{2}=9$
Hence, Required equation is
$\frac{x^{2}}{25}+\frac{y^{2}}{9}=1$

Topic: Mathematics, Ellipse, NDA, NDA 2013 (Phase 1)
The equation of the line joining the origin to the point of
interesection of the lines $\frac{x}{a}+\frac{y}{b}=1$ and $\frac{x}{b}+\frac{y}{a}=1$ is

Options:
A $x-y=0$
B $x+y=0$
C $\quad x=0$
D $y=0$

The correct answer is:
$x-y=0$

Solution:
$\frac{x}{a}+\frac{y}{b}=1$ ...(i)
and $\frac{x}{b}+\frac{y}{a}=1$ ...(ii)
From solving equations (i) and (ii), we get the intersection point. $b x+a y=a x+b y$
$\Rightarrow(b-a) x=(b-a) y$
$\therefore \quad x=y$ ...(iii)
$\Rightarrow \frac{x}{a}+\frac{x}{b}=1$
$\therefore \quad x(a+b)=a b$
$\therefore \quad \mathrm{x}=\frac{\mathrm{ab}}{\mathrm{a}+\mathrm{b}}$ and $\mathrm{y}=\frac{\mathrm{ab}}{\mathrm{a}+\mathrm{b}}$ from equation (iii)
Now, equation of line joining $(0,0)$ and $\left(\frac{a b}{a+b}, \frac{a b}{a+b}\right)$
Here, slope of line $=1$
$\therefore \quad y=x$
$\therefore \quad x-y=0$
$\therefore \quad$ Option (a) is correct.

Topic: Mathematics, Straight Lines, NDA, NDA 2015 (Phase 1)
The equation of the line passing through the point $(2,3)$ and the point of intersection of lines $2 \mathrm{x}-3 \mathrm{y}+7=0$ and $7 x+4 y+2=0$ is

Options:
A $21 \mathrm{x}+46 \mathrm{y}-180=0$
B $21 x-46 y+96=0$
C $46 \mathrm{x}+21 \mathrm{y}+155=0$
D $46 \mathrm{x}-21 \mathrm{y}-29=0$

The correct answer is:
$21 x-46 y+96=0$

Solution:
Required line is $(2 x-3 y+7)+k(7 x+4 y+2)=0$. Given that this line passes through $(2,3)$ $\therefore(2(2)-3(3)+7)+\mathrm{k}(7(2)+4(3)+2)=0$
$\Rightarrow 4-9+7+\mathrm{k}(14+12+2)=0$
$\Rightarrow 2+\mathrm{k}(28)=0$
$\Rightarrow 28 \mathrm{k}=-2 \Rightarrow \mathrm{k}=\frac{-1}{14}$
Required line
$\Rightarrow(2 x-3 y+7)+\left(\frac{-1}{14}\right)(7 x+4 y+2)=0$
$\Rightarrow 28 x-42 y+98-7 x-4 y-2=0$
$\Rightarrow 21 \mathrm{x}-46 \mathrm{y}+96=0 .$

Topic: Mathematics, Straight Lines, NDA, NDA 2017 (Phase 2)
The equation of the line, the reciprocals of whose intercepts on the axes are $m$ and $n$, is given by

Options:
A $n x+m y=m n$
B $m x+n y=1$
C $\quad m x+n y=m n$
D $\quad m x-n y=1$

The correct answer is:
$m x+n y=1$

Solution:
Let line be $\frac{\mathrm{x}}{\mathrm{a}}+\frac{\mathrm{y}}{\mathrm{b}}=1$
given that $\frac{1}{\mathrm{a}}=\mathrm{m}$ and $\frac{1}{\mathrm{~b}}=\mathrm{n}$
$\mathrm{a}=\frac{1}{\mathrm{~m}}, \mathrm{~b}=\frac{1}{\mathrm{n}}$
equation of line, $m x+n y=1$

Topic: Mathematics, Straight Lines, NDA, NDA 2014 (Phase 1)
The equation of the line, when the portion of it intercepted between the axes is divided by the point $(2,3)$ in the ratio of $3: 2$, is

Options:
A Either $\mathrm{x}+\mathrm{y}=4$ or $9 \mathrm{x}+\mathrm{y}=12$
B Either $\mathrm{x}+\mathrm{y}=5$ or $4 \mathrm{x}+9 \mathrm{y}=30$
C Either $\mathrm{x}+\mathrm{y}=4$ or $\mathrm{x}+9 \mathrm{y}=12$
D Either $\mathrm{x}+\mathrm{y}=5$ or $9 \mathrm{x}+4 \mathrm{y}=30$

The correct answer is:
Either $\mathrm{x}+\mathrm{y}=5$ or $9 \mathrm{x}+4 \mathrm{y}=30$

Solution:
Intercept form of line is $\frac{x}{a}+\frac{y}{b}=1$.
We know, the point which divides a line joining two points $\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)$ and $\left(\mathrm{x}_{2}, \mathrm{y}_{2}\right)$ in the ratio $\mathrm{m}: \mathrm{n}$ is
$\left(\frac{m x_{2}+n x_{1}}{m+n}, \frac{m y_{2}+n y_{1}}{m+n}\right)$
Case $1: \mathrm{m}: \mathrm{n}=2: 3$
$\therefore(2,3)=\left(\frac{(2)(\mathrm{a})+3(0)}{2+3}, \frac{2(0)+3(\mathrm{~b})}{2+3}\right)$
$\Rightarrow(2,3)=\left(\frac{2 \mathrm{a}}{5}, \frac{3 \mathrm{~b}}{5}\right)$
$\Rightarrow \frac{2 \mathrm{a}}{5}=2 \quad ; \frac{3 \mathrm{~b}}{5}=3$
$\Rightarrow \mathrm{a}=5 ; \mathrm{b}=5$
Equation of line is $\frac{x}{5}+\frac{y}{5}=1 \Rightarrow x+y=5$
Case $2: \mathrm{m}: \mathrm{n}=3: 2$
$\therefore(2,3)=\left(\frac{3(a)+2(0)}{3+2} ; \frac{3(0)+2(b)}{3+2}\right)$
$\Rightarrow(2,3)=\left(\frac{3 \mathrm{a}}{5}, \frac{2 \mathrm{~b}}{5}\right)$
$\Rightarrow \frac{3 a}{5}=2 ; \frac{2 b}{5}=3$
$\Rightarrow \mathrm{a}=\frac{10}{3}, \mathrm{~b}=\frac{15}{2}$
Equation of line is $\frac{\frac{x}{10}}{3}+\frac{y}{\frac{15}{2}}=1$
$\Rightarrow \frac{3 x}{10}+\frac{2 y}{15}=1$
$\Rightarrow 9 \mathrm{x}+4 \mathrm{y}=30$

Topic: Mathematics, Straight Lines, NDA, NDA 2018 (Phase 1)
The equation of the locus of a point which is equidistant from the axes is

Options:
A $y=2 x$
B $x=2 y$
C $y=\pm x$
D $2 y+x=0$

The correct answer is:
$y=\pm x$

Solution:
$\sqrt{(x-0)^{2}+(y+y)^{2}}=\sqrt{(x-x)^{2}+(y-0)^{2}}$
$\Rightarrow \sqrt{(\mathrm{x}-0)^{2}}=\sqrt{(\mathrm{y}-0)^{2}}$
$\Rightarrow \sqrt{\mathrm{x}^{2}}=\sqrt{\mathrm{y}^{2}}$
$\Rightarrow \mathrm{y}=\pm \mathrm{x}$

Topic: Mathematics, Straight Lines, NDA, NDA 2013 (Phase 2)
The equation of the plane passing through the intersection of the planes $2 \mathrm{x}+\mathrm{y}+2 z=9,4 \mathrm{x}-5 \mathrm{y}-4 \mathrm{z}=1$ and the point $(3,2,1)$ is

Options:
A $10 x-2 y+2 z=28$
B $10 x+2 y+2 z=28$
C $10 x+2 y-2 z=28$
D $10 x-2 y-2 z=24$

The correct answer is:
$10 x-2 y+2 z=28$

Solution:
The equation of plane passing through the intersection of planes $2 x+y+2 z=9$ and $4 x-5 y-4 z=1$ is $(2 x+y+2 z-9)+\lambda(4 x-5 y-4 z-1)=0$
Given that this plane passes through $(3,2,1)$ $\Rightarrow 2(3)+2+2(1)-9+\lambda[4(3)-5(2)-4(1)-1]=0$
$\Rightarrow 1+\lambda(-3)=0$
$\Rightarrow \lambda=\frac{1}{3}$
$\therefore$ Equation is $(2 x+y+2 z-9)$
$+\frac{1}{3}(4 x-5 y-4 z-1)=0$
$\Rightarrow 6 x+3 y+6 z-27+4 x-5 y-4 z-1=0$
$\Rightarrow 10 x-2 y+2 z-28=0$
$\Rightarrow 10 x-2 y+2 z=28$

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2019 (Phase 1)
The equation of the plane passing through the line of intersection of the planes $x+y+z=1,2 x+3 y+4 z=7$, and perpendicular to the plane $x-5 y+3 z=5$ is given by

Options:
A $x+2 y+3 z-6=0$
B $x+2 y+3 z+6=0$
C $3 x+4 y+5 z-8=0$
D $3 x+4 y+5 z+8=0$

The correct answer is:
$x+2 y+3 z-6=0$

Solution:
Given planes, $\mathrm{p}_{1}: \mathrm{x}+\mathrm{y}+\mathrm{z}=1$
$\mathrm{p}_{2}: 2 \mathrm{x}+3 \mathrm{y}+4 \mathrm{z}=7$
So, equation of plane passing through intersection of planes $\mathrm{p}_{1}$ and $\mathrm{p}_{2}$ is $x+y+z-1+k(2 x+3 y+4 z-7)=0$
$\Rightarrow x+y+z-1+2 k x+3 k y+4 k z-7 k=0$
$\Rightarrow x(1+2 k)+y(1+3 k)+z(1+4 k)-1-7 k=0 .$
This is perpendicular to $x-5 y+3 z=5$. $\Rightarrow x-5 y+3 z-5=0$
$\Rightarrow 1(1+2 k)-5(1+3 k)+3(1+4 k)=0$
$\Rightarrow 1+2 k-5-15 k+3+12 k=0$
$\Rightarrow-k-1=0 \Rightarrow k=-1$
$\therefore$ Equation of plane is $x+y+z-1-1(2 x+3 y+4 z-7)=0$
$\Rightarrow x+y+z-1-2 x-3 y-4 z+7=0$
$\Rightarrow-x-2 y-3 z+6=0$
$\Rightarrow x+2 y+3 z-6=0 .$

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2017 (Phase 2)
The equation of trajectory of a projectile is $y=10 x-\left(\frac{5}{9}\right) x^2$.
If we assume $g=10 \mathrm{~ms}^{-2}$, the range of projectile (in metre) is

Options:
A 36
B 24
C 18
D 9

The correct answer is:
18

Solution:
Equation of projectile is
$y=10 x-\left(\frac{5}{9}\right) x^2$
Standard equation is
$y=x \tan \theta-\frac{g}{2 u^2 \cos ^2 \theta} \cdot x^2$
On comparing,
$\tan \theta=10$
and
$\begin{array}{rlrl}
& \frac{g}{2 u^2 \cos ^2 \theta} & =\frac{9}{9} \\
\Rightarrow & 10 u^2 \cos ^2 \theta & =9 \mathrm{~g} \\
& \ddots & g & =10 \mathrm{~m} / \mathrm{s}^2 \\
& \therefore & u^2 \cos ^2 \theta & =9
\end{array}$
(given)
$\therefore$ Range of projectile $R=\frac{2 u^2 \sin \theta \cos \theta}{g}$
$\begin{aligned}
& =\frac{2 u^2 \tan \theta \cdot \cos ^2 \theta}{g} \\
& =\frac{2\left(u^2 \cos ^2 \theta\right) \cdot \tan \theta}{g} \\
& =\frac{2 \times 9 \times 10}{10}=18 \mathrm{~m}
\end{aligned}$

Topic: Physics, Motion In Two Dimensions, AP EAMCET, AP EAMCET 2005