datavorous/entrance-exam-dataset huggingface dataset:
What is the equation of circle which touches the lines $\mathrm{x}=0$, $\mathrm{y}=0$ and $\mathrm{x}=2 ?$

Options:
A $x^{2}+y^{2}+2 x+2 y+1=0$
B $x^{2}+y^{2}-4 x-4 y+1=0$
C $x^{2}+y^{2}-2 x-2 y+1=0$
D None of these

The correct answer is:
$x^{2}+y^{2}-2 x-2 y+1=0$

Solution:
Refer to the figure it is clear that coordinates of centre of circle are $(1,1)$ and diameter of circle $=2$ and hence radius of circle is 1 .
- Equation of circle with centre $(1,1)$ and radius $=1$ is $(x-1)^{2}+(y-1)^{2}=1$
$\Rightarrow x^{2}-2 x+1+y^{2}-2 y+1=1$
$\Rightarrow \quad x^{2}+y^{2}-2 x-2 y+1=0$

Topic: Mathematics, Circle, NDA, NDA 2007 (Phase 2)
What is the equation of line passing through $(0,1)$ and making an angle with the y-axis equal to the inclination of the line $\mathrm{x}-\mathrm{y}=4$ with $\mathrm{x}$ -axis?

Options:
A $y=x+1$
B $x=y+1$
C $2 x=y+2$
D None of the above

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

Solution:
Given line is $\mathrm{x}-\mathrm{y}=4$
slope $=1$ i.e. $\mathrm{m}=1$ Since required line passes through $(0,1)$ $\therefore \mathrm{y}-1=\mathrm{m}(\mathrm{x}-0)$
$\Rightarrow \mathrm{y}-1=1(\mathrm{x}) \Rightarrow \mathrm{y}=\mathrm{x}+1$

Topic: Mathematics, Straight Lines, NDA, NDA 2012 (Phase 1)
What is the equation of parabola whose verted is at $(0,0)$ and focus is at $(0,-2)$ ?

Options:
A $y^{2}+8 x=0$
B $y^{2}-8 x=0$
C $x^{2}+8 y=0$
D $x^{2}-8 y=0$

The correct answer is:
$x^{2}+8 y=0$

Solution:
Focus is $(0,-2)$ $-2$ and parabola is
$x^{2}=-8 y$
$x^{2}=4 a y$
$x^{2}+8 y=0$

Topic: Mathematics, Parabola, NDA, NDA 2014 (Phase 1)
What is the equation of straight line passing through the point $(4,3)$ and making equal intercepts on the coordinate axes?

Options:
A $x+y=7$
B $3 x+4 y=7$
C $x-y=1$
D None of these

The correct answer is:
$x+y=7$

Solution:
Let equation of line be $\frac{x}{a}+\frac{y}{a}=1$ or $x+y=a$ line passing through $(4,3)$, then $\mathrm{a}=0$ Required equation, $x+y=7$

Topic: Mathematics, Straight Lines, NDA, NDA 2014 (Phase 1)
What is the equation of the circle which passes through the points $(3,-2)$ and $(-2,0)$ and having its centre on the line $2 x-y-3=0 ?$

Options:
A $x^{2}+y^{2}+3 x+2=0$
B $x^{2}+y^{2}+3 x+12 y+2=0$
C $x^{2}+y^{2}+2 x=0$
D $x^{2}+y^{2}=5$

The correct answer is:
$x^{2}+y^{2}+3 x+12 y+2=0$

Solution:
Given, centre of circle lies on line $2 x-y-3=0$ Let $x=h$ $2 h-y-3=0 \Rightarrow y=2 n-3$
$\therefore$ centre of circle $=(\mathrm{h}, 2 \mathrm{~h}-3)$
We know, OA $=$ OB $\Rightarrow(\mathrm{h}-3)^{2}+(2 \mathrm{~h}-3+2)^{2}=(\mathrm{h}+2)^{2}+(2 \mathrm{~h}-3)^{2}$
$\Rightarrow \mathrm{h}^{2}-6 \mathrm{~h}+9+(2 \mathrm{~h}-1)^{2}=\mathrm{h}^{2}+4 h+4+4 \mathrm{~h}^{2}-12 \mathrm{~h}+9$
$\Rightarrow y^{2}-6 h+9+4 h^{2}-4 h+1$
$=h^{2}+4 h+4+\not 4 h^{2}-12 h+9$
$\Rightarrow-10 h+10=-8 h+13$
$\Rightarrow-2 h=3 \Rightarrow h=\frac{-3}{2}$
$\therefore$ centre $=\left(\frac{-3}{2}, 2\left(\frac{-3}{2}\right)-3\right)=\left(\frac{-3}{2},-6\right)$
$\therefore$ radius $=(\mathrm{h}+2)^{2}+(2 \mathrm{~h}-3)^{2}$
$=\left(\frac{-3}{2}+2\right)^{2}+\left[2\left(\frac{-3}{2}\right)-3\right]^{2}=\frac{1}{4}+36$
$\therefore$ Equation of circle
$\Rightarrow\left(x+\frac{3}{2}\right)^{2}+(y+6)^{2}=\frac{1}{4}+36$
$\Rightarrow x^{2}+y^{2}+3 x+12 y+2=0$

Topic: Mathematics, Circle, NDA, NDA 2017 (Phase 1)
What is the equation of the curve passing through the origin and satisfying the differential equation $d y=(y \tan x+\sec x) d x ?$

Options:
A $\mathrm{y}=\mathrm{x} \cos \mathrm{x}$
B $y \cos x=x$
C $x y=\cos x \quad$
D $y \sin x=x \quad$

The correct answer is:
$\mathrm{y}=\mathrm{x} \cos \mathrm{x}$

Solution:
The differential equation $\mathrm{dy}=(\mathrm{y} \tan \mathrm{x}+\sec \mathrm{x}) \mathrm{d} \mathrm{x}$ can be written as
$\frac{d y}{d x}=y \tan x+\sec x$
or, $\quad \frac{d y}{d x}-y \tan x=\sec x$
which is of the form $\frac{d y}{d x}+P(x) \cdot y=Q(x)$ Here $\mathrm{P}(\mathrm{x})=-\tan \mathrm{x}$ and $\mathrm{Q}(\mathrm{x})=\sec \mathrm{x}$
Integrating factor $\mathrm{IF}=\mathrm{e}^{\int \mathrm{P}(\mathrm{x}) \mathrm{dx}}$
$I F=e^{\int-\tan x d x}=e^{\int-\frac{\sin x}{\cos x} d x}$
Putting $\cos \mathrm{x}=\mathrm{t}$
$-\sin x d x=d t$
$I f=\mathrm{e}^{\int \frac{\mathrm{dt}}{\mathrm{t}}}=\mathrm{e}^{\log _{\mathrm{e}} \mathrm{t}}=\mathrm{t}=\cos \mathrm{x}$
The solution is
$\mathrm{y} \cdot \mathrm{Q}(\mathrm{x})=\int \mathrm{I} \mathrm{F} \cdot \mathrm{Q}(\mathrm{x}) \mathrm{d} \mathrm{x}+\mathrm{c}$
or, $\mathrm{y} \cdot \sec \mathrm{x}=\int \cos \mathrm{x} \cdot \sec \mathrm{x} \mathrm{d} \mathrm{x}+\mathrm{c}$
or, $\mathrm{y} \cdot \sec \mathrm{x}=\int \mathrm{d} \mathrm{x}+\mathrm{c}$
or, $y \cdot \sec x=x+c$
Since the curve passes through the origin. $0=0+\mathrm{c} \Rightarrow \mathrm{c}=0$
and $y \sec x=x$
or, $\mathrm{y}=\mathrm{x} \cos \mathrm{x}$

Topic: Mathematics, Differential Equations, NDA, NDA 2007 (Phase 1)
What is the equation of the curve passing through the point $\left(0, \frac{\pi}{3}\right)$ satisfying the differential equation $\sin x \cos y d x+\cos x \sin y d y=0 ?$

Options:
A $\cos x \cos y=\frac{\sqrt{3}}{2}$
B $\sin x \sin y=\frac{\sqrt{3}}{2}$
C $\sin x \sin y=\frac{1}{2}$
D $\cos x \cos y=\frac{1}{2}$

The correct answer is:
$\cos x \cos y=\frac{1}{2}$

Solution:
Given differential equation is $\sin x \cos y d x+\cos x \sin y d y=0$
$\Rightarrow \sin x \cos y d x=-\cos x \sin y d y$
$\Rightarrow \frac{\sin x}{\cos x} d x=-\frac{\sin y}{\cos y} d y$
Integrate on both side
$\int \frac{\sin x}{\cos x} d x=-\int \frac{\sin y}{\cos y} d y$
$\Rightarrow-\log (\cos x)=\log (\cos y)+\log c$
where $\log c$ is constant of integration. $\Rightarrow-\log c=\log (\cos y)+\log (\cos x)$
$\frac{1}{c}=\cos y \cos x$
(1)
Since, this curve passing through $\left(0, \frac{\pi}{3}\right)$ $\therefore$ it satisfies equation (1)
So, $\frac{1}{c}=\cos \frac{\pi}{3} \cdot \cos 0$
$\frac{1}{c}=\frac{1}{2} \times 1 \Rightarrow c=2$
Hence, required equation of curve is $\cos x \cos y=\frac{1}{2}$

Topic: Mathematics, Differential Equations, NDA, NDA 2011 (Phase 2)
What is the equation of the curve whose slope at any point is equal to $2 x$ and which passes through the origin?

Options:
A $y(1-x)=x^{2}$
B $y^{2}\left(1+x^{2}\right)=x^{4}$
C $y^{2}=(x+1)^{2}$
D $y=x^{2}$

The correct answer is:
$y=x^{2}$

Solution:
Slope at the point $(x, y)=2 x$ $\therefore \frac{d y}{d x}=2 x$
$\Rightarrow \int d y=\int 2 x d x$
$\Rightarrow y=x^{2}+c$
Given that, it passes through the origin. So, $0=(0)^{2}+c$
$\Rightarrow c=0$
Then, $y=x^{2}$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2008 (Phase 2)
What is the equation of the ellipse having foci $(\pm 2,0)$ and
the eccentricity $\frac{1}{4} ?$

Options:
A $\frac{x^{2}}{64}+\frac{y^{2}}{60}=1$
B $\frac{x^{2}}{60}+\frac{y^{2}}{64}=1$
C $\frac{x^{2}}{20}+\frac{y^{2}}{24}=1$
D $\frac{x^{2}}{24}+\frac{y^{2}}{20}=1$

The correct answer is:
$\frac{x^{2}}{64}+\frac{y^{2}}{60}=1$

Solution:
foci: $(\pm 2,0), \mathrm{e}=\frac{1}{4}$
$\mathrm{c}=2, \mathrm{e}=\frac{1}{4}=\frac{\mathrm{c}}{\mathrm{a}} \Rightarrow \frac{2}{\mathrm{a}}=\frac{1}{4} \Rightarrow \mathrm{a}=8$
We know, $a^{2}-b^{2}=c^{2}$
$\Rightarrow b^{2}=a^{2}-c^{2}=8^{2}-2^{2}=64-4=60$
Eqn of ellipse $\Rightarrow \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$
$\Rightarrow \frac{x^{2}}{64}+\frac{y^{2}}{60}=1$

Topic: Mathematics, Ellipse, NDA, NDA 2017 (Phase 1)
What is the equation of the ellipse whose vertices are $(\pm 5,0)$ and foci are at $(\pm 4,0) ?$

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

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

Solution:
Given.
vertices, $(5,0)$ and $(-5,0)$ foci $(4,0)$ and $(-4,0)$
i.e., $\mathrm{ae}=4$
$\Rightarrow \mathrm{e}=\frac{4}{\mathrm{a}}=\frac{4}{5}$
we know, $b^{2}=a^{2}\left(1-e^{2}\right)$
$=5^{2}\left(1-\left(\frac{4}{5}\right)^{2}\right)$
$=25\left(\frac{25-16}{25}\right)=9$
$\therefore \mathrm{a}^{2}=25, \mathrm{~b}^{2}=9$
Equation of ellipse is $\frac{x^{2}}{25}+\frac{y^{2}}{9}=1$.

Topic: Mathematics, Ellipse, NDA, NDA 2018 (Phase 1)
What is the equation of the hyperbola having latus rectum
and eccentrieity 8 and $\frac{3}{\sqrt{5}}$ respectively?

Options:
A $\frac{x^{2}}{25}-\frac{y^{2}}{20}=1$
B $\frac{x^{2}}{40}-\frac{y^{2}}{20}=1$
C $\frac{x^{2}}{40}-\frac{y^{2}}{30}=1$
D $\frac{x^{2}}{30}-\frac{y^{2}}{25}=1$

The correct answer is:
$\frac{x^{2}}{25}-\frac{y^{2}}{20}=1$

Solution:
Let the equation of hyperbola be $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$
Latus rectum $=8=\frac{2 b^{2}}{a} \Rightarrow b^{2}=4 a$...(i)
$\begin{aligned} & \text { Also, } b^{2}=a^{2}\left(e^{2}-1\right) \\ \Rightarrow & 4 a=a^{2}\left(e^{2}-1\right) \end{aligned} \quad[$ From(i)]
$\Rightarrow 4 a=a^{2}\left[\left(\frac{3}{\sqrt{5}}\right)^{2}-1\right]$
$\Rightarrow \quad a=5 \& b^{2}=20$
Equation is $\frac{x^{2}}{25}-\frac{y^{2}}{20}=1$

Topic: Mathematics, Hyperbola, NDA, NDA 2016 (Phase 2)
What is the equation of the line joining the origin with the point of intersection of the lines $4 x+3 y=12$ and $3 x+4 y=12$ ?

Options:
A $x+y=1$
B $x-y=1$
C $3 y=4 x$
D $x=y$

The correct answer is:
$x=y$

Solution:
The equation of given lines are
$$
4 x+3 y=12 ...(i)
$$
and $3 x+4 y=12$ ...(ii)
On simplifying (i) and (ii), we get
$x=\frac{12}{7}$ and $y=\frac{12}{7}$
Point of intersection of given line is $\left(\frac{12}{7}, \frac{12}{7}\right)$. Hence, the equation of line passing through $(0,0)$ and $\left(\frac{12}{7}, \frac{12}{7}\right)$ is
$\frac{y-0}{x-0}=\frac{\frac{12}{7}-0}{\frac{12}{7}-0} \Rightarrow y=x$

Topic: Mathematics, Straight Lines, NDA, NDA 2011 (Phase 1)
What is the equation of the line mid-way between the lines $3 x-4 y+12=0$ and $3 x-4 y=6$ ?

Options:
A $3 x-4 y-9=0$
B $3 x-4 y+9=0$
C $\quad 3 x-4 y-3=0$
D $3 x-4 y+3=0$

The correct answer is:
$3 x-4 y+3=0$

Solution:
$3 x-4 y+12=0$ or $y=\frac{3}{4} x+3$
$3 x-4 y=6 \quad$ or $\quad y=\frac{3}{4} x-\frac{3}{2}$
Equation of line mid-way between these two lines $\mathrm{y}=\frac{3}{4} \mathrm{x}+\left(\frac{3-\frac{3}{2}}{2}\right)$
$\mathrm{y}=\frac{3}{4} \mathrm{x}+\frac{3}{4}$
$4 y=3 x+3$
$3 x-4 y+3=0$

Topic: Mathematics, Pair of Lines, NDA, NDA 2014 (Phase 1)
What is the equation of the line passing through $\begin{array}{ll}(2,-3) \text { and parallel to Y-axis? } & {}\end{array}$

Options:
A $\quad Y=-3$
B $\quad Y=2$
C $X=2$
D $X=-3$

The correct answer is:
$X=2$

Solution:
Since required line is parallel to $y$ -axis therefore its equation is $x=2 .(\because$ Required line passes through $(2,-3))$

Topic: Mathematics, Straight Lines, NDA, NDA 2011 (Phase 1)
What is the equation of the line passing through the point of intersection of the lines $\mathrm{x}+2 \mathrm{y}-3=0$ and $2 \mathrm{x}-\mathrm{y}+5=0$ and parallel to the line $\mathrm{y}-\mathrm{x}+10=0 ?$

Options:
A $7 x-7 y+18=0$
B $5 x-7 y+18=0$
C $5 x-5 y+18=0$
D $x-y+5=0$

The correct answer is:
$5 x-5 y+18=0$

Solution:
Given lines,
$x+2 y-3=0$ ...(i)
$2 x-y+5=0$ ...(ii)
(1) $\Rightarrow \mathrm{x}+2\left(\frac{11}{5}\right)-3=0$
$\Rightarrow x+\frac{22}{5}-3=0 \Rightarrow x+\frac{7}{5}=0 \Rightarrow x=\frac{-7}{5}$
Given, the required line is parallel to $\mathrm{y}-\mathrm{x}+10=0$. $\Rightarrow \mathrm{y}=\mathrm{x}-10$
$\Rightarrow \mathrm{y}=(1) \mathrm{x}+(-10)$
$\therefore$ slope $(\mathrm{m})=1$
- Required line with slope 1 and passing through
$\left(\frac{-7}{5}, \frac{11}{5}\right)$ is
$\mathrm{y}-\mathrm{y}_{1}=\mathrm{m}\left(\mathrm{x}-\mathrm{x}_{1}\right)$
$\Rightarrow \mathrm{y}-\frac{11}{5}=1\left(\mathrm{x}+\frac{7}{5}\right)$
$\Rightarrow 5 \mathrm{y}-11=5 \mathrm{x}+7 \Rightarrow 5 \mathrm{x}-5 \mathrm{y}+18=0$

Topic: Mathematics, Straight Lines, NDA, NDA 2018 (Phase 1)
What is the equation of the line through $(1,2)$ so that the segment of the line intercepted between the axes is bisected at this point $?$

Options:
A $\quad 2 x-y=4$
B $2 x-y+4=0$
C $2 x+y=4$
D $2 x+y+4=0$

The correct answer is:
$2 x+y=4$

Solution:
$\frac{0+\mathrm{x}}{2}=1 ; \frac{0+\mathrm{y}}{2}=2$
$x=2 ; y=4$
Equation of line passing through $(2,0)$ and $(0,4)$
$\mathrm{y}-0=\frac{4-0}{0-2}(\mathrm{x}-2)$
$y=-2 x+4$
$2 x+y=4$

Topic: Mathematics, Straight Lines, NDA, NDA 2014 (Phase 1)
What is the equation of the line which passes through $(4,-5)$ and is perpendicular to $3 x+4 y+5=0 ?

Options:
A $\quad 4 x-3 y-31=0$
B $3 x-4 y-41=0$
C $\quad 4 x+3 y-1=0$
D $3 x+4 y+8=0$

The correct answer is:
$\quad 4 x-3 y-31=0$

Solution:
$3 \mathrm{x}+4 \mathrm{y}+5=0$ or $\mathrm{y}=\frac{-3}{4} \mathrm{x}+\frac{-5}{4}$
Slope $=\frac{-3}{4}$
Slope of required line, $\mathrm{m}=\frac{-1}{-\frac{3}{4}}=\frac{4}{3}$
Also line passes through $(4,-5)$
Equation of line, $\mathrm{y}+5=\frac{4}{3}(\mathrm{x}-4)$
$\Rightarrow 3 y+15=4 x-16$
$\Rightarrow 4 x-3 y-31=0$

Topic: Mathematics, Straight Lines, NDA, NDA 2013 (Phase 2)
What is the equation of the parabola, whose vertex and focus are on the $x$ -axis at distance a and b from the origin respectively? ( $\mathrm{b}>\mathrm{a}>0$ )

Options:
A $y^{2}=8(b-a)(x-a)$
B $y^{2}=4(b+a)(x-a)$
C $y^{2}=4(b-a)(x+a)$
D $y^{2}=4(b-a)(x-a)$

The correct answer is:
$y^{2}=4(b-a)(x-a)$

Solution:
The parabola's vertex and focus lie on $x$ -axis at points $(\mathrm{a}, 0)$ and $(\mathrm{b}, 0)$. Vertex and focus lie on the $\mathrm{x}$ -axis hence. the axis of parabola is $x$ -axis. Equation of parabola Vertex whose is a point $\left(x_{1}, y_{1}\right)$ then is $\left(y-y_{1}\right)^{2}=4 k\left(x-x_{1}\right)$
$\mathrm{So}, \mathrm{y}_{1}=0$ and $\mathrm{x}_{1}=\mathrm{a}$ and $\mathrm{k}=$ distance between focus and vertex $=(\mathrm{b}-\mathrm{a})$ so the equation is $(y-0)^{2}=4(b-a) \cdot(x-a)$
i.e., $y^{2}=4(b-a)(x-a)$

Topic: Mathematics, Parabola, NDA, NDA 2007 (Phase 1)
What is the equation of the plane passing through the point $(1,-1,-1)$ and perpendicular to each of the planes $x-2 y-8 z=0$ and $2 x+5 y-z=0 ?$

Options:
A $\quad 7 x-3 y+2 z=14$
B $2 x+5 y-3 z=12$
C $x-7 y+3 z=4$
D $14 x-5 y+3 z=16$

The correct answer is:
$14 x-5 y+3 z=16$

Solution:
Since, the required plane passing through $(1,-1,-1)$ therefore only equation given in option 'd' satisfied by the point $(1,-1,-1)$
Hence, Required equation of plane is $14 x-5 y+3 z=16$

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2011 (Phase 2)
What is the equation of the plane passing through the points $(-2,6,-6),(-3,10,-9)$ and $(-5,0,-6) ?$

Options:
A $\quad 2 x-y-2 z=2$
B $2 x+y+3 z=3$
C $x+y+z=6$
D $x-y-z=3$

The correct answer is:
$\quad 2 x-y-2 z=2$

Solution:
We know, the equation of plane passing thr
points $\left(x_{1}, y_{1}, z_{1}\right),\left(x_{2}, y_{2}, z_{2}\right)$ and $\left(x_{3}, y_{3}, z_{3}\right)$ is
So, the plane passing through points $(-2,6,-6),(-3$
$10,-9)$ and $(-5,0,-6)$ is $\left|\begin{array}{ccc}x+2 & y-6 & z+6 \\ -1 & 4 & -3 \\ -3 & -6 & 0\end{array}\right|=0$
$\Rightarrow(x+2)(-18)-(y-6)(-9)+(z+6)(6+12)=10$

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2018 (Phase 1)
What is the equation of the plane passing through $\left(\mathrm{x}_{1}, \mathrm{y}_{1}, \mathrm{z}_{1}\right)$ and normal to the line with $ < \mathrm{a}, \mathrm{b}, \mathrm{c}>$ as direction
ratios?

Options:
A $a x+b y+c z=a x_{1}+b y_{1}+c z_{1}$
B $a\left(x+x_{1}\right)+b\left(y+y_{1}\right)+c\left(z+z_{1}\right)=0$
C $a x+b y+c z=0$
D $a x+b y+c z=x_{1}+y_{1}+z_{1}=0$

The correct answer is:
$a x+b y+c z=a x_{1}+b y_{1}+c z_{1}$

Solution:
The equation of the plane passing through $\left(\mathrm{x}_{1}, \mathrm{y}_{1}, \mathrm{z}_{1}\right)$ and normal to the line with $ < \mathrm{a}, \mathrm{b}, \mathrm{c}>$ as direction ratios,
is given by $a\left(x-x_{1}\right)+b\left(y-y_{1}\right)+c\left(z-z_{1}\right)=0$
$a x-a x_{1}+b y-b y_{1}+c z-c z_{1}=0$
$\Rightarrow \quad a x+b y+c z=a x_{1}+b y_{1}+c z_{1}$
which is required equation of plane.

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2007 (Phase 2)
What is the equation of the plane through $z$ -axis and parallel

Options:
A $x \cot \theta+y=0$
B $x \tan \theta-y=0$
C $x+y \cot \theta=0$
D $x-y \tan \theta=0$

The correct answer is:
$x \tan \theta-y=0$

Solution:
Let equation of plane through z-axis is
$a x+b y=0$
It is given that this plane is parallel to the line
$\frac{x-1}{\cos \theta}=\frac{y+2}{\sin \theta}=\frac{z-3}{0}$
Since the plane parallel to the line
$a \cos \theta+b \sin \theta=0$
$\Rightarrow a \cos \theta=-b \sin \theta \Rightarrow a=-b \tan \theta$
$-b \tan \theta x+b y=0$
$\Rightarrow x \tan \theta-y=0(\because b \neq 0)$
which is required equation of plane.

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2010 (Phase 1)
What is the equation of the right bisector of the line segment joining $(1,1)$ and $(2,3)$ ?

Options:
A $2 x+4 y-11=0$
B $2 x-4 y-5=0$
C $2 x-4 y-11=0$
D $x-y+1=0$

The correct answer is:
$2 x+4 y-11=0$

Solution:
Equation of given line is $(y-3)=\left(\frac{3-1}{2-1}\right)(x-2)$
$\Rightarrow y=2 x-1$
Slope $m_{1}=2$
and slope of perpendicular $=-\frac{1}{2}$ The perpendicular is also bisector, therefore it will pass through its mid-point. $\Rightarrow$ Coordinates of mid-point of given line are :
$\left(\frac{2+1}{2}, \frac{3+1}{2}\right)$ or $\left(\frac{3}{2}, 2\right)$
So, equation of perpendicular bisector is:
$(y-2)=-\frac{1}{2}\left(x-\frac{3}{2}\right)$
$\Rightarrow 2 x+4 y-11=0$

Topic: Mathematics, Pair of Lines, NDA, NDA 2016 (Phase 2)
What is the equation of the sphere which has its centre at $(6,-1,2)$ and touches the plane $2 x-y+2 z-2=0 ?$

Options:
A $x^{2}+y^{2}+z^{2}+12 x-2 y+4 z+16=0$
B $x^{2}+y^{2}+z^{2}+12 x-2 y+4 z-16=0$
C $x^{2}+y^{2}+z^{2}-12 x+2 y-4 z+16=0$
D $x^{2}+y^{2}+z^{2}-12 x+2 y-4 z+25=0$

The correct answer is:
$x^{2}+y^{2}+z^{2}-12 x+2 y-4 z+16=0$

Solution:
Given centre of sphere is $(6,-1,2)$ and $e q^{n}$ of plane is $2 x-y+2 z-2=0$
Since, sphere touches the plane therefore $\perp$ distance from centre to the plane is radius of the sphere.
$\therefore$ Radius $=\frac{2(6)-1(-1)+2(2)-2}{\sqrt{4+1+4}}=\frac{15}{3}=5$
$\therefore$ Required equation of sphere is $(x-6)^{2}+(y+1)^{2}+(z-2)^{2}=5^{2}$
$\Rightarrow x^{2}+y^{2}+z^{2}-12 x+2 y-4 z+16=0$

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2009 (Phase 2)
What is the equation of the sphere with unit radius having centre at the origin?

Options:
A $x^{2}+y^{2}+z^{2}=0$
B $x^{2}+y^{2}+z^{2}=1$
C $x^{2}+y^{2}+z^{2}=2$
D $x^{2}+y^{2}+z^{2}=3$

The correct answer is:
$x^{2}+y^{2}+z^{2}=1$

Solution:
$\left(x-x_{1}\right)^{2}+\left(y-y_{1}\right)^{2}+\left(z-z_{1}\right)^{2}=r^{2}$
Centre $(0,0,0)$ and radius $=1$ $(x-0)^{2}+(y-0)^{2}+(z-0)^{2}=(1)^{2}$
$x^{2}+y^{2}+z^{2}=1$

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2013 (Phase 2)
What is the equation of the straight line cutting off an intercept 2 from the negative direction of $y$ -axis and inclined at $30^{\circ}$ with the positive direction of $\mathrm{x}$ -axis?

Options:
A $x-2 \sqrt{3} y-3 \sqrt{2}=0$
B $x+2 \sqrt{3} y-3 \sqrt{2}=0$
C $x+\sqrt{3} y-2 \sqrt{3}=0$
D $x-\sqrt{3} y-2 \sqrt{3}=0$

The correct answer is:
$x-\sqrt{3} y-2 \sqrt{3}=0$

Solution:
Given, angle $(\theta)=30^{\circ}$
$\therefore \mathrm{m}=\tan 30^{\circ}=\frac{1}{\sqrt{3}}$
Point $=(-2,0)$ i.e., $\mathrm{C}=-2$. Slope-intercept form is $\mathrm{y}=\mathrm{mn}+\mathrm{c}$
$\Rightarrow \mathrm{y}=\frac{1}{\sqrt{3}} \mathrm{x}-2$
$\Rightarrow \sqrt{3} \mathrm{y}=\mathrm{x}-2 \sqrt{3}$
$\Rightarrow x-\sqrt{3} y-2 \sqrt{3}=0$

Topic: Mathematics, Straight Lines, NDA, NDA 2018 (Phase 1)
What is the equation of the straight line parallel to $2 \mathrm{x}+3 \mathrm{y}$ $+1=0$ and passes through the point $(-1,2) ?$

Options:
A $2 x+3 y-4=0$
B $2 x+3 y-5=0$
C $x+y-1=0$
D $3 x-2 y+7=0$

The correct answer is:
$2 x+3 y-4=0$

Solution:
The equation of line parallel to $2 x+3 y+1=0$ is $2 x+3 y+K=0$
It is passing through point $(-1,2)$ $\therefore 2(-1)+3(2)+\mathrm{K}=0$
$\Rightarrow-2+6+\mathrm{K}=0 \Rightarrow \mathrm{K}=-4$
Eqn. is $2 \mathrm{x}+3 \mathrm{y}-4=0$

Topic: Mathematics, Straight Lines, NDA, NDA 2017 (Phase 1)
What is the equation of the straight line passing through the point $(2,3)$ and making an intercept on the positive y-axis equal to twice its intercept on the positive x-axis?

Options:
A $\quad 2 x+y=5$
B $2 x+y=7$
C $x+2 y=7$
D $2 x-y=1$

The correct answer is:
$2 x+y=7$

Solution:
Given line passes through $(2,3)$
Intercept form: $\frac{x}{a}+\frac{y}{b}=1$
$\Rightarrow \frac{x}{a}+\frac{y}{2 a}=1$
$\Rightarrow 2 \mathrm{x}+\mathrm{y}=2 \mathrm{a}$
But this passes through $(2,3)$ $\therefore 2 \mathrm{a}=2(2)+3=7$
$\Rightarrow a=\frac{7}{2}$
$\therefore$ Equation of line is $2 \mathrm{x}+\mathrm{y}=2\left(\frac{7}{2}\right)$
$\Rightarrow 2 x+y=7$

Topic: Mathematics, Straight Lines, NDA, NDA 2018 (Phase 1)
What is the equation of the straight line which is perpendicular to $\mathrm{y}=\mathrm{x}$ and passes through $(3,2) ?$

Options:
A $x-y=5$
B $x+y=5$
C $x+y=1$
D $x-y=1$

The correct answer is:
$x+y=5$

Solution:
Given line, $\mathrm{y}=\mathrm{x}$ $\Rightarrow x-y=0$
Slope of this line $=\frac{-1}{-1}=1$ Slope of line perpendicular to this line $=-1$ The perpendicular line passes through $(3,2)$ $\therefore$ Equation is $\mathrm{y}-2=-1(\mathrm{x}-3) \Rightarrow \mathrm{y}-2=-\mathrm{x}+3$
$\Rightarrow x+y-5=0 \Rightarrow x+y=5$

Topic: Mathematics, Straight Lines, NDA, NDA 2019 (Phase 1)
What is the equation of the straight line which passes through the point of intersection of the straight lines $\mathrm{x}+2 \mathrm{y}=5$ and $3 \mathrm{x}+7 \mathrm{y}=17$ and is perpendicular to the straight line $3 \mathrm{x}+4 \mathrm{y}=10 ?$

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

The correct answer is:
$4 x-3 y+2=0$

Solution:
Intersecting lines are $: x+2 y=5 \& 3 x+7 y=17$ On solving these we get: $x=1$ \& $y=2$ Equation of perpendicular line is
$3 x+4 y=10$ or $y=\frac{-3}{4} x+10$
So, slope $=\frac{-3}{4}$
$\Rightarrow$ Slope of required line $=\frac{4}{3}$ Equation of given line is
$(y-2)=\frac{4}{3}(x-1)$ or $4 x-3 y+2=0$

Topic: Mathematics, Straight Lines, NDA, NDA 2016 (Phase 2)
What is the equation to circle which touches both the axes and has centre on the line $x+y=4$ ?

Options:
A $x^{2}+y^{2}-4 x+4 y+4=0$
B $x^{2}+y^{2}-4 x-4 y+4=0$
C $x^{2}+y^{2}+4 x-4 y-4=0$
D $x^{2}+y^{2}+4 x+4 y-4=0$

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

Solution:
The equation of circle, which touches both the axes, is given by
$$
x^{2}+y^{2}-2 r x-2 r y+r^{2}=0
$$
$\ldots \ldots . .(\mathrm{i})$
Now, the centre $(r, r)$ of this circle lies on the line
$$
\begin{array}{l}
x+y=4 \\
r+r=4 \Rightarrow r=2
\end{array}
$$
$\therefore \quad$ Put value of $r$ in Eq. (i), we get
$$
x^{2}+y^{2}-4 x-4 y+4=0
$$
which is required equation of circle,

Topic: Mathematics, Circle, NDA, NDA 2010 (Phase 2)
What is the equation to the plane through $(1,2,3)$ parallel to $3 x+4 y-5 z=0 ? \quad$

Options:
A $3 x+4 y+5 z+4=0$
B $\quad 3 x+4 y-5 z+14=0$
C $\quad 3 x+4 y-5 z+4=0$
D $3 x+4 y-5 z-4=0$

The correct answer is:
$\quad 3 x+4 y-5 z+4=0$

Solution:
The equation of a plane parallel to the plane $3 x+4 y-5 z=0$ is given by,
$$
3 x+4 y-5 z=d
$$
....(i) Sinceplane (i) passes through $(1,2,3)$ then, $3+8-15=\mathrm{d}$ $\Rightarrow \mathrm{d}=-4$
$\therefore$ from(i)
$$
3 x+4 y-5 z+4=0
$$

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2012 (Phase 2)
What is the equation to the sphere whose centre is at $(-2,3,4)$ and radius is 6 units?

Options:
A $x^{2}+y^{2}+z^{2}+4 x-6 y-8 z=7$
B $x^{2}+y^{2}+z^{2}+6 x-4 y-8 z=7$
C $x^{2}+y^{2}+z^{2}+4 x-6 y-8 z=4$
D $x^{2}+y^{2}+z^{2}+4 x+6 y+8 z=4$

The correct answer is:
$x^{2}+y^{2}+z^{2}+4 x-6 y-8 z=7$

Solution:
Given, centre of sphere $(\mathrm{h}, \mathrm{k}, 1)=(-2,3,4)$ radius $(\mathrm{r})=6$ units. Equation of sphere is $(\mathrm{x}-\mathrm{h})^{2}+(\mathrm{y}-\mathrm{k})^{2}+(\mathrm{z}-\ell)^{2}=\mathrm{r}^{2}$
$\Rightarrow(x+2)^{2}+(y-3)^{2}+(z-4)^{2}=6^{2}$
$\Rightarrow x^{2}+4 x+4+y^{2}-6 y+9+z^{2}-8 z+16=36$
$\Rightarrow x^{2}+y^{2}+z^{2}+4 x-6 y-8 z=7$

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2018 (Phase 1)