datavorous/entrance-exam-dataset huggingface dataset:
The line $A A^{\prime}$ is on charged infinite $A$ conducting plane which is perpendicular to the plane of the paper. The plane has a surface density of charge $\sigma$ and $B$ is ball of mass $m$ with a like charge of magnitude $q . B$ is connected by string from a point on the line $A A^{\prime}$. The tangent of angle (0) formed between the line $A A^{\prime}$ and the string is

Options:
A $\frac{q \sigma}{2 \varepsilon_{0} m g}$
B $\frac{q \sigma}{4 \pi \varepsilon_{g} m g}$
C $\frac{q \sigma}{2 \pi e_{0} m g}$
D $\frac{q \sigma}{\varepsilon_{0} m g}$

The correct answer is:
$\frac{q \sigma}{\varepsilon_{0} m g}$

Solution:
The diagram is as follows

The electric field due to charged infinite conducting sheet is $E=\frac{\sigma}{\varepsilon_{0}}$

Now, force (electric force) on the charged ball is $F=q E=\frac{q \sigma}{\varepsilon_{0}}$
The resultant of electric force and $m g$ balance the tension produced in the string. So, tan $\theta=\frac{F_{e}}{m g}=\frac{q \sigma}{\frac{\varepsilon_{0}}{m g}}=\frac{q \sigma}{\varepsilon_{0} m g}$

Topic: Physics, Electrostatics, JEE Main
The line joining the points $(-2,1,-8)$ and $(a, b, c)$ is parallel to the line whose direction ratios are $6,2,3$. The values of $a, b, c$ are

Options:
A $4,3,-5$
B $1,2,-13 / 2$
C $10,5,-2$
D None of these

The correct answer is:
$4,3,-5$

Solution:
$\begin{aligned}
& \text {According to question, } \frac{a+2}{6}=\frac{b-1}{2}=\frac{c+8}{3} \\
& \Rightarrow a=4, b=3, c=-5
\end{aligned}$

Topic: Mathematics, Three Dimensional Geometry, JEE Main
The line joining the points $(3,5,-7)$ and $(-2,1,8)$ meets the $y z$-plane at point

Options:
A $\left(0, \frac{13}{5}, 2\right)$
B $\left(2,0, \frac{13}{5}\right)$
C $\left(0,2, \frac{13}{5}\right)$
D $(2,2,0)$

The correct answer is:
$\left(0, \frac{13}{5}, 2\right)$

Solution:
Line joining the points $(3,5,-7)$ and $(-2,1,8)$ is, $\frac{x-3}{(-2)-(3)}=\frac{y-5}{(1)-(5)}=\frac{z-(-7)}{8-(-7)}$
$\begin{gathered}
\frac{x-3}{-5}=\frac{y-5}{-4}=\frac{z+7}{15}=K \\
\therefore x=-5 K+3, y=-4 K+5, \quad z=15 K-7
\end{gathered}$
L. Line (i) meets the $y z$-plane
$\therefore \quad-5 K+3=0 \Rightarrow K=3 / 5$
Put the value of $K$ in $x, y, z$
So the required point is $(0,13 / 5,2)$.

Topic: Mathematics, Three Dimensional Geometry, JEE Main
The line joining two points $A(2,0), B(3,1)$ is rotated about $A$ in anti-clockwise direction through an angle of $15^{\circ}$. The equation of the line in the now position, is

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

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

Solution:
$\begin{array}{l}
\text { Slope of } \mathrm{AB}=\frac{1}{1} \\
\Rightarrow \tan \theta=\mathrm{m}_{1}=1 \text { or } \theta=45^{\circ}
\end{array}$
Thus, slope of new line is $\tan \left(45^{\circ}+15^{\circ}\right)$
$=\tan 60^{\circ}=\sqrt{3}$
$(\because$ it is rotated anti-clockwise, so the angle will be $45^{\circ}+15^{\circ}=60^{\circ}$ )
Hence, the equation is $y=\sqrt{3} x+c$ But it passes through $(2,0)$
So, $c=-2 \sqrt{3}$
Thus, required equation is $y=\sqrt{3} x-2 \sqrt{3}$

Topic: Mathematics, Straight Lines, JEE Main
The line K+12x+Ky-2K2-2=0 passes through the point m,n for all real values of K, then

Options:
A m+n=2
B m-n=6
C mn=2
D mn=12

The correct answer is:
m-n=6

Solution:
K+12m+Kn-2K2-2=0, âKâR
âm-2K2+2m+nK+m-2=0, âKâR
âm-2=0 & 2m+n=0
âm=2 & n=-4
âm+n=-2,m-n=6,mn=-12

Topic: Mathematics, Quadratic Equation, JEE Main
The line L1≡3x-4y+1=0 touches the circles C1 and C2. Centers of C1 and C2 are A1(1,2) and A23,1 respectively Then, the length (in units) of the transverse common tangent of C1 and C2 is equal to

Options:

The correct answer is:
1

Solution:
Note that A1 and A2 lie on different sides of L1,

hence L1 is a transverse common tangent
Radius of C1=3-8+15=45
Radius of C2=9-4+15=65
A1A2=4+1=5
⇒ Length of the transverse common tangent is A1A22-r1+r22=5-22=1 unit

Topic: Mathematics, Circle, JEE Main
The line L1:x5+yb=1 passes through the point 13,32 and is parallel to L2:xc+y3=1. Then, the distance between L1 and L2 is

Options:
A 17 units
B 1715 units
C 2317 units
D 2315 units

The correct answer is:
2317 units

Solution:
Slope of the line L1=-b5 and slope of the line L2=-3c.

The line L1 is parallel to the line L2

⇒b5=3c⇒bc=15
As 13,32 lies on L1, so

135+32b=1
⇒32b=-85⇒b=-20
So, c=-34

∴ Equation of L2:y-4x=3
Hence, distance between L1 and L2=52-32+317=2317 units

Topic: Mathematics, Straight Lines, JEE Main
The line $L$ given by $\frac{x}{5}+\frac{y}{b}=1$ passes through the point $(13,32)$. The line $K$ is parallel to $L$ and has the equation $\frac{x}{c}+\frac{y}{3}=1$. Then the distance between $L$ and $K$ is

Options:
A
$\sqrt{17}$

B
$\frac{17}{\sqrt{15}}$

C
$\frac{23}{\sqrt{17}}$

D
$\frac{23}{\sqrt{15}}$

The correct answer is:
$\frac{23}{\sqrt{17}}$

Solution:
Slope of line $L=-\frac{b}{5}$
Slope of line $\mathrm{K}=-\frac{3}{\mathrm{c}}$
Line $L$ is parallel to line $k$.
$$
\Rightarrow \frac{\mathrm{b}}{5}=\frac{3}{\mathrm{c}} \quad \Rightarrow \mathrm{bc}=15
$$
$(13,32)$ is a point on $\mathrm{L}$.
$$
\begin{aligned}
& \Rightarrow \frac{13}{5}+\frac{32}{b}=1 \quad \Rightarrow \frac{32}{b}=-\frac{8}{5} \\
& \Rightarrow \mathrm{b}=-20 \quad \Rightarrow \mathrm{c}=-\frac{3}{4} \\
&
\end{aligned}
$$
Equation of $K: y-4 x=3$
Distance between $\mathrm{L}$ and $\mathrm{K}=\frac{|52-32+3|}{\sqrt{17}}=\frac{23}{\sqrt{17}}$

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2010
The line L given by x5+yb=1 passes through the point (13,32). If the line K is parallel to L and has the equation xc+y3=1, then the distance between L and K is

Options:
A 2315 units
B 17 units
C 1715 units
D 2317 units

The correct answer is:
2317 units

Solution:
Since, line L passes through (13,32)
∴135+32b=1
⇒32b=1-135=-85
⇒b=-32×58=-20
⇒L:x5-y20=1
Given, K:xc+y3=1 is a line parallel to L=0
∴ The line K must have equation
x5-y20=a
or x5a-y20a=1
On comparing with xc+y3=1, we get,
-20a=3, c=5a
⇒a=-320, c=-1520
∴ Distance between the two lines is
a-1125+1400=-320-117400=2317 units

Topic: Mathematics, Straight Lines, JEE Main
The line $l x+m y+n=0$ will be a tangent to the circle $x^2+y^2=a^2$ iff

Options:
A $n^2\left(l^2+m^2\right)=a^2$
B $a^2\left(l^2+m^2\right)=n^2$
C $n(l+m)=a$
D $a(l+m)=n$

The correct answer is:
$a^2\left(l^2+m^2\right)=n^2$

Solution:
Line $y=m x+c$ is tangent, if $c= \pm a \sqrt{1+m^2}$.
Now $1 x+m y+n=0$ or $y=-\frac{1}{m} x-\frac{n}{m}$ is tangent, if
$-\frac{n}{m}= \pm a \sqrt{1+\left(\frac{l}{m}\right)^2} \text { or } n^2=a^2\left(m^2+l^2\right) \text {. }$

Topic: Mathematics, Circle, JEE Main
The line $\ lx+m y-n=0$ will be tangent to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, if

Options:
A $a^2 l^2+b^2 m^2=n^2$
B $a l^2+b m^2=n^2$
C $a^2 l+b^2 m=n$
D None of these

The correct answer is:
$a^2 l^2+b^2 m^2=n^2$

Solution:
$y=\frac{-l}{m} x+\frac{n}{m}$ is tangent to $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, if $\frac{n}{m}= \pm \sqrt{b^2+a^2\left(\frac{l}{m}\right)^2}$ or $n^2=m^2 b^2+l^2 a^2$.

Topic: Mathematics, Ellipse, JEE Main
The line of intersection of the planes r→ ⋅3i^-j^+ k^=1 and r→ ⋅i^+4j^-2k^=2, is,

Options:
A x-6132=y-5137=z-13
B x-472=y-7=z+5713
C x-6132=y-513-7=z-13
D x-47-2=y7=z-5713

The correct answer is:
x-6132=y-513-7=z-13

Solution:
n→= n→1×n→2
= i^j^k^3-1114-2
= i^ -2-j^-7+k^13
n→=-2i+7j^+13k^
Now equations of plane are
3x-y+z=1 and
x+4y-2z=2
Let, z=0
⇒ 3x-y=1
⇒ x+4y=2
⇒ 13x=6
x=613 and
y=513
∴ Equation of required line is,
x-613-2=y-5137=z-013
or
x-6132=y-513-7=z-13

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2017 (08 Apr Online)
The line parallel to the $x$-axis and passing through the intersection of the lines ax $+$ $2 b y+3 b=0$ and $b x-2 a y-3 a=0$, where $(a, b) \neq(0,0)$ is

Options:
A
below the $x$-axis at a distance of $\frac{3}{2}$ from it

B
below the $x$-axis at a distance of $\frac{2}{3}$ from it

C
above the $x$-axis at a distance of $\frac{3}{-}$ from it

D
above the $x$-axis at a distance of $\frac{2}{3}$ from it

The correct answer is:
below the $x$-axis at a distance of $\frac{3}{2}$ from it

Solution:
$$
\begin{aligned}
& a x+2 b y+3 b+\lambda(b x-2 a y-3 a)=0 \\
& \Rightarrow(a+b \lambda) x+(2 b-2 a \lambda) y+3 b-3 \lambda a=0 \\
& a+b \lambda=0 \Rightarrow \lambda=-a / b \\
& \Rightarrow a x+2 b y+3 b-\frac{a}{b}(b x-2 a y-3 a)=0 \\
& \Rightarrow a x+2 b y+3 b-a x+\frac{2 a^2}{b} y+\frac{3 a^2}{b}=0 \\
& y\left(2 b+\frac{2 a^2}{b}\right)+3 b+\frac{3 a^2}{b}=0 \\
& y\left(\frac{2 b^2+2 a^2}{b}\right)=-\left(\frac{3 b^2+3 a^2}{b}\right) \\
& y=\frac{-3\left(a^2+b^2\right)}{2\left(b^2+a^2\right)}=\frac{-3}{2} \\
&
\end{aligned}
$$
$y=-\frac{3}{2}$ so it is $3 / 2$ units below $x$-axis.

Topic: Mathematics, Three Dimensional Geometry, JEE Main
The line parallel to the $x$-axis and passing through the intersection of the lines ax $+$ $2 b y+3 b=0$ and $b x-2 a y-3 a=0$, where $(a, b) \neq(0,0)$ is

Options:
A
below the $x$-axis at a distance of $\frac{3}{2}$ from it

B
below the $x$-axis at a distance of $\frac{2}{3}$ from it

C
above the $x$-axis at a distance of $\frac{3}{-}$ from it

D
above the $x$-axis at a distance of $\frac{2}{3}$ from it

The correct answer is:
below the $x$-axis at a distance of $\frac{3}{2}$ from it

Solution:
$$
\begin{aligned}
& a x+2 b y+3 b+\lambda(b x-2 a y-3 a)=0 \\
& \Rightarrow(a+b \lambda) x+(2 b-2 a \lambda) y+3 b-3 \lambda a=0 \\
& a+b \lambda=0 \Rightarrow \lambda=-a / b \\
& \Rightarrow a x+2 b y+3 b-\frac{a}{b}(b x-2 a y-3 a)=0 \\
& \Rightarrow a x+2 b y+3 b-a x+\frac{2 a^2}{b} y+\frac{3 a^2}{b}=0 \\
& y\left(2 b+\frac{2 a^2}{b}\right)+3 b+\frac{3 a^2}{b}=0 \\
& y\left(\frac{2 b^2+2 a^2}{b}\right)=-\left(\frac{3 b^2+3 a^2}{b}\right) \\
& y=\frac{-3\left(a^2+b^2\right)}{2\left(b^2+a^2\right)}=\frac{-3}{2} \\
&
\end{aligned}
$$
$y=-\frac{3}{2}$ so it is $3 / 2$ units below $x$-axis.

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2005
The line parallel to $x$-axis and passing through the point of intersection of lines $a x+2 b y+3 b=0$ and $b x-2 a y-3 a=0$, where $(a, b) \neq(0,0)$ is

Options:
A
above $x$-axis at a distance $2 / 3$ from it

B
above $x$-axis at a distance $3 / 2$ from it

C
below $x$-axis at a distance $3 / 2$ from it

D
below $x$-axis at a distance $2 / 3$ from it

The correct answer is:
below $x$-axis at a distance $3 / 2$ from it

Solution:
Given lines are
$$
a x+2 b y+3 b=0 \text { and }
$$
$$
b x-2 a y-3 a=0
$$
Since, required line is $\|$ to $x$-axis
$$
\therefore x=0
$$
We put $x=0$ in given equation, we get
$$
2 b y=-3 b \Rightarrow y=-\frac{3}{2}
$$
This shows that the required line is below $x$-axis at a distance of $\frac{3}{2}$ from it.

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2012 (26 May Online)
The line passes through $(1,0)$ and $(-2, \sqrt{3})$ makes an angle of ..... with $x$-axis

Options:
A $60^{\circ}$
B $120^{\circ}$
C $150^{\circ}$
D $135^{\circ}$

The correct answer is:
$150^{\circ}$

Solution:
Gradient of the line which passes through $(1,0)$ and $(-2, \sqrt{3})$ is $m=\frac{\sqrt{3}-0}{-2-1}=-\frac{1}{\sqrt{3}}$
$\Rightarrow=\tan ^{-1}\left(-\frac{1}{\sqrt{3}}\right)=150^{\circ}$

Topic: Mathematics, Straight Lines, JEE Main
The line passing through $\left(-1, \frac{\pi}{2}\right)$ and perpendicular to $\sqrt{3} \sin \theta+2 \cos \theta=\frac{4}{r}$ is :

Options:
A $2=\sqrt{3} r \cos \theta-2 r \sin \theta$
B $5=-2 \sqrt{3} r \sin \theta+4 r \cos \theta$
C $2=\sqrt{3} r \cos \theta+2 r \sin \theta$
D $5=2 \sqrt{3} r \sin \theta+4 r \cos \theta$

The correct answer is:
$2=\sqrt{3} r \cos \theta-2 r \sin \theta$

Solution:
Given,
$\sqrt{3} \sin \theta+2 \cos \theta=\frac{4}{r}$ ...(i)
Any line perpendicular to Eq. (i) is
$\sqrt{3} \sin \left(\frac{\pi}{2}+\theta\right)+2 \cos \left(\frac{\pi}{2}+\theta\right)=\frac{k}{r}$
$\sqrt{3} \cos \theta-2 \sin \theta=\frac{k}{r}$
It passes through $\left(-1, \frac{\pi}{2}\right)$, then
$\sqrt{3} \cos \frac{\pi}{2}-2 \sin \frac{\pi}{2}=\frac{k}{-1}$
$-2=\frac{k}{-1} \Rightarrow k=2$
Thus, the equation is
$\sqrt{3} \cos \theta-2 \sin \theta=\frac{2}{r}$
$\therefore \quad \sqrt{3} r \cos \theta-2 r \sin \theta=2$

Topic: Mathematics, Straight Lines, JEE Main
The line passing through the points $(3,-4)$ and $(-2,6)$ and a line passing through $(-3,6)$ and $(9,-18)$ are

Options:
A Perpendicular
B Parallel
C Makes an angle $60^{\circ}$ with each other
D None of these

The correct answer is:
Parallel

Solution:
$m_1=\frac{6+4}{-2-3}=\frac{10}{-5}=-2 \text { and } m_2=\frac{-18-6}{9-(-3)}=-2$
Hence the lines are parallel.

Topic: Mathematics, Straight Lines, JEE Main
The line passing through the points $(5,1, a)$ and $(3, b, 1)$ crosses the $y z-$ plane at the point $\left(0, \frac{17}{2}, \frac{-13}{2}\right)$. Then

Options:
A
$a=2, b=8$

B
$a=4, b=6$

C
$a=6, b=4$

D
$a=8, b=2$

The correct answer is:
$a=6, b=4$

Solution:
Equation of line passing through $(5,1, a)$ and $(3, b, 1)$ is
$$
\frac{x-5}{2}=\frac{y-1}{1-b}=\frac{z-a}{a-1}=\lambda
$$
If line crosses $y z-$ plane i.e., $x=0$
$$
\begin{aligned}
& x=2 \lambda+5=0 \\
& \Rightarrow \lambda=-5 / 2,
\end{aligned}
$$
Since, $y=\lambda(1-b)+1=\frac{17}{2}$
$$
\begin{aligned}
& -\frac{5}{2}(1-b)+1=\frac{17}{2} \\
& b=4
\end{aligned}
$$
Also, $z=\lambda(a-1)+a=-\frac{13}{2}$
$$
\begin{aligned}
& -\frac{5}{2}(a-1)+a=-\frac{13}{2} \\
& \Rightarrow a=6
\end{aligned}
$$

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2008
The line segment joining A5,0 and B10cos⁡θ,10sin⁡θ is divided internally in the ratio 2:3 at P. If θ varies, then the perimeter of locus of P is

Options:
A 4π units
B 16π units
C 8π units
D 6π units

The correct answer is:
8π units

Solution:
Let Ph,k be the point dividing te line segment in the ratio 2:3, then
h=210cos⁡θ+352+3=4cos⁡θ+3 and k=210sin⁡θ+302+3=4sin⁡θ
∴h-32+k2=16
⇒ Locus of Ph,k is x-32+y2=16, which is a circle.

Hence, the perimeter of the circle is 2π4=8π units

Topic: Mathematics, Straight Lines, JEE Main
The line segment joining the points \( (1,2) \) and \( (k, 1) \) is divided by the line \( 3 x+4 y-7=0 \) in the ratio \( 4: 9 \), then \( k \) is

Options:
A \( -2 \)
B \( 2 \)
C \( -3 \)
D \( 3 \)

The correct answer is:
\( -2 \)

Solution:
Using section formula, we can write the point of division has the coordinates

4k+94+9, 4+9×24+9=4k+913, 2213

This point must lie in the given line.

So,

3×4k+913+4×2213-7=0

⇒3×4k+913=313

⇒ k=- 2

Topic: Mathematics, Straight Lines, JEE Main
The line, that is coplanar to the line x+3-3=y-11=z-55, is

Options:
A x+1-1=y-22=z-54
B x+1-1=y-22=z-55
C x-1-1=y-22=z-55
D x+11=y-22=z-55

The correct answer is:
x+1-1=y-22=z-55

Solution:
We know that two lines x-x1a1=y-y1b1=z-z1c1 and x-x2a2=y-y2b2=z-z2c2 are coplanar if x2-x1y2-y1z2-z1a1b1c1a2b2c2=0

Let us verify this option.

The given lines are x+3-3=y-11=z-55 and x+1-1=y-22=z-55.

=-1--32-15-5-315-125

=210-315-125

=25-10-1-15--5+0

=2-5+10=0

Therefore, this is the correct option.

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2023 (13 Apr Shift 2)
The line through the points m,-9 and 7,m has slope m. Then, the x-intercept of this line is

Options:
A -18
B -6
C 6
D 18

The correct answer is:
6

Solution:
Given, m=m+97-m⇒7m-m2=m+9
⇒m2-6m+9=0⇒m=3

So, the equation of line is
y+9=3x-3

y=0⇒x=6

Hence, the x-intercept of the line is 6

Topic: Mathematics, Straight Lines, JEE Main
The line $x-2 y+4 z+4=0, x+y+z-8=0$ intersect the plane $x-y+2 z+1=0$ at the point

Options:
A $(-2,5,1)$
B $(2,-5,1)$
C $(2,5,-1)$
D $(2,5,1)$

The correct answer is:
$(2,5,1)$

Solution:
Solving the \(3 \mathrm{eq}^n\) s will give as the required point of intersection
\(\begin{aligned} & x-2 y+4 z=-4 \\ & x+y+z=8 \\ & x-y+2 z=-1 \\ & D=\left|\begin{array}{ccc} 1 & -2 & 4 \\ 1 & 1 & 1 \\ 1 & -1 & 2 \end{array}\right|=1(3)+2(1)+4(-2)=-3 \\ & \begin{aligned} D_1=\left|\begin{array}{rrr} -4 & -2 & 4 \\ 8 & 1 & 1 \\ -1 & -1 & 2 \end{array}\right|= & -4(3)+2(17)+4(-7) \\ & =-6 \end{aligned} \\ & D_2=\left|\begin{array}{ccc} 1 & -4 & 4 \\ 1 & 8 & 1 \\ 1 & -1 & 2 \end{array}\right|=-15 \\ & D_3=\left|\begin{array}{ccc} 1 & -2 & -4 \\ 1 & 1 & 8 \\ 1 & -1 & -1 \end{array}\right|=-3 \\ & x=\frac{D_1}{D}=2, \quad y=\frac{D_2}{D}=5, \quad z=\frac{D_3}{D}=1 \\ & \end{aligned}\)
\((2, S, 1)\) is the required point.

Topic: Mathematics, Three Dimensional Geometry, JEE Main
The line $x=\frac{\pi}{4}$ divides the area of the region bounded by $y=\sin x, y=\cos x$ and $x$-axis $\left(0 \leq x \leq \frac{\pi}{2}\right)$ into two regions of areas $A_1$ and $A_2$. Then $A_1, A_2$ equals

Options:
A 4: 1
B 3: 1
C 2: 1
D 1: 1

The correct answer is:
1: 1

Solution:
Area,
$\begin{aligned}
A_1 & =\int_0^{\pi / 4} \sin x d x \\
& =-[\cos x]_0^{\pi / 4} \\
& =1-\frac{1}{\sqrt{2}}=\frac{\sqrt{2}-1}{\sqrt{2}}
\end{aligned}$
and area, $A_2=\int_{\pi / 4}^{\pi / 2} \cos x d x$
$\begin{aligned} & =[\sin x]_{\pi / 4}^{\pi / 2}=\left[1-\frac{1}{\sqrt{2}}\right]=\frac{\sqrt{2}-1}{\sqrt{2}} \\ & \therefore \quad A_1: A_2=\frac{\sqrt{2}-1}{\sqrt{2}}: \frac{\sqrt{2}-1}{\sqrt{2}}=1: 1 \\ & \end{aligned}$

Topic: Mathematics, Area Under Curves, JEE Main
The line xa+yb=1 touches the curve y=be-xa at the point

Options:
A 0,0
B 0,a
C 0,b
D b,0

The correct answer is:
0,b

Solution:
Let the point be x1, y1
∴ y1=be-x1a ......(i)
Also, the slope of the tangent to the curve is dydxx1, y1=-ba e-x1a =-y1a (by (i))
Now, the equation of tangent of the given curve at point x1, y1 is y-y1=-y1a x-x1
⇒ xa+yy1=x1a+1
Comparing with xa+yb=1 ,
We get, y1=b and 1+x1a=1
⇒ x1=0
Hence, the point is (0, b).

Topic: Mathematics, Application of Derivatives, JEE Main
The line x=c2 divides the triangle with vertices 0,0,1,1 and 9,1 into two regions. If the area of the two regions is the same, then c is equal to

Options:
A 30
B 6
C 72
D 15

The correct answer is:
6

Solution:
2⋅ 1 2 ⋅ 9 1 1 c 2 1 1 c 2 c 18 1 = 1 2 9 1 1 1 1 1 0 0 1 = 1 2 ⋅8=4
⇒c2=3 or c2=15 (not possible)

⇒c=6

Topic: Mathematics, Straight Lines, JEE Main
The line $x \cos \alpha+y \sin \alpha=p$ will be a tangent to the circle $x^2+y^2-2 a x \cos \alpha-2 a y \sin \alpha=0$, if $p=$

Options:
A $0$ or a
B $0$
C $2 a$
D $0$ or $2 a$

The correct answer is:
$0$ or $2 a$

Solution:
$x \cos \alpha+y \sin \alpha-p=0$ is a tangent, if perpendicular from centre on it is equal to radius of the circle. Here centre is $(a \cos \alpha, a \sin \alpha)$ and radius is $a$.
$\therefore\left|\frac{a \cos ^2 \alpha+a \sin ^2 \alpha-p}{\sqrt{1}}\right|=a$
i.e. $|a-p|=a \Rightarrow p=0$ or $p=2 a$

Topic: Mathematics, Circle, JEE Main
The line $x \cos \alpha+y \sin \alpha=p$ will touch the parabola $y^2=4 a(x+a)$, if

Options:
A $p \cos \alpha+a=0$
B $p \cos \alpha-a=0$
C $a \cos \alpha+p=0$
D $a \cos \alpha-p=0$

The correct answer is:
$p \cos \alpha+a=0$

Solution:
$x \cos \alpha+y \sin \alpha-p=0 \ldots(i)$
$2 a x-y y_1+2 a\left(x_1+2 a\right)=0 \ldots(ii)$
From (i) and (ii), $\frac{\cos \alpha}{2 a}=\frac{\sin \alpha}{-y}=\frac{-p}{2 a(x+2 a)}$
$\Rightarrow y=-2 a \tan \alpha$ and $x=-p \sec \alpha-2 a$
$\therefore y^2=4 a(x+a) \Rightarrow 4 a^2 \tan ^2 \alpha=-4 a(p \sec \alpha+a)$
$\Rightarrow p \cos \alpha+a=0$

Topic: Mathematics, Parabola, JEE Main
The line xk=y2=z-12 form an isosceles triangle with the planes 2x+y+3z-1=0 and x+2y-3z-1=0, then the value of k is equal to

Options:

The correct answer is:
74

Solution:
If given line makes an isosceles triangle with the planes, then the angle that the line makes with the normal vectors to the planes are equal
A vector parallel to the line is bâ=<k,2,-12>
A normal vector to the first plane n1â=<2,1,3>
A normal vector to the second plane n2â=<1,2,-3>
cosÎ¸= b â â n i â b â n 1 â = b â â n 2 â b â n 2 â â2k+2â36=k+4+36
âk=74

Topic: Mathematics, Three Dimensional Geometry, JEE Main
The line $x+m y+n=0$ will be a tangent to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$, if

Options:
A $a^2 \ell^2+b^2 m^2=n^2$
B $a^2 \ell^2-b^2 m^2=n^2$
C $a m^2-b^2 n^2=a^2 l^2$
D None of these

The correct answer is:
$a^2 \ell^2-b^2 m^2=n^2$

Solution:
If $y=M x+C$ is tangent to hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$, then $C^2=a^2 M^2-b^2$.
$l x+m y+n=0 \Rightarrow y=-\frac{1}{m} x-\frac{n}{m}$
Comparing this equation with $\mathrm{y}=\mathrm{Mx}+\mathrm{C}$, we get
$M=-\frac{1}{m}, C=-\frac{n}{m}$
Now, $\mathrm{C}^2=\mathrm{a}^2 \mathrm{M}^2-\mathrm{b}^2$
$\begin{aligned}
& \Rightarrow \frac{n^2}{m^2}=a^2 \frac{l^2}{m^2}-b^2 \\
& \Rightarrow n^2=a^2 l^2-b^2 m^2
\end{aligned}$

Topic: Mathematics, Hyperbola, JEE Main
The line $x+y=2$ is tangent to the curve $x^2=3-2 y$ at its point

Options:
A $(1,1)$
B $(-1,1)$
C $(\sqrt{3}, 0)$
D $(3,-3)$

The correct answer is:
$(1,1)$

Solution:
Given curve $x^2=3-2 y...(i)$
Differentiate w.r.t. $x, \quad 2 x=0-2 \frac{d y}{d x} \Rightarrow \frac{d y}{d x}=-x$
Slope of the tangent of the curve $=-x$
From the given line, slope $=-1, \therefore x=1$ and from equation (i), $y=1$
$\therefore$ Co-ordinate of the point is $(1,1)$.

Topic: Mathematics, Application of Derivatives, JEE Main
The line x=y touches a circle at the point (1,1). If the circle also passes through the point (1,-3), then its radius is

Options:
A 32
B 3
C 2
D 22

The correct answer is:
22

Solution:
Equation of family of circles touches line x–y=0 at point (1,1) is
(x-1)2+(y-1)2+λx–y=0, λ∈R
If a member of this family, also passes through point (1,–3) then
1-12+-3-12+λ1--3=0⇒λ=-4
⇒ Required circle is
x-12+y-12-4x-y=0
⇒ x2+y2-6x+2y+2=0
Where radius is r=32+12-2=22

Topic: Mathematics, Circle, JEE Main, JEE Main 2019 (10 Apr Shift 1)
The line $\mathrm{y}=2 \mathrm{t}^2$ intersects the ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$ in real points if

Options:
A $|t| \leq 1$
B $|t| < 1$
C $|t|>1$
D $|t| \geq 1$

The correct answer is:
$|t| \leq 1$

Solution:
$$
\begin{aligned}
& \text { Hints: } \frac{x^2}{9}+\frac{y^2}{4}=1 ; \mathrm{y}=2 \mathrm{t}^2 \\
& \frac{x^2}{9}+\frac{4 t^4}{4}=1 \Rightarrow \frac{x^2}{9}+t^4=1 \Rightarrow x^2=9\left(1-t^4\right) \\
& x^2 \geq 0 \Rightarrow 9\left(1-t^4\right) \geq 0 \Rightarrow t^4-1 \leq 0 \\
& \Rightarrow\left(t^2-1\right)\left(t^2+1\right) \leq 0 \\
& \Rightarrow t^2-1 \leq 0 \quad\left(\because t^2+1>0\right) \\
& \Rightarrow|t| \leq 1
\end{aligned}
$$

Topic: Mathematics, Ellipse, JEE Main
The line y=2x+C is a tangent to the parabola y 2 −4y−8x=4 at a point whose abscissa is α, then the ordered pair α,C is

Options:
A − 1 2 ,4
B â 1 2 ,5
C 4, 5
D − 1 2 , 1 2

The correct answer is:
− 1 2 ,5

Solution:
Equation of parabola is y−2 2 =8 x+1
So, equation of tangent with slope 2 is
y−2=2 x+1 + 2 2 ⇒y=2x+5⇒C=5
Point of contact is − 1 2 ,4 ⇒α=− 1 2

Topic: Mathematics, Parabola, JEE Main
The line $y=2 x+c$ is a tangent to the parabola $y^2=16 x$, if $c$ equals

Options:
A $-2$
B $-1$
C $0$
D $2$

The correct answer is:
$2$

Solution:
According to the condition,
$c=\frac{4}{2}=2$\(=\frac{a}{m}\), where \(\mathrm{a}=4, \mathrm{~m}=2\).

Topic: Mathematics, Parabola, JEE Main
The line $y=\sqrt{3}$ meets the graph $y=\tan x$, where
$x \in\left(0, \frac{\pi}{2}\right)$, in k points. What is k equal to?

Options:
A One
B Two
C Three
D Infinity

The correct answer is:
One

Solution:
Line $\mathrm{y}=\sqrt{3}$ and graph $\mathrm{y}=\tan \mathrm{x}$
Now, we have $\sqrt{3}=\tan \mathrm{x}$
$\Rightarrow \tan x=\tan 60^{\circ}$
$\Rightarrow x=60^{\circ} \quad\left[\because x \in\left(0, \frac{\pi}{2}\right)\right]$
Hence, one intersecting point is possible in the given domain i.e., $\mathrm{k}=1$.

Topic: Mathematics, Trigonometric Equations, NDA, NDA 2014 (Phase 2)
The line $y=m x+c$ intercepts the circle $x^2+y^2=r^2$ in two distinct points, if

Options:
A $-r \sqrt{1+m^2} < c < r \sqrt{1+m^2}$
B $c < -r \sqrt{1+m^2}$
C $c < r \sqrt{1+m^2}$
D None of the above

The correct answer is:
$-r \sqrt{1+m^2} < c < r \sqrt{1+m^2}$

Solution:
Equation of the circle is
$$
x^2+y^2=r^2
$$
and the line is $m x-y+c=0$
The line (ii) intersect (i) in two distinct points, if the length of the $\perp^r$ from the centre $(0,0)$ to the line (ii) is less then $r$
Thus, $\left|\frac{0-0+c}{\sqrt{m^2+1}}\right| < r$
$$
\begin{array}{ccc}
\Rightarrow & |c| < r \sqrt{m^2+1} \\
\Rightarrow & -r \sqrt{m^2+1} < c < r \sqrt{m^2+1}
\end{array}
$$

Topic: Mathematics, Circle, JEE Main