datavorous/entrance-exam-dataset huggingface dataset:
The plots of radial distribution functions for various orbitals of hydrogen atom against ' r' are given below.

The correct plot for 3s orbital is :

Options:
A
B
C
D

The correct answer is:

Solution:
Calculate the number of radial nodes using the

Number of radial nodes =n-ℓ-1

=3-0-1=2

Radial nodes= number of cuts on x-axis ( r).
The curve should start from zero. It can be positive or negative.
Size of the curve increases after touching the x- axis every time .

By keeping all this, we get the following curve:

Therefor corresponding graph is (C)

Hence, answer is (3)

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2021 (25 Feb Shift 1)
The plots of the extent of adsorption (x/m) Vs pressure at different temperature are as follows;

The correct order of increasing temp for curves I, II, III, IV are ;

Options:
A T 1 > T 2 > T 3 > T 4
B T 4 > T 3 > T 2 > T 1
C T 3 > T 2 > T 1 > T 4
D can't be predicted

The correct answer is:
T 4 > T 3 > T 2 > T 1

Solution:
Physisorption is exothermic in nature. Therefore, in accordance with Le-Chatliers principle, it decreases with increase in temperature. This means physisorption occurs at low temperature more readily. So, The correct order of temperature for curves I, II, III, IV are :
T 4 > T 3 > T 2 > T 1

Topic: Chemistry, Surface Chemistry, JEE Main
The pOH value of a solution whose hydroxide ion concentration is $6.2 \times 10^{-9} \mathrm{~mol} /$ litre is

Options:
A 8.21
B 6.21
C 7.75
D 7.21

The correct answer is:
8.21

Solution:
$-\log (\mathrm{OH})=\mathrm{pOH} ;-\log 6.2 \times 10^{-9}=\mathrm{pOH}$
$\therefore \mathrm{pOH}=8.21$

Topic: Chemistry, Ionic Equilibrium, JEE Main
The point ${ }^{(0,5)}$ is closest to the curve $x^2=2 y$ at

Options:
A $(2 \sqrt{2}, 0)$
B $(0,0)$
C $(2,2)$
D None of these

The correct answer is:
None of these

Solution:
Let a point on the curve by $(h, k)$
Then $h^2=2 k...(i)$
$\begin{aligned}& \text { Distance }=D=\sqrt{h^2+(k-5)^2} \\
& \text { By (i); } D=\sqrt{2 k+(k-5)^2} \\
& \frac{d D}{d k}=\frac{1}{2 \sqrt{2 k+(k-5)^2}} \times 2(k-5)+2=0 \quad \Rightarrow k=4\end{aligned}$
So, at $k=4$ function $D$ must be minimum.
Then point will be $( \pm 2 \sqrt{2}, 4)$.

Topic: Mathematics, Application of Derivatives, JEE Main
The point $(0.1,3.1)$ with respect to the circle $x^2+y^2-2 x-4 y+3=0$, is

Options:
A At the centre of the circle
B Inside the circle but not at the centre
C On the circle
D Outside the circle

The correct answer is:
Outside the circle

Solution:
$x^2+y^2-2 x-4 y+3=0$
$x=0.1, y=3.1$
$(0.1)^2+(3.1)^2-2 \times 0.1-4 \times(3.1)+3=0$
$=0.01+9.61-0.2-12.4+3$
$={0.02\gt0}$
Outside of the circle

Topic: Mathematics, Circle, JEE Main
The point 2,1 is translated parallel to the line L:x-y=4 by 23 units. If the new point Q lies in the third quadrant, then the equation of the line passing through Q and perpendicular to L is

Options:
A x+y=2-6
B 2x+2y=1-6
C x+y=3-36
D x+y=3-26

The correct answer is:
x+y=3-26

Solution:
Equation of line QR perpendicular to L:x-y=4 is
x+y-c=0 …1
Distance of QR from 2,1 is 23
⇒ 23=2+1-c2
⇒ 26=3-c
⇒ c-3=±26 ⇒c=3±26
Line can be x+y=3±26
x+y=3-26 (since Q is in the third quadrant)

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2016 (09 Apr Online)
The point $(2,3)$ is a limiting point of a coaxial system of circles of which $x^2+y^2=9$ is a member. The co-ordinates of the other limiting point is given by

Options:
A $\left(\frac{18}{13}, \frac{27}{13}\right)$
B $\left(\frac{9}{13}, \frac{6}{13}\right)$
C $\left(\frac{18}{13},-\frac{27}{13}\right)$
D $\left(-\frac{18}{13},-\frac{9}{13}\right)$

The correct answer is:
$\left(\frac{18}{13}, \frac{27}{13}\right)$

Solution:
$(x-2)^2+(y-3)^2=0$ or $\left(x^2+y^2-9\right)-4 x-6 y+22=0$
or $\left(x^2+y^2-9\right)-\lambda(2 x+3 y-11)=0$ represents the family of co-axial circles.
$C=\left(\lambda, \frac{3 \lambda}{2}\right), r=\sqrt{\lambda^2+\frac{9 \lambda^2}{4}-11 \lambda+9}$
For limiting points $r=0$
$\Rightarrow 13 \lambda^2-44 \lambda+36=0 \Rightarrow \lambda=\frac{18}{13}, 2$
$\therefore$ The limiting points are $(2,3)$ and $\left[\frac{18}{13}, \frac{3}{2}\left(\frac{18}{13}\right)\right]$ or $\left(\frac{18}{13}, \frac{27}{13}\right)$

Topic: Mathematics, Circle, JEE Main
The point $(3,-4)$ lies on both the circles
$x^{2}+y^{2}-2 x+8 y+13=0$
and $\mathrm{x}^{2}+\mathrm{y}^{2}-4 x+6 y+11=0$
Then, the angle between the circles is

Options:
A $60^{\circ}$
B $\tan ^{-1}\left(\frac{1}{2}\right)$
C $\tan ^{-1}\left(\frac{3}{5}\right)$
D $135^{\circ}$

The correct answer is:
$135^{\circ}$

Solution:
Given circles are
$$
x^{2}+y^{2}-2 x+8 y+13=0
$$
and $x^{2}+y^{2}-4 x+6 y+11=0$
Here $C_{1}=(1,-4), C_{2}=(2,-3)$
$$
\Rightarrow r_{1}=\sqrt{1+16-13}=2
$$
and $r_{2}=\sqrt{4+9-11}=\sqrt{2}$
So, $d=C_{1} C_{2}=\sqrt{(2-1)^{2}+(-3+4)^{2}}=\sqrt{2}$
$$
\therefore \cos \theta=\frac{d^{2}-r_{1}^{2}-r_{2}^{2}}{2 r_{1} r_{2}}=\frac{2-4-2}{2 \times 2 \times \sqrt{2}}=-\frac{1}{\sqrt{2}}
$$
$$
\Rightarrow \theta=135^{\circ}
$$

Topic: Mathematics, Circle, JEE Main
The point $(4,-3)$ with respect to the ellipse $4 x^2+5 y^2=1$

Options:
A Lies on the curve
B Is inside the curve
C Is outside the curve
D Is focus of the curve

The correct answer is:
Is outside the curve

Solution:
Using the condition the point $\left(x_1, y_1\right)$ lies
(i) On the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}-1=0$ if
$\frac{x_1^2}{a^2}+\frac{y_1^2}{b^2}-1=0$
(ii) Outside the ellipse if $\frac{x_1^2}{a^2}+\frac{y_1^2}{b^2}-1\gt0$
(iii) Inside the ellipse if $\frac{x_1^2}{a^2}+\frac{y_1^2}{b^2}-1 \lt 0$
Given ellipse is $\frac{x^2}{1 / 4}+\frac{y^2}{1 / 5}=1$
$\therefore \frac{16}{1 / 4}+\frac{9}{1 / 5}-1=64+45-1\gt0$
Point $(4,-3)$ lies outside the ellipse.

Topic: Mathematics, Ellipse, JEE Main
The point $(-4,5)$ is the vertex of a square and one of its diagonals is $7 x-y+8=0$. The equation of the other diagonal is

Options:
A $7 x-y+23=0$
B $7 y+x=30$
C $7 y+x=31$
D $\quad x-7 y=30$

The correct answer is:
$7 y+x=31$

Solution:
$$
\begin{aligned}
\text { Hints } & : x+7 y=k \\
& -4+35=k \\
& 31=k \\
& x+7 y-31=0
\end{aligned}
$$

Topic: Mathematics, Straight Lines, JEE Main
The point a2,a+1 lies in the angle between the lines 3x+y+1=0 and x+2y-5=0 containing the origin. If a is an integer, then the sum of all possible values of a is

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

The correct answer is:
-3

Solution:
a2,a+1 & 0,0 lie on same side of line 3x+y+1=0
⇒0+0+13a2+a+1+1>0
⇒3a2+a+2>0⇒a∈R … 1
a2,a+1 & 0,0 also lie on same side of x+2y-5=0
⇒0+0-5a2+2a+2-5>0
⇒a2+2a-3<0⇒a+3a-1<0
⇒a∈-3,1 … 2
From 1 & 2, we get,

a∈-3,1
⇒ sum of all the integral values of ‘a’ is -2+-1+0=-3

Topic: Mathematics, Straight Lines, JEE Main
The point A divides the line segment joining the points (-5,1) and (3,5) in the ratio k:1 internally and the coordinates of points B and C are 1,5 and 7,-2 respectively. If the area of âABC is 2 sq. units, then the sum of all the values of k is equal to

Options:
A 329
B 7
C 949
D 319

The correct answer is:
949

Solution:
A≡3k-5k+1,5k+1k+1
Area of ∆ABC=2 sq.units
⇒123k-5k+15+2+1-2-5k+1k+1+75k+1k+1-5=±2
⇒14k-66=±4k+1⇒k=7 or 319

Topic: Mathematics, Straight Lines, JEE Main
The point at which the line segment joining A1,1 and B5,5 subtends an obtuse angle is

Options:
A 7,7
B 0,5
C 2,4
D 1,5

The correct answer is:
2,4

Solution:
Equation of the circle assuming AB as the diameter is x-1x-5+y-1y-5=0
⇒x2+y2-6x-6y+10=0
Now, for point Ph,k and circle S=0,

if S1<0, then the point h,k lies inside the circle S=0 and at this point diameter of the circle subtends an obtuse angle.
Now, from options, we get,

7-17-5+7-17-5>0
0-10-5+5-15-5>0
2-12-5+4-14-5<0
1-11-5+5-15-0=0
Hence, 2,4 is the required point

Topic: Mathematics, Circle, JEE Main
The point diametrically opposite to the point $P(1,0)$ on the circle $x^2+y^2+2 x+4 y-3=0$ is

Options:
A
$(3,-4)$

B
$(-3,4)$

C
$(-3,-4)$

D
$(3,4)$

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

Solution:
Centre $(-1,-2)$
Let $(\alpha, \beta)$ is the required point $\frac{\alpha+1}{2}=-1$ and $\frac{\beta+0}{2}=-2$.

Topic: Mathematics, Circle, JEE Main, JEE Main 2008
The point moves such that the area of the triangle formed by it with the points $(1,5)$ and $(3,-7)$ is $21 s q$. unit. The locus of the point is

Options:
A $6 x+y-32=0$
B $6 x-y+32=0$
C $x+6 y-32=0$
D $6 x-y-32=0$

The correct answer is:
$6 x+y-32=0$

Solution:
Let the point be $(x, y)$. Area of triangle with points $(x, y),(1,5)$ and $(3,-7)$ is 21 sq. units $\therefore \quad \frac{1}{2}\left|\begin{array}{ccc}x & y & 1 \\ 1 & 5 & 1 \\ 3 & -7 & 1\end{array}\right|=21$
Solving; locus of point $(x, y)$ is $6 x+y-32=0$

Topic: Mathematics, Straight Lines, JEE Main
The point of contact of the given circles $x^2+y^2-6 x-6 y+10=0$ and $x^2+y^2=2$, is

Options:
A $(0,0)$
B $(1,1)$
C $(1, -1)$
D $(-1,-1)$

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

Solution:
$x^2+y^2-6 x-6 y+10=0...(i)$
$x^2+y^2=2...(ii)$
$\Rightarrow-6 x-6 y+12=0$ or $x+y-2=0...(iii)$
$\Rightarrow x^2+y^2+2 x y=4...\{from (iii)\}$
or $2 x y=2...\{from (ii)\}$
and $x-y=\sqrt{(x+y)^2-4 x y}=\sqrt{4-4}=0$
or $x=y$ and $x+y=2 \Rightarrow x=1, y=1$
Trick : Required point must satisfy both the circles. Obviously $(1,1)$ satisfies both the equations simultaneously.

Topic: Mathematics, Circle, JEE Main
The point of contact of the line $y=x-1$ with $3 x^2-4 y^2=12$ is

Options:
A $(4,3)$
B $(3,4)$
C $(4,-3)$
D None of these

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

Solution:
The equation of the line and hyperbola are
$\begin{aligned}
& y=x-1 ....(i)\\
& 3 x^2-4 y^2=12 ....(ii)
\end{aligned}$
From (i) and (ii), we get
$\begin{aligned}
& 3 x^2-4(x-1)^2=12 \\
& \Rightarrow 3 x^2-4\left(x^2-2 x+1\right)=12
\end{aligned}$
or $x^2-8 x+16=0 \Rightarrow x=4$
From (i), $y=3$. So point of contact is $(4,3)$.

Topic: Mathematics, Hyperbola, JEE Main
The point of contact of the tangent $18 x-6 y+1=0$ to the parabola $y^2=2 x$ is

Options:
A $\left(\frac{-1}{18}, \frac{-1}{3}\right)$
B $\left(\frac{-1}{18}, \frac{1}{3}\right)$
C $\left(\frac{1}{18}, \frac{-1}{3}\right)$
D $\left(\frac{1}{18}, \frac{1}{3}\right)$

The correct answer is:
$\left(\frac{1}{18}, \frac{1}{3}\right)$

Solution:
Let point of contact be $(h, k)$, then tangent at this point is $k y=x+h$.
$x-k y+h=0 \equiv 18 x-6 y+1=0$
or $\frac{1}{18}=\frac{k}{6}=\frac{h}{1} \quad$ or $\quad k=\frac{1}{3}, \quad h=\frac{1}{18}$

Topic: Mathematics, Parabola, JEE Main
The point of contact of the tangent $y=x+2$ to the hyperbola $5 x^2-9 y^2=45$ is

Options:
A $(9 / 2,5 / 2)$
B $(5 / 2,9 / 2)$
C $(-9 / 2,-5 / 2)$
D None of these

The correct answer is:
$(-9 / 2,-5 / 2)$

Solution:
Hyperbola is $\frac{x^2}{9}-\frac{y^2}{5}=1$.
Hence point of contact is $\left[\frac{-9(1)}{\sqrt{9-5}}, \frac{-5}{\sqrt{9-5}}\right] \equiv\left[\frac{-9}{2}, \frac{-5}{2}\right]$.
Trick : Since the point $\left(-\frac{9}{2},-\frac{5}{2}\right)$ satisfies both the equations.

Topic: Mathematics, Hyperbola, JEE Main
The point of dissimilarity between lanthanides and actinides is

Options:
A three outermost shells are partially filled
B they show oxidation state of $+3$ (common)
C they are called inner transition elements
D they are radioactive in nature

The correct answer is:
they are radioactive in nature

Solution:
Lanthanides and actinides both are not $\begin{array}{lll}\text { radioactive. } & \text { Except } & \text { promethium, } & \text { all }\end{array}$
lanthanides are non-radioactive. While all actinides are radioactive.

Topic: Chemistry, d and f Block Elements, JEE Main
The point of intersection of lines $\frac{x-4}{5}=\frac{y-1}{2}=\frac{z}{1}$ and $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ is

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

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

Solution:
The general point on the planes are given by $(5 \lambda+4,2 \lambda+1, \lambda)$ and $(2 \mu+1,3 \mu+2,4 \mu+3)$ for some constants $\lambda, \mu$.
Substitute $\lambda=4 \mu+3$ in the equation $5 \lambda+4=2 \mu+1$ we get $\mu=-1$. Hence $(2 \mu+1,3 \mu+2,4 \mu+3)=(-1,-1,-1)$.

Topic: Mathematics, Three Dimensional Geometry, JEE Main
The point of intersection of tangents at the ends of the latus-rectum of the parabola $y^2=4 x$ is equal to

Options:
A $(1,0)$
B $(-1,0)$
C $(0,1)$
D $(0,-1)$

The correct answer is:
$(-1,0)$

Solution:
Equation of the tangent at $\left(x_1, y_1\right)$ on the parabola $y^2=4 a x$ is $y y_1=2 a\left(x+x_1\right)$
$\therefore$ In this case, $a=1$
The co-ordinates at the ends of the latus rectum of the parabola $y^2=4 x \quad$ are $\quad L(1,2)$ and $L_1(1,-2)$

Topic: Mathematics, Parabola, JEE Main
The point of intersection of the line $\frac{x}{1}=\frac{y-1}{2}=\frac{z+2}{3}$ and the plane $2 x+3 y+z=0$ is

Options:
A $(0,1,-2)$
B $(1,2,3)$
C $(-1,9,-25)$
D $\left(\frac{-1}{11}, \frac{9}{11} \frac{-25}{11}\right)$

The correct answer is:
$\left(\frac{-1}{11}, \frac{9}{11} \frac{-25}{11}\right)$

Solution:
$\frac{x}{1}=\frac{y-1}{2}=\frac{z+2}{3}=r \text {, (say) }$
So, $x=r, y=2 r+1, z=3 r-2$
$\begin{aligned}
& \therefore 2 r+3(2 r+1)+(3 r-2)=0 \Rightarrow r=\frac{-1}{11} \\
& x=\frac{-1}{11}, y=\frac{9}{11}, z=-\frac{25}{11} \\
&
\end{aligned}$

Topic: Mathematics, Three Dimensional Geometry, JEE Main
The point of intersection of the line joining the points $(3,4,1)$ and $(5,1,6)$ and the $x y$-plane is

Options:
A $(13,23,0)$
B $\left(\frac{13}{5}, \frac{23}{5}, 0\right)$
C $(-13,23,0)$
D $\left(-\frac{13}{5}, \frac{23}{5}, 0\right)$

The correct answer is:
$\left(\frac{13}{5}, \frac{23}{5}, 0\right)$

Solution:
For $x y$-plane,
$\begin{aligned}
& \qquad z=0 \Rightarrow \frac{6 \lambda+1}{\lambda+1}=0 \Rightarrow \lambda=-\frac{1}{6} \\
& \text { For } x y \text {-plane, } \\
& \therefore x=\frac{-5+18}{5}=\frac{13}{5}, y=\frac{-1+24}{5}=\frac{23}{5}
\end{aligned}$

Topic: Mathematics, Three Dimensional Geometry, JEE Main
The point of intersection of the line joining the points \( (3,4,1) \) and \( (5,1,6) \) and the \( x y \)-plane is

Options:
A \( (13,23,0) \)
B \( \left(\frac{13}{5}, \frac{23}{5}, 0\right) \)
C \( (-13,23,0) \)
D \( \left(-\frac{13}{5}, \frac{23}{5}, 0\right) \)

The correct answer is:
\( \left(\frac{13}{5}, \frac{23}{5}, 0\right) \)

Solution:
Let the intersection point x, y, 0 divides the line segment joining the given points in ratio Î» :1 and lies in xy -plane,
z=0â6Î»+1Î»+1=0
âÎ»=-16
â´ x=-5+185=135, y=-1+245=235 .

Topic: Mathematics, Three Dimensional Geometry, JEE Main
The point of intersection of the lines $\frac{x-5}{3}=\frac{y-7}{-1}=\frac{z+2}{1}, \frac{x+3}{-36}=\frac{y-3}{2}=\frac{z-6}{4}$ is

Options:
A $21, \frac{5}{3}, \frac{10}{3}$
B $(2,10,4)$
C $(-3,3,6)$
D $(5,7,-2)$

The correct answer is:
$21, \frac{5}{3}, \frac{10}{3}$

Solution:
Given lines are,
$\begin{aligned}
& \qquad \frac{x-5}{3}=\frac{y-7}{-1}=\frac{z+2}{1}=r_1, \text { (say) } \\
& \text { and } \frac{x+3}{-36}=\frac{y-3}{2}=\frac{z-6}{4}=r_2, \text { (say) } \\
& \therefore x=3 r_1+5=-36 r_2-3, y=-r_1+7=3+2 r_2 \text { and } z=r_1-2=4 r_2+6 \\
& \text { On solving, we get } x=21, y=\frac{5}{3}, z=\frac{10}{3} .
\end{aligned}$
On solving, we get
Trick: Check through options.

Topic: Mathematics, Three Dimensional Geometry, JEE Main
The point of intersection of the lines $\left(a^3+3\right) x+a y+a-3=0$ and $\left(a^5+2\right) x+(a+2) y+2 a+3=0$ (a real) lies on the $y$-axis for

Options:
A
no value of $a$

B
more than two values of $a$

C
exactly one value of $a$

D
exactly two values of $a$

The correct answer is:
no value of $a$

Solution:
Given equation of lines are $\left(a^3+3\right) x+a y+a-3=0$ and $\left(a^5+2\right) x+(a+2) y+2 a+3=0$ (a real)
Since point of intersection of lines lies on $\mathrm{y}$-axis.
$\therefore$ Put $x=0$ in each equation, we get $a \mathrm{y}+a-3=0$ and $(a+2) y+2 a+3=0$
On solving these we get
$$
\begin{aligned}
& (a+2)(a-3)-a(2 a+3)=0 \\
& \Rightarrow a^2-a-6-2 a^2-3 a=0 \\
& \Rightarrow \quad-a^2-4 a-6=0 \Rightarrow a^2+4 a+6=0 \\
& \Rightarrow \quad a=\frac{-4 \pm \sqrt{16-24}}{2}=\frac{-4 \pm \sqrt{-8}}{2} \\
& (\text { not real) }
\end{aligned}
$$
This shows that the point of intersection of the lines lies on the $y$-axis for no value of ' $a$ '.

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2012 (07 May Online)
The point of intersection of the lines râ=7iË^+10jË^+13k^+s(2iË^+3jË^+4k^) and râ=3iË^+5jË^+7k^+t(iË^+2jË^+3k^) is

Options:
A i˙^+j˙^-k^
B 2i˙^-j˙^+4k^
C i˙^-j˙^+k^
D i˙^+j˙^+k^

The correct answer is:
i˙^+j˙^+k^

Solution:
Two given lines intersect, if
7i˙^+10j˙^+13k^+s2i˙^+3j˙^+4k^=3i˙^+5j˙^+7k^+t(i˙^+2j˙^+3k^)
⟹7+2si˙^+10+3sj˙^+(13+4s)k^=3+ti˙^+5+2tj˙^+(7+3t)k^
⟹7+2s=3+t
⟹2s-t=-4 ...(i)
10+3s=5+2t
⟹3s-2t=-5 ...(ii)
and 13+4s=7+3t
⟹4s-3t=-6 ...(iii)
On solving Eqs. (i) and (iii), we get
s=-3, t=-2
∴ Required point is
7i˙^+10j˙^+13k^+-3[2i˙^+3j˙^+4k^]
= i˙^+j˙^+k^

Topic: Mathematics, Vector Algebra, JEE Main
The point of intersection of the lines which are common tangent to both the curves y=x2 and x2+y+1=0 is

Options:
A 0,-14
B 0,-13
C 0,-15
D 0,-12

The correct answer is:
0,-12

Solution:
For y=x2,

equation of the tangent is y=mx-m24
For this line to be tangent to x2+y+1=0
⇒x2+mx-m24+1=0 will have D=0
⇒m2+4m24-1=0⇒m=±2
⇒ Tangents are y=2x-12 and y=-2x-12
⇒ Point of intersection of the tangents is 0,-12

Topic: Mathematics, Parabola, JEE Main
The point of intersection of the normals to the parabola $y^2=4 x$ at the ends of its latus rectum is :

Options:
A
$(0,2)$

B
$(3,0)$

C
$(0,3)$

D
$(2,0)$

The correct answer is:
$(3,0)$

Solution:
We know that point of intersection of the normal to the parabola $y^2=4 a x$ at the ends of its latus rectum is $(3 a, 0)$
Hence required point of intersection $=(3,0)$

Topic: Mathematics, Parabola, JEE Main, JEE Main 2013 (23 Apr Online)
The point of intersection of the plane 3x-5y+2z=6 with the straight line passing through the origin and perpendicular to the plane 2x-y-z=4 is

Options:
A 1,-1,-1
B -1,-1,2
C 4,2,2
D 43,-23,-23

The correct answer is:
43,-23,-23

Solution:
A vector parallel to line is 2i^-j^-k^
So equation of line is
x-02=y-0-1=z-0-1
Any point on the line is 2λ,-λ,-λ
It must satisfy the given plane
6λ+5λ-2λ=6
9λ=6⇒λ=23
Point is 43,-23,-23

Topic: Mathematics, Three Dimensional Geometry, JEE Main
The point of intersection of the tangent to the parabola y2=4x which also touches x2+y2=12 is

Options:
A -1,0
B -12,0
C -2,0
D -32,0

The correct answer is:
-1,0

Solution:
Equation of the tangent to the parabola is y=mx+1m
Equation of the tangent to the circle is y=mx±121+m2
For common tangent
1m2=121+m2⇒m4+m2-2=0⇒m2=1⇒m=±1
⇒ Equation of the tangents are y=x+1 and y=-x-1
⇒ Point of intersection of these tangents is -1,0

Topic: Mathematics, Parabola, JEE Main
The point of intersection of the tangents drawn to the hyperbola x2a2-y2b2=1 at the points where it is intersected by the line lx+my+n=0 is

Options:
A a2ln,b2mn
B a2ln,-b2mn
C -a2ln,b2mn
D -a2ln,-b2mn

The correct answer is:
-a2ln,b2mn

Solution:
Let, x1,y1 be the required point. Then, the equation of the chord of contact of the tangents drawn from x1,y1 to the given hyperbola is
xx1a2-yy1b2=1 â¦(i)
The given line is lx+my+n=0 â¦(ii)
Equations (i) and (ii) represent the same line
â´x1a2l=-y1b2m=1-n
âx1=-a2ln,y1=b2mn.
Hence, the required point is -a2ln,b2mn

Topic: Mathematics, Hyperbola, JEE Main
The point of lines represented by $3 a x^2+5 x y+\left(a^2-2\right) y^2=0$ and perpendicular to each other for

Options:
A
two values of a

B
$\forall a$

C
for one value of a

D
for no values of a

The correct answer is:
two values of a

Solution:
$3 a+a^2-2=0 \Rightarrow a^2+3 a-2=0 \Rightarrow a=\frac{-3 \pm \sqrt{9+8}}{2}=\frac{-3 \pm \sqrt{17}}{2}$

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2002
The point of the curve $y^2=2(x-3)$ at which the normal is parallel to the line $y-2 x+1=0$ is

Options:
A $(5,2)$
B $\left(-\frac{1}{2},-2\right)$
C $(5,-2)$
D $\left(\frac{3}{2}, 2\right)$

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

Solution:
Given $y^2=2(x-3)...(i)$
Differentiate w.r.t. $x$,
$2 y \cdot \frac{d y}{d x}=2 \Rightarrow \frac{d y}{d x}=\frac{1}{y}$
Slope of the normal $=\frac{-1}{\left(\frac{d y}{d x}\right)}=-y$
Slope of the given line $=2$
$\therefore y=-2$
From equation (i), $x=5$
$\therefore$ Required point is $(5,-2)$

Topic: Mathematics, Application of Derivatives, JEE Main
The point of the curve $y^{2}=2(x-3)$ at which the normal is parallel to the line $y-2 x+1=0$ is

Options:
A $(5,2)$
B $\left(-\frac{1}{2},-2\right)$
C $(5,-2)$
D $\left(\frac{3}{2}, 2\right)$

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

Solution:
Given, $y^{2}=2(x-3) \quad$...(i)
On differentiating w.r.t. $x$, we get $2 y \frac{d y}{d x}=2$
$\Rightarrow$
$\frac{d y}{d x}=\frac{1}{y}$
Slope of the normal $=\frac{-1}{(d y / d x)}=-y$ Slope of the given line $=2$ $\therefore \quad y=-2$
From Eq. (i), $x=5$
$\therefore$ Required point is $(5,-2)$.

Topic: Mathematics, Application of Derivatives, JEE Main
The point on stress-strain graph at which the plastic flow starts is known as,

Options:
A elastic limit
B neck formation
C yield point
D breaking point

The correct answer is:
yield point

Solution:
Refer above diagram.

A is elastic limit, B is upper yield stress point, C is lower yield stress point, D is ultimate stress point and E is the fracture point.

So, till the point C, the material will have elastic property but once crossed, the material would not be able to come back to its original condition. Hence, it will behave as a plastic.

Topic: Physics, Mechanical Properties of Solids, JEE Main
The point on the curve 6y=4x3-3x2, the tangent at which makes an equal angle with the coordinate axes is

Options:
A 1,-16
B -1,-76
C -12,-524
D 12,-124

The correct answer is:
-12,-524

Solution:
6y=4x3-3x2
6dydx=12x2-6x
dydx=2x2-x=±1
2x2-x-1=0 or 2x2-x+1=0
x-12x+1=0 or x∈ϕ
x=1,x=-12
Hence, points are 1,16 and -12,-524

Topic: Mathematics, Application of Derivatives, JEE Main
The point on the curve \( y=x^{2} \) which is closest to \( \left(4,-\frac{1}{2}\right) \), is

Options:
A \( (1,1) \)
B \( (2,4) \)
C \( \left(\frac{2}{3}, \frac{4}{9}\right) \)
D \( \left(\frac{4}{3}, \frac{16}{9}\right) \)

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

Solution:
y=x2 Let any point on this parabola is t,t2

Equation of normal at this point is
x+2ty=t+2t3
It passes through 4,-12
4-t=t+2t3
2t3+2t-4=0
t3+t-2=0
t=1

So point is 1,1

Topic: Mathematics, Application of Derivatives, JEE Main
The point on the ellipse 16x2+9y2=400, where the ordinate decreases at the same rate at which the abscissa increases is a,b, then a+3b can be

Options:
A 16
B 19
C 6
D 9

The correct answer is:
19

Solution:
Differentiating the equation of ellipse, we get
32x dx dt +18y dy dt =0
â32xdxdt=-18ydydt
As dxdt=-dydt
â´32x=18y
â16x=9y
Putting in the equation of the ellipse, we get,
16 x 2 + 256 81 â9 x 2 =400
âx2+169x2=25
âx=Â±3
ây=Â±163
â´a=3,b=163
âa+3b=3+16=19

Topic: Mathematics, Application of Derivatives, JEE Main
The point on the ellipse x26+y23=1 which is nearest to the line x+y=7 is

Options:
A 2,2
B -2,-1
C 5,12
D 2,1

The correct answer is:
2,1

Solution:
Shortest distance always lies along the common normal
⇒ Slope of common normal is 1
Now, the equation of the normal at the point acos⁡θ,bsin⁡θ on x2a2+y2b2=1 is
axcosθ-bysinθ=a2-b2, where slope is abtan⁡θ
⇒63tan⁡θ=1⇒tan⁡θ=12
⇒sin⁡θ=13 and cos⁡θ=23
or sin⁡θ=-13, cos⁡θ=-23 (rejected because it gives farthest point)
⇒ Nearest point is 623,313=2,1

Topic: Mathematics, Ellipse, JEE Main
The point on the line \( \frac{x-2}{1}=\frac{y+3}{-2}=\frac{z+5}{-2} \) at a distance of \( 6 \) from the point \( (2,-3,-5) \) is

Options:
A \( (3,-5,-3) \)
B \( (4,-7,-9) \)
C \( (0,2,-1) \)
D \( (-3,5,3) \)

The correct answer is:
\( (4,-7,-9) \)

Solution:
Let point is x-21=y+3-2=z+5-2=λ
P≡ λ+2, -2λ-3,-2λ-5
Let, Q≡2, -3,-5
Given, PQ=6
⇒PQ2=36
⇒λ+2-22+-2λ-3+32+2λ-5+52=36
⇒9λ2=36
⇒λ2=4
⇒λ=± 2
∴ From λ=2, P≡4,-7,- 9.
And from λ=-2, P ≡0, 1, -1.

Topic: Mathematics, Three Dimensional Geometry, JEE Main
The point on the parabola $y^2=8 x$ at which the normal is inclined at $60^{\circ}$ to the $x$-axis has the co-ordinates

Options:
A $(6,-4 \sqrt{3})$
B $(6,4 \sqrt{3})$
C $(-6,-4 \sqrt{3})$
D $(-6,4 \sqrt{3})$

The correct answer is:
$(6,-4 \sqrt{3})$

Solution:
Normal at $(h, k)$ to the parabola $y^2=8 x$ is
$y-k=-\frac{k}{4}(x-h)$
Gradient $=\tan 60^{\circ}=\sqrt{3}=-\frac{k}{4}$ $\Rightarrow k=-4 \sqrt{3}$ and $h=6$
Hence required point is $(6,-4 \sqrt{3})$.

Topic: Mathematics, Parabola, JEE Main