datavorous/entrance-exam-dataset huggingface dataset:
If the length of a stretched string is shortened by $40 \%$ and the tension is increased by $44 \%$, then the ratio of the final and initial fundamental frequencies is

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

The correct answer is:
$2: 1$

Solution:
Fundamental frequency in case of string is
$n=\frac{1}{2 l} \sqrt{\frac{T}{m}} \Rightarrow n \propto \frac{\sqrt{T}}{1} \Rightarrow \frac{n^{\prime}}{n}=\sqrt{\frac{T^{\prime}}{T}} \times \frac{1}{r}$
putting $T^{\prime}=T+0.44 T=\frac{144}{100} T$ and $\left.l^{\prime}=|-0.4|=\frac{3}{5} \right\rvert\,$
We get $\frac{n^{\prime}}{n}=\frac{2}{1}$.

Topic: Physics, Waves and Sound, JEE Main
If the length of a stretched string is shortened by $40 \%$ and the tension is increased by $44 \%$, then the ratio of the final and initial fundamental frequencies is :

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

The correct answer is:
$2: 1$

Solution:
Fundamental frequency, $n=\frac{1}{2 l} \sqrt{\frac{T}{m}}$
Initial frequency, $n_1=\frac{1}{2 l} \sqrt{\frac{T}{m}}$
New length, $l^{\prime}=l-\frac{l \times 40}{100}=0.6 l$
New tension, $T^{\prime}=T+\frac{T \times 44}{100}$
$=1.44 \mathrm{~T}$
New frequency $n_2=\frac{1}{2 l^{\prime}} \sqrt{\frac{T^{\prime}}{m}}$
$\therefore \quad \frac{n_2}{n_1}=\frac{\frac{1}{2 l^{\prime}} \sqrt{\frac{T^{\prime}}{m}}}{\frac{1}{2 l} \sqrt{\frac{T}{m}}}$
$\frac{n_2}{n_1}=\frac{l}{l^{\prime}} \sqrt{\left(\frac{T^{\prime}}{T}\right)}$
$=\frac{l}{0.6 l} \sqrt{\frac{1.44 T}{T}}$
$=\frac{1}{0.6} \times 1.2$
$=2$
$\therefore \quad n_2: n_1=2: 1$

Topic: Physics, Waves and Sound, JEE Main
If the length of a vector be 21 and direction ratios be $2,-3,6$ then its direction cosines are

Options:
A $\frac{2}{21}, \frac{-1}{7}, \frac{2}{7}$
B $\frac{2}{7}, \frac{-3}{7}, \frac{6}{7}$
C $\frac{2}{7}, \frac{3}{7}, \frac{6}{7}$
D None of these

The correct answer is:
$\frac{2}{7}, \frac{-3}{7}, \frac{6}{7}$

Solution:
D.c.'s are $\frac{2}{\sqrt{2^2+(-3)^2+6^2}}, \frac{-3}{\sqrt{49}}$ and $\frac{6}{\sqrt{49}}$ or $\frac{2}{7}, \frac{-3}{7}, \frac{6}{7}$

Topic: Mathematics, Three Dimensional Geometry, JEE Main
If the length of a wire is made double and radius is halved of its respective values. Then, the Young's modulus of the material of the wire will :

Options:
A Remain same
B Become 8 times its initial value
C Become 1th 4 of its initial value
D Become 4 times its initial value

The correct answer is:
Remain same

Solution:
Young's modulus is the ratio of stress to strain.

Young's modulus is a property of a material, and it does not depend on the geometry i.e. length and the radius of the wire. Therefore, the Young's modulus of the wire will remain same.

Topic: Physics, Mechanical Properties of Solids, JEE Main, JEE Main 2022 (29 Jul Shift 1)
If the length of direct common tangent and transverse common tangent of two circles with integral radii are 3 units and 1 unit respectively, then the reciprocal of the square of the distance between the centres of the circles is equal to

Options:

The correct answer is:
0.1

Solution:
Let two circles are of radii r1 & r2 and with centres C1 & C2 respectively, then
C1C22-r1-r22=9 &
C1C22-r1+r22=1
⇒4r1r2=8⇒r1r2=2
Assume, r1=2 & r2=1
Then, C1C22=1+32=10
1C1C22=0.1

Topic: Mathematics, Circle, JEE Main
If the length of perpendicular drawn from origin on a plane is 7 units and its direction ratios are $-3,2,6$, then that plane is

Options:
A $-3 x+2 y+6 z-7=0$
B $-3 x+2 y+6 z-49=0$
C $3 x-2 y+6 z+7=0$
D $-3 x+2 y-6 z-49=0$

The correct answer is:
$-3 x+2 y+6 z-49=0$

Solution:
Equation of a plane, when direction ratio and length of perpendicular is given, $a x+b y+c z=p \sqrt{a^2+b^2+c^2}$
Given, $(a, b, c) \rightarrow(-3,2,6)$
$\begin{aligned}
& -3 x+2 y+6 z=7 \sqrt{(-3)^2+2^2+6^2} \\
& -3 x+2 y+6 z=49
\end{aligned}$

Topic: Mathematics, Three Dimensional Geometry, JEE Main
If the length of potentiometer wire is increased, then the length of the previously obtained balance point will

Options:
A Increase
B Decrease
C Remain unchanged
D Become two times

The correct answer is:
Increase

Solution:
When the length of potentiometer wire is increased, the potential gradient decreases and the length of previous balance point is increased.

Topic: Physics, Current Electricity, JEE Main
If the length of tangent drawn from the point $(5,3)$ to the circle $x^2+y^2+2 x+k y+17=0$ be 7 , then $k=$

Options:
A $4$
B $-4$
C $-6$
D $13 / 2$

The correct answer is:
$-4$

Solution:
According to the condition,
$\begin{aligned}& \sqrt{(5)^2+(3)^2+2(5)+k(3)+17}=7 \\& \Rightarrow 61+3 k=49 \Rightarrow k=-4 .\end{aligned}$

Topic: Mathematics, Circle, JEE Main
If the length of the chord of the circle, x2+y2=r2(r>0) along the line, y-2x=3 is r, then r2 is equal to:

Options:
A 95
B 12
C 245
D 125

The correct answer is:
125

Solution:
AB=r,AD=r2

CD=rsin60°=3r2
⇒0+0-312+22=3r2⇒r=235⇒r2=125

Topic: Mathematics, Circle, JEE Main, JEE Main 2020 (05 Sep Shift 2)
If the length of the latus rectum of an ellipse is 4 units and the distance between a focus and its nearest vertex on the major axis is 32 units, then its eccentricity is

Options:
A 23
B 12
C 19
D 13

The correct answer is:
13

Solution:
Given length of latus-rectum of an ellipse is 2b2a=4⇒b2=2a and a-ae=32⇒a1-e=32.

Now we know, b2=a21-e2

⇒b2=a1-ea1+e

⇒2a=32a1+e ∵ a1-e=32

⇒43=1+e

⇒e=13.

Topic: Mathematics, Ellipse, JEE Main, JEE Main 2018 (16 Apr Online)
If the length of the latus rectum of the ellipse x2+4y2+2x+8y-λ=0 is 4, and l is the length of its major axis, then λ+l is equal to _____.

Options:

The correct answer is:
75

Solution:
Given,

x2+4y2+2x+8y-λ=0 is a ellipse

Now on rearranging the terms we get,

⇒x+12λ+5+y+12λ+54=1

We know that length of latus rectum is 2b2a=4

So, 2λ+54λ+5=4

⇒2λ+54=4λ+5

⇒λ+5=8λ+5

⇒λ+52=64λ+5

⇒λ2+25+10λ=64λ+5

⇒λ2-54λ-295=0

⇒ λ=59 or -5

But λ≠-5

So, l=2a=2λ+5=264=16

⇒ λ+l=59+16=75

Topic: Mathematics, Ellipse, JEE Main, JEE Main 2022 (27 Jul Shift 1)
If the length of the major axis of the ellipse x2a2+y2b2=1 is three times the length of minor axis, its eccentricity is

Options:
A 13
B 13
C 23
D 223

The correct answer is:
223

Solution:
We have

x2a2+y2b2=1

Given,

Length of the major axis=3Length of the minor axis

⇒2a=32b

⇒a=3b

Now,

e=1-b2a2

⇒e=1-b29b2

⇒ e=1-19=83=223

Topic: Mathematics, Ellipse, JEE Main
If the length of the minor axis of ellipse is equal to half of the distance between the foci, then the eccentricity of the ellipse is :

Options:
A 53
B 32
C 13
D 25

The correct answer is:
25

Solution:
The length of minor axis is given by 2b and the distance between foci is given by 2ae

Given: 2b=12×2ae

⇒2b=ae

⇒4b2=a2e2

⇒4b2a2=e2

⇒41-e2=e2

⇒4-4e2=e2

⇒e=25

Topic: Mathematics, Ellipse, JEE Main, JEE Main 2024 (30 Jan Shift 1)
If the length of the pendulum in pendulum clock increases by 0.1%, then the error in time per day is:

Options:
A 43.2 s
B 8.64 s
C 86.4 s
D 4.32 s

The correct answer is:
43.2 s

Solution:
T=2πlg

T∝l12

ΔTT=12Δll

ΔT24×3600=12×0.1100

ΔT=12×0.1100×24×3600

ΔT=43.2 s

Topic: Physics, Mathematics in Physics, JEE Main, JEE Main 2021 (26 Aug Shift 2)
If the length of the perpendicular from the point β,0,β,β≠0 to the line, x1=y-10=z+1-1 is 32 , then β is equal to

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

The correct answer is:
-1

Solution:
Let Q=λ,1,-λ-1

Direction ratio's of PQ≡λ-β,1-0, -λ-1-β

∴PQ is perpendicular to given line, hence

1λ-β+01-1-λ-1-β=0

⇒λ-β+λ+1+β=0

⇒λ=-12

∴ PQ= λ-β2+12+-λ-1-β2=32

⇒-12-β2+12+12-1-β2=32

⇒β2+14+β+1+β2+14+β=32

⇒2β2+2β=0⇒β=0 or -1

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2019 (10 Apr Shift 1)
If the length of the potentiometer wire is increased, then the accuracy in the determination of the null point

Options:
A decreases
B increases
C remains unaffected
D none of these

The correct answer is:
increases

Solution:
As the length of the potentiometer wire increases, the potential gradient decreases, effectively increasing the accuracy.

Topic: Physics, Current Electricity, NEET
If the length of the projection of the line segment joining the points 1,2,-1 and 3,5,5 on the plane 3x-4y+12z=5 is equal to d units, then the value of 169 d2 is equal to

Options:

The correct answer is:
3925

Solution:
Let, point A be 1,2,-1, B be 3,5,5
ABâ=2i^+3j^+6k^
If Î¸ is the angle between the normal vector to the plane and ABâ then, the length of the projection =ABâsinâ¡Î¸
Normal vector =3i^-4j^+12k^
So, cosâ¡Î¸=ABâânâABânâ=6-12+72713=6691
Length of the projection =d=7Ã392591=515713 units

Topic: Mathematics, Three Dimensional Geometry, JEE Main
If the length of the simple pendulum is increased by $44 \%$, then what is the change in time period of pendulum

Options:
A $22 \%$
B $20 \%$
C $33 \%$
D $44 \%$

The correct answer is:
$20 \%$

Solution:
$\begin{gathered}T=2 \pi \sqrt{\frac{l}{g}} \Rightarrow \frac{T_2}{T_1}=\sqrt{\frac{I_2}{l_1}}=\sqrt{\frac{144}{100}}=\frac{12}{10} \\ \Rightarrow T_2=1.2 T_1\end{gathered}$
Hence $\%$ increase $=\frac{T_2-T_1}{T_1} \times 100=20 \%$

Topic: Physics, Oscillations, JEE Main
If the lengths of the sides of a right-angled triangle ABC, right angled at C, are in arithmetic progression, then the value of 5sin⁡A+sin⁡B is

Options:

The correct answer is:
7

Solution:
a+d2=a-d2+a2 (Pythagoras theorem)
a2+d2+2ad=a2+d2-2ad+a2⇒4ad=a2⇒a=4d
sin⁡A=a-da+d=4d-d4d+d=35
sin⁡B=aa+d=4d5d=45
⇒5sin⁡A+5sin⁡B=3+4=7
⇒5sin⁡A+sin⁡B=7

Topic: Mathematics, Sequences and Series, JEE Main
If the lengths of the sides of a triangle are decided by the three throws of a single fair die, then the probability that the triangle is of maximum area given that it is an isosceles triangle, is:

Options:
A 169
B 126
C 121
D 115

The correct answer is:
169

Solution:
In a Δ sum of two sides should be greater then the third side
It is given that Δ is isosceles
∴ Total cases in which isosceles Δ will come is
1,1,1,2,2,3,2,2,1,2,2,2,3,3,4,3,3,1,3,3,2,3,3,3,3,3,5,4,4,1,4,4,2,4,4,3,4,4,4,4,4,5,4,4,6,5,5,1,5,5,2,5,5,3,5,5,4,5,5,5,5,5,6,6,6,1,6,6,2,6,6,3,6,6,4,6,6,5,6,6,6
Total cases =27×3−12=69
Favorable cases =1
Probability = 169.

Topic: Mathematics, Probability, JEE Main, JEE Main 2015 (11 Apr Online)
If the lengths of the sides of a triangle are in A.P and the greatest angle is double the smallest, then a ratio of lengths of the sides of this triangle is:

Options:
A 3:4:5
B 5:6:7
C 5:9:13
D 4:5:6

The correct answer is:
4:5:6

Solution:
Let a<b<c are three sides and θ is the smallest angle. Hence three angles are θ, π-3θ, 2θ
Given, 2b=a+c⇒2sin⁡B=sin⁡A+sin⁡C (using sine rule)
⇒2sin 3θ=sin θ+sin 2θ
⇒23sin θ-4sin3θ=sin θ1+2cos θ
⇒6-81-cos2θ=1+2cos θ
⇒cos θ=34, -12 ( -12 is rejected since θ is acute)
∴ a:b:c=sinA:sin B:sin C=sin θ:sin 3θ:  sin 2θ
=1:3-4 sin2θ:2cos θ=1:4cos2θ-1:2 cos θ
=4:5:6

Topic: Mathematics, Properties of Triangles, JEE Main, JEE Main 2019 (08 Apr Shift 2)
If the lengths of the tangents from P1,3 and Q3,7 to a circle are 2 units and 18 units respectively, then the length of the tangent from R7,15 to the same circle is

Options:
A 98 units
B 170 units
C 50 units
D None of these

The correct answer is:
170 units

Solution:
Let, circle is x2+y2+2gx+2fy+C=0

and the length of the tangent from 7,15 as l units, then
1+9+2g+6f+C=2 … 1
9+49+6g+14f+C=18 … 2
49+225+14g+30f+C=l2 … 3
Now, equation 21+3 implies
2+18+49+225+18g+42f+3C=4+l2
⇒69+225+36g+14f+C=4+l2
⇒69+225+318-58=4+l2 (from 2)
⇒l2=225+69-4-120
⇒l2=170⇒l=170
⇒ length of the tangent from R is 170 units

Topic: Mathematics, Circle, JEE Main
If the lengths of the transverse axis and the latus rectum of a hyperbola are 6 and 83 respectively, then, the equation of the hyperbola is ________

Options:
A 4x2-9y2=72
B 4x2-9y2=36
C 9x2-4y2=72
D 9x2-4y2=36

The correct answer is:
4x2-9y2=36

Solution:
Given: 2a=6 and 2b2a=83
Hence, a=3,b=2
Then, equation of hyperbola is x29-y24=1

Topic: Mathematics, Hyperbola, JEE Main
If the letters of the word ALGORITHM are arranged a random in a row what is the probability the letters GOR must remain together as a unit?

Options:

None

Solution:
So, If we take GOR as one letter then the No. of favourable cases is 7 !
Required probability $=\frac{7 !}{9 !}=\frac{7 !}{(9)(8) 7 !}=\frac{1}{72}$

Topic: Mathematics, Probability
If the letters of the word CORONA are arranged in all possible ways and these words are written in order of a dictionary, then the word CORONA appears at the serial number

Options:
A 108
B 110
C 106
D 112

The correct answer is:
108

Solution:
ACNOOR
Number of words starting with
A→5!2!=60
CA→4!2!=12
CN→4!2!=12
COA→3!=6
CON→3!=6
COO→3!=6
CORA→2!=2
CORN→2!=2
COROAN→1=1
CORONA→1=1
108¯
So, the serial number of the word CORONA is 108.

Topic: Mathematics, Permutation Combination, JEE Main
If the letters of the word MATHS are permuted and all possible words so formed are arranged as in a dictionary with serial numbers, then the serial number of the word THAMS is

Options:
A

103

B 102
C 101
D 104

The correct answer is:

103

Solution:
The given word is MATHS

Arranging the letters alphabetically, we get

AHMST

When the word starts with any of the letters A/H/M/S, the number of possibilities=4!×4=96

Now when the word starts with TA, then the number of possibilities=3!=6

Now when the word starts with THAMS, then the number of possibilities =1

Rank =96+6+1=103

Hence, rank of the work THAMS is 103.

Topic: Mathematics, Permutation Combination, JEE Main, JEE Main 2023 (11 Apr Shift 2)
If the letters of the word 'MOTHER' be permuted and all the words so formed (with or without meaning) be listed as in a dictionary, then the position of the word 'MOTHER' is.....

Options:

The correct answer is:
309

Solution:
'MOTHER'

Rearrange in alphabetic order EHMORT

Number of words start with E=5!

Number of words start with H=5!

Number of words start with ME=4!

Number of words start with MH=4!

Number of words start with MOE=3!

Number of words start with MOH=3!

Number of words start with MOR=3!

Number of words start with MOTE=2!

Number of word MOTHER=1

2×5!+2×4!+3×3!+2!+1=240+48+18+2+1=309

Topic: Mathematics, Permutation Combination, JEE Main, JEE Main 2020 (02 Sep Shift 1)
If the letters of the word REGULATIONS be arranged in such a way that relative positions of the letters of the word GULATIONS remain the same then the probability that there are exactly 4 letters between R and E is

Options:
A 355
B 655
C 955
D 755

The correct answer is:
655

Solution:
The totals letters in the word REGULATIONS are 11.

There are 11 places corresponding to 11 letters and in these 11 places R and E can be arranged in P211 ways.

Now, Total possible cases where only 4 letters are between R and E are 6 as 1, 6, 2, 7, 3, 8, 4, 9, 5, 10, 6, 11 where the numbers represent the place number and on reversing the order, we get total 12 cases.

Hence, the required probability is 12P211

=12×9!11!

=655

Topic: Mathematics, Probability, AP EAMCET, AP EAMCET 2021 (19 Aug Shift 1)
If the letters of word SACHIN are arranged in all possible ways and these words are written out as in dictionary, then the word SACHIN appears at serial number

Options:
A
601

B
600

C
603

D
602

The correct answer is:
601

Solution:
Alphabetical order is
A, C, H, I, N, S
No. of words starting with $A-5$ !
No. of words starting with $C-5$ !
No. of words starting with $\mathrm{H}-5$ !
No. of words starting with I $-5$ !
No. of words starting with $N-5$ !
$\mathrm{SACHIN}-1$
$601$

Topic: Mathematics, Permutation Combination, JEE Main, JEE Main 2005
If the line 2x+6y=2 touches the hyperbola x2-2y2=a2, then a2 is equal to

Options:

The correct answer is:
4

Solution:
The tangent to the hyperbola is y=-26x+26

Also, tangent to hyperbola in slope form is y=mx±a2m2-b2

i.e. 26=a2-262-a22

On squaring, we get,
46=a2⋅46−a22⇒23=a26⇒a2=4

Topic: Mathematics, Hyperbola, JEE Main
If the line, 2 x-y+3=0 is at a distance 15 and 25 from the lines 4x-2y+α=0 and 6x-3y+β=0 respectively, then the sum of all possible values of α and β is ____________.

Options:

The correct answer is:
30

Solution:
2 x-y+3=0……(i)

4x-2y+α=0 ⇒ 2x-y+α2=0……(ii)6x-3y+β=0 ⇒ 2x-y+β3=0……(iii)

d1=α2-322+12=15⇒|α-6|=2

⇒ α-6=2,-2⇒ α=8,4

d2=β3-322+12=25 ⇒ |β-9|=6

⇒ β-9=6,-6 ⇒ β=15,3

Sum of all values of α and β=30.

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2020 (05 Sep Shift 1)
If the line $2 x+y+k=0$ is normal to the parabola $y^2=-8 x$, then the value of $k$ will be

Options:
A $-16$
B $-8$
C $-24$
D $24$

The correct answer is:
$24$

Solution:
$y=-2 x-k$ is normal to $y^2=-8 x$
or $-k=-\left\{-4(-2)-2(-2)^3\right\}=-(8+16) \Rightarrow k=24$.

Topic: Mathematics, Parabola, JEE Main
If the line $2 x+y=k$ passes through the point which divides the line segment joining the points $(1,1)$ and $(2,4)$ in the ratio $3: 2$, then $k$ equals

Options:
A
$\frac{29}{5}$

B
$5$

C
$6$

D
$\frac{11}{5}$

The correct answer is:
$6$

Solution:
Point $p=\left(\frac{6+2}{5}, \frac{12+2}{5}\right)$
$\mathrm{p}=\left(\frac{8}{5}, \frac{14}{5}\right)$
$p\left(\frac{8}{5}, \frac{14}{5}\right)$ lies on $2 x+y=k \quad \Rightarrow \frac{16}{5}+\frac{14}{5}=k \quad \Rightarrow k=\frac{30}{5}=6$

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2012 (Offline)
If the line $3 x-4 y-7=0$ and $2 x-3 y-5=0$ pass through diameters of a circle of area $49 \pi$ square units, then the equation of the circle is

Options:
A $x^2+y^2-2 x+2 y-47=0$
B $x^2+y^2+2 x-2 y-51=0$
C $x^2+y^2-2 x+2 y+51=0$
D $x^2+y^2+2 x+2 y+47=0$

The correct answer is:
$x^2+y^2-2 x+2 y-47=0$

Solution:
Centre of the circle is the point of intersection of the diameters
$3 x-4 y-7=0$ and $2 x-3 y-5=0$
which is $(1,-1)$ and $r=7$
$\begin{aligned} & (x-1)^2+(y+1)^2=7^2 \\ & \Rightarrow x^2+y^2-2 x+2 y-47=0\end{aligned}$

Topic: Mathematics, Circle, JEE Main
If the line 3x+4y-24=0 intersects the x-axis is at the point A and the y-axis at the point B, then the incentre of the triangle OAB, where O is the origin, is:

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

The correct answer is:
2, 2

Solution:
The x intercept of the line 3x+4y-24=0 can be obtained by putting y=0 i.e. 3x-24=0

⇒x=8

Hence, the point A is 8, 0.

Now, the y intercept of the line 3x+4y-24=0 can be obtained by putting x=0 i.e. 4y-24=0

⇒y=6

Hence, the point B is 0, 6.

In ∆OAB, we have OA=b=8, OB=a=6 and AB=c=8-02+0-62=64+36=10 units.

And, we know that the co-ordinates of incentre I of a triangle with vertices x1, y1, x2, y2 and x3, y3 and having length of the sides opposite to these vertices as a, b and c, respectively, is

I≡ax1+bx2+cx3a+b+c, ay1+by2+cy3a+b+c

⇒I≡68+80+10024, 60+86+10024

⇒I=2, 2.

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2019 (10 Jan Shift 1)
If the line $3 x-m y+5=0$ is a tangent to the hyperbola $3 x^2-4 y^2=300$ then the square of the $Y-$ intercept made by this tangent line is

Options:
A $\frac{25}{3}$
B $\frac{35}{3}$
C $\frac{45}{7}$
D $\frac{15}{7}$

The correct answer is:
$\frac{15}{7}$

Solution:
$3 x^2-4 y^2=300$
$$
\frac{x^2}{100}-\frac{y^2}{75}=1
$$
and $3 x-m y+5=0 \Rightarrow y=\frac{3}{m} x+\frac{5}{m}$ Now, for condition of tangents:
$$
\begin{aligned}
& \mathrm{c}^2=\mathrm{a}^2 \mathrm{~m}^2-\mathrm{n}^2 \Rightarrow \frac{25}{\mathrm{~m}^2}=100 \times \frac{9}{\mathrm{~m}^2}-75 \\
& \Rightarrow \mathrm{m}^2=\frac{35}{3}
\end{aligned}
$$
Square of $y$ - intercept of $x$-radius
$$
\mathrm{c}=\frac{25}{\mathrm{~m}^2}=25 \times \frac{3}{35}=\frac{15}{7}
$$

Topic: Mathematics, Hyperbola, JEE Main
If the line $\mathrm{a} x+\mathrm{b} y+\mathrm{c}=0$ is a normal to the curve $x y=1$, then

Options:
A $a>0, b>0$
B $a>0, b < 0$
C $\mathrm{a} < 0, \mathrm{~b} < 0$
D $a=0, b=0$

The correct answer is:
$a>0, b < 0$

Solution:
$\begin{aligned}
& x y=1 \\
\therefore \quad & y=\frac{1}{x} \\
\therefore & y^{\prime}=\frac{-1}{x^2}
\end{aligned}$
$\therefore \quad$ Slope of the normal $=x^2$
Slope of the line $a x+b y+c=0$ is $\frac{-a}{b}$.
Since the line $a x+b y+c=0$ is a normal to the curve $x y=1$,
$x^2=-\frac{\mathrm{a}}{\mathrm{b}}$
For this condition to hold true, either $\mathrm{a} < 0, \mathrm{~b}>0$ or $\mathrm{b} < 0, \mathrm{a}>0$

Topic: Mathematics, Hyperbola, JEE Main
If the line ax+y=c, touches both the curves x2+y2=1 and y2=42x, then c is equal to:

Options:
A 12
B 2
C 12
D 2

The correct answer is:
2

Solution:
The equation of the tangent to the parabola y2=4ax is y=mx+am.

Hence, the equation of tangent to the parabola y2=42x is y=mx+2m ...1

A line y=mx+c is a tangent to the circle x2+y2=r2 if c2=r21+m2.

Hence, the line 1 is also a tangent to the circle x2+y2=1, if 2m2=1m2+1

⇒m2m2+1-2=0

⇒m4+m2-2=0

⇒m2+2m2-1=0

⇒m2=1 since m2+2≠0

⇒m=±1.

Thus, the equation of the common tangent is y=±x±2, ⇒±x-y=∓2.

Given, the equation of the common tangent is ax+y=c, hence, c=±2 and c=2.

Topic: Mathematics, Parabola, JEE Main, JEE Main 2019 (10 Apr Shift 2)
If the line joining the points \( (0,3) \) and \( (5,-2) \) is a tangent to the curve \( y=\frac{c}{x+1} \), then find the value of \( c \)

Options:

The correct answer is:
4.00

Solution:
Given that

line joining (0, 3)(5,-2)
y-3=-2-35-0x-0
y-3=-x
y=3-x is tangent to curve i.e
touches y=cx+1

3-x=cx+1
3x+3-x2-x=c
x2+2x+(c-3)=0
D=0
(-2)2-4×1(c-3)=0
4-4 c+12=0
c=4

Topic: Mathematics, Application of Derivatives, JEE Main
If the line joining the points $\hat{\mathbf{i}}+\hat{\mathbf{j}}$ and $3 \hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}}$ meets the plane that passes through the point $2 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}$ and parallel to the vectors $3 \hat{\mathbf{j}}+5 \hat{\mathbf{k}}$ and $3 \hat{\mathbf{i}}-\hat{\mathbf{k}}$ at, $P$, then the position vector of the point $P$ is

Options:
A $-27 \hat{\mathbf{i}}+\hat{\mathbf{j}}+14 \hat{\mathbf{k}}$
B $29 \hat{\mathbf{i}}+\hat{\mathbf{j}}-14 \hat{\mathbf{k}}$
C $-14 \hat{\mathbf{i}}+89 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}$
D $2 \hat{\mathbf{i}}+5 \hat{\mathbf{j}}-7 \hat{\mathbf{k}}$

The correct answer is:
$29 \hat{\mathbf{i}}+\hat{\mathbf{j}}-14 \hat{\mathbf{k}}$

Solution:
Equation of line joining points $(\hat{\mathbf{i}}+\hat{\mathbf{j}})$ and $(3 \hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}})$ in cartesian form is
$$
\frac{x-1}{2}=\frac{y-1}{0}=\frac{z}{-1}=r(\text { let })
$$
Then, point on line Eq. (i) is $p(2 r+1,1,-r)$.
Now, equation of plane passes through the point $(2 \hat{\mathbf{i}}+4 \hat{\mathbf{j}})$ and parallel to the vectors $(3 \hat{\mathbf{j}}+5 \hat{\mathbf{k}})$ and $(3 \hat{\mathbf{i}}-\hat{\mathbf{k}})$ in cartesian form is
$$
\begin{aligned}
& \left|\begin{array}{ccc}
x-2 & y-4 & z \\
0 & 3 & 5 \\
3 & 0 & -1
\end{array}\right| & =0 \\
\Rightarrow & -3(x-2)+15(y-4)-9 z & =0 \\
\Rightarrow & 3 x-15 y+9 z+54 & =0
\end{aligned}
$$
Let the point $P(2 r+1,1,-r)$ on the plane Eq. (ii) itself, so $r=14$
So, position vector of the point $P$ is
$$
\mathbf{O P}=29 \hat{\mathbf{i}}+\hat{\mathbf{j}}-14 \hat{\mathbf{k}} .
$$

Topic: Mathematics, Three Dimensional Geometry, JEE Main
If the line l1:3y-2x=3 is the angular bisector of the lines l2:x-y+1=0 and l3:αx+βy+17=0, then α2+β2-α-β is equal to ............

Options:

The correct answer is:
348

Solution:
Given,

L1:3y-2x=3 is angular bisector of L2:x-y+1=0 and L3:αx+βy+17=0

Now finding point of intersection of L1 & L2 we get, 0, 1

And point will lie on L3, so α×0+β×1+17=0

⇒β=-17
Any point, say -32,0 on L1 should be equidistant from lines L2 & L3

Now using the formula of distance of a point from the line we get,
⇒-32-0+112+12=-3α2+0+17α2+(-17)2

⇒α-7 α-17=0

Now, for α=17, L2 & L3 coincides

So, α=7

Now putting the value of α & β in given expression we get,

α2+β2-α-β=-172+72-7+17=348

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2023 (11 Apr Shift 2)