datavorous/entrance-exam-dataset huggingface dataset:
The equation of the circle which touches the X-axis and Y-axis at the points (1, 0) and (0, 1) respectively is

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

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

Solution:
On plotting the diagram based on the given information, we find centre = (1,1) and radius = 1
$\therefore$ Equation of circle, $(x-1)^2+(y-1)^2=1^2$
$$
x^2+y^2-2 x-2 y+1=0
$$

Topic: Mathematics, Circle, JEE Main
The equation of the circle which touches $x$-axis at $(3,0)$ and passes through $(1,4)$ is given by

Options:
A $x^2+y^2-6 x-5 y+9=0$
B $x^2+y^2+6 x+5 y-9=0$
C $x^2+y^2-6 x+5 y-9=0$
D $x^2+y^2+6 x-5 y+9=0$

The correct answer is:
$x^2+y^2-6 x-5 y+9=0$

Solution:
$k^2=4+(k-4)^2 \Rightarrow k=\frac{5}{2}$
Hence required equation of circle is
$(x-3)^2+\left(y-\frac{5}{2}\right)^2=\left(\frac{5}{2}\right)^2 \Rightarrow x^2+y^2-6 x-5 y+9=0$
Trick : Only (a) passes through $(1,4)$.

Topic: Mathematics, Circle, JEE Main
The equation of the circle whose radius is 5 units and which touches the circle x2+y2-2x-4y-20=0 externally at the point 5,5 is

Options:
A x-92+y+82=25
B x-92+y-82=25
C x+82+y+82=25
D x+82+y-92=25

The correct answer is:
x-92+y-82=25

Solution:
The equation of the given circle is
x2+y2-2x-4y-20=0
Its centre is C11,2 and radius =5.
The circle touches another circle of radius 5 externally at P5,5. Let its centre be C2Î±,Î².
Clearly, P5,5 is the midpoint of C1C2. Therefore,
Î±+12=5 and Î²+22=5âÎ±=9,Î²=8.
Hence, the equation of the required circle is
x-Î±2+y-Î²2=52
or, x-92+y-82=52

Topic: Mathematics, Circle, JEE Main
The equation of the circumcircle of the square formed by the lines x2-8x+12=0 and y2-14y+45=0 is

Options:
A x2+y2-4x-7y+57=0
B x2+y2-8x-14y+57=0
C x2+y2-8x-14y+5=0
D 2x2+y2-8x-14y+57=0

The correct answer is:
x2+y2-8x-14y+57=0

Solution:
The given equation of lines are x-2x-6=0, y-5y-9=0

i.e. x=2, x=6, y=5, y=9

Circle with segment joining 2,5 and 6,9 as the diameter is x-2x-6+y-5y-9=0

Topic: Mathematics, Circle, JEE Main
The equation of the common chord of the circles $(x-a)^2+(y-b)^2=c^2$ and $(x-b)^2+(y-a)^2=c^2$ is

Options:
A $x-y=0$
B $x+y=0$
C $x+y=a^2+b^2$
D $x-y=a^2-b^2$

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

Solution:
We know that the equation of common chord is $S_1-S_2=0$, where $S_1$ and $S_2$ are the equations of given circles, therefore
$\begin{aligned} & (x-a)^2+(y-b)^2+c^2-(x-b)^2-(y-a)^2-c^2=0 \\ & \Rightarrow 2 b x-2 a x+2 a y-2 b y=0 \\ & \Rightarrow 2(b-a) x-2(b-a) y=0 \Rightarrow x-y=0\end{aligned}$

Topic: Mathematics, Circle, JEE Main
The equation of the common tangent drawn to the curves $y^2=8 x$ and $x y=-1$ is

Options:
A $y = 2x + 1$
B $2y = x + 6$
C $y = x + 2$
D $3y = 8x + 2$

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

Solution:
Let $P\left(t, \frac{-1}{t}\right)$ be a point on $x y=-1$ Equation of tangent of $P$
$\begin{aligned}
& =\frac{1}{2}\left[x\left(\frac{-1}{t}\right)+x y\right]=-1-x+t^2 y=-2 t \\
& y=\frac{x}{t^2}+\frac{2}{t}
\end{aligned}$
This is also at tangent of $y^2=8 x$
$\begin{aligned}
& \text { i.e. } c=\frac{a}{m} \\
& \frac{2}{t}=\frac{2}{\frac{1}{t^2}} \\
& \Rightarrow t^3=1 \\
& \Rightarrow t=1
\end{aligned}$
Hence, the equation of common tangent
$\begin{aligned}
& y=\frac{x}{1}+\frac{2}{1} \\
& y=x+2
\end{aligned}$

Topic: Mathematics, Hyperbola, JEE Main
The equation of the common tangent to the curves y2=4x and x2+32y=0 is x+by+c=0. The value of |sin–1(sin 1)+sin–1(sin b)+sin–1(sin c)| is equal to

Options:

The correct answer is:
1

Solution:
y = mx + 1 m is tangent to y2=4x
Now, x 2 + 32 mx + 1 m = 0 ⇒ x 2 + 3 2 mx + 3 2 m = 0
⇒ 3 2 m 2 - 4 × 3 2 m = 0 ⇒ m = 1 2
Equation of the tangent is x–2y+4=0
⇒|sin-1(sin1)+sin-1(sin(-2))+sin-1(sin4)|
=|1+2–π+π–4|=1

Topic: Mathematics, Parabola, JEE Main
The equation of the common tangent to the parabola \( y^{2}=8 x \) and rectangular hyperbola \( x y=-1 \) is

Options:
A \( x-y+2=0 \)
B \( 9 x-3 y+2=0 \)
C \( 2 x-y+1=0 \)
D \( x+2 y-1=0 \)

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

Solution:
Equation of tangent of a parabola y2=8x is y=mx+2m.

Let, y=mx+2m is a common tangent of parabola y2=8x and hyperbola xy=-1.

⇒xmx+2m=-1

⇒mx2+2mx+1=0

Apply the discriminant =0

⇒2m2-4m·1=0

⇒m3=1

⇒m=1

Hence, the common tangent of the parabola y2=8x and the hyperbola xy=-1 is y=x+2.

Topic: Mathematics, Hyperbola, JEE Main
The equation of the common tangent to the parabolas y2=2x and x2=16y is

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

The correct answer is:
x+2y+2=0

Solution:
Any tangent to y2=2x in slope form is
y=mx+12m which is also tangent to x2=16y
âx2=16mx+12m will have equal roots.
â Discriminant of x2-16mx-8m=0 is zero
â16m2-4Ã1-8m=0

âm3=-18âm=-12
â Equation of the common tangent is y=-x2-1
âx+2y+2=0

Topic: Mathematics, Parabola, JEE Main
The equation of the common tangent touching the circle $(x-3)^2+y^2=9$ and the parabola $y^2=4 x$ above the $x$-axis, is

Options:
A $\sqrt{3} y=3 x+1$
B $\sqrt{3} y=-(x+3)$
C $\sqrt{3} y=x+3$
D $\sqrt{3} y=-(3 x+1)$

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

Solution:
Any tangent to $y^2=4 x$ is
$y=m x+\frac{1}{m}$ It touches the circle, if $3=\left|\frac{3 m+\frac{1}{m}}{\sqrt{1+m^2}}\right|$
or $\quad 9\left(1+m^2\right)=\left(3 m+\frac{1}{m}\right)^2$
or $\quad \frac{1}{m^2}=3 \quad \therefore m= \pm \frac{1}{\sqrt{3}}$.
For the common tangent to be above the $x$-axt
$\therefore$ Common tangent is, $\quad y=\frac{1}{\sqrt{3}} x+\sqrt{3}$ $\Rightarrow \sqrt{3} y=x+3$.

Topic: Mathematics, Parabola, JEE Main
The equation of the conic with focus at $(1,-1)$ directrix along $x-y+1=0$ and with eccentricity $\sqrt{2}$ is

Options:
A $x^{2}-y^{2}=1$
B $x y=1$
C $2 x y-4 x+4 y+1=0$
D $2 x y+4 x-4 y-1=0$

The correct answer is:
$2 x y-4 x+4 y+1=0$

Solution:
Let $\mathrm{P}(\mathrm{x}, \mathrm{y})$ be any point on the conic. Then,
$$
\begin{array}{l}
\sqrt{(x-1)^{2}+(y+1)^{2}}=\sqrt{2}\left(\frac{x-y+1}{\sqrt{2}}\right) \\
\Rightarrow(x-1)^{2}+(y+1)^{2}=(x-y+1)^{2} \\
\Rightarrow 2 x y-4 x+4 y+1=0
\end{array}
$$

Topic: Mathematics, Parabola, JEE Main
The equation of the curve for which the slope of the tangent at any point is given by x+y+1dydx=1 is (where, c is an arbitrary constant)

Options:
A xy=ex-c
B xy=cey+2
C x=cey-y-2
D x=ey+y-c

The correct answer is:
x=cey-y-2

Solution:
The given equation is dxdy+x-1=y+1
Integrating Factor=e-∫dy=e-y
Thus, the curve is x⋅ e −y =∫ y+1 e −y dy
⇒xe-y=-y+1e-y-e-y+c
or x=cey-y-2

Topic: Mathematics, Differential Equations, JEE Main
The equation of the curve lying in the first quadrant, such that the portion of the x-axis cut-off between the origin and the tangent at any point P is equal to the ordinate of P, is (where, c is an arbitrary constant)

Options:
A y=cexy
B yexy=c
C yeyx=c
D y=ceyx

The correct answer is:
yexy=c

Solution:
Equation of the tangent at Px,y
is Y-y=dydxX-x
Putting Y=0, we get,

X=x-yy'=y
â´dxdy-xy=-1 is a linear differential equation.
Î.F.=e-lnâ¡y=1y
Thus, the solution is x1y=â«-11ydy
or xy=-lnâ¡y+lnâ¡c
âcy=exy
Hence, yexy=c is the equation of the curve

Topic: Mathematics, Differential Equations, JEE Main
The equation of the curve passing through the origin and satisfying the differential equation $\left(1+x^2\right) \frac{d y}{d x}+2 x y=4 x^2$ is

Options:
A
$\left(1+x^2\right) y=x^3$

B
$3\left(1+x^2\right) y=2 x^3$

C
$\left(1+x^2\right) y=3 x^3$

D
$3\left(1+x^2\right) y=4 x^3$

The correct answer is:
$3\left(1+x^2\right) y=4 x^3$

Solution:
Given differential equation is
$$
\begin{aligned}
&\left(1+x^2\right) \frac{d y}{d x}+2 x y=4 x^2 \\
&\Rightarrow \frac{d y}{d x}+\left(\frac{2 x}{1+x^2}\right) y=\frac{4 x^2}{1+x^2}
\end{aligned}
$$
This is linear diff. equation
$$
\mathrm{I} . \mathrm{F}=e^{\int \frac{2 x}{1+x^2} d x}=e^{\log \left(1+x^2\right)}=1+x^2
$$
Solution is
$$
\begin{aligned}
&y\left(1+x^2\right)=\int \frac{4 x^2}{1+x^2} \times 1+x^2+\mathrm{C} \\
&\Rightarrow y\left(1+x^2\right)=\frac{4 x^3}{3}+\mathrm{C} \\
&\Rightarrow \text { Required curve is } \\
&3 y\left(1+x^2\right)=4 x^3(\because \mathrm{C}=0)
\end{aligned}
$$

Topic: Mathematics, Differential Equations, JEE Main
The equation of the curve passing through the origin and satisfying the differential equation $\left(1+x^2\right) \frac{d y}{d x}+2 x y=4 x^2$ is

Options:
A
$\left(1+x^2\right) y=x^3$

B
$3\left(1+x^2\right) y=2 x^3$

C
$\left(1+x^2\right) y=3 x^3$

D
$3\left(1+x^2\right) y=4 x^3$

The correct answer is:
$3\left(1+x^2\right) y=4 x^3$

Solution:
Given differential equation is
$$
\begin{aligned}
&\left(1+x^2\right) \frac{d y}{d x}+2 x y=4 x^2 \\
&\Rightarrow \frac{d y}{d x}+\left(\frac{2 x}{1+x^2}\right) y=\frac{4 x^2}{1+x^2}
\end{aligned}
$$
This is linear diff. equation
$$
\mathrm{I} . \mathrm{F}=e^{\int \frac{2 x}{1+x^2} d x}=e^{\log \left(1+x^2\right)}=1+x^2
$$
Solution is
$$
\begin{aligned}
&y\left(1+x^2\right)=\int \frac{4 x^2}{1+x^2} \times 1+x^2+\mathrm{C} \\
&\Rightarrow y\left(1+x^2\right)=\frac{4 x^3}{3}+\mathrm{C} \\
&\Rightarrow \text { Required curve is } \\
&3 y\left(1+x^2\right)=4 x^3(\because \mathrm{C}=0)
\end{aligned}
$$

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2013 (25 Apr Online)
The equation of the curve passing through the point 1,1 and satisfying the differential equation dydx=x+2y-3y-2x+1 is

Options:
A x2-4xy-y2+6x+2y-4=0
B x2+4xy-y2-6x+2y+4=0
C x2+4xy-y2-6x-2y+4=0
D x2+4xy+y2-6x-2y-4=0

The correct answer is:
x2+4xy-y2-6x-2y+4=0

Solution:
dydx=x+2y-3y-2x+1
ydy-2xdy+dy=xdx+2ydx-3dx
ydy-2xdy+ydx+dy-xdx+3dx=0
ydy-2dxy+dy-xdx+3dx=0

On integrating, we get,
y22-2xy+y-x22+3x+C=0
Since, the curve passes through 1,1

⇒12-2+1-12+3+C=0
C=-2
y2-4xy+2y-x2+6x-4=0 or x2+4xy+y2-6x-2y+4=0

Topic: Mathematics, Differential Equations, JEE Main
The equation of the curve satisfying the differential equation dydx+2yx2=2x2 and passing through 12,e4+1 is

Options:
A y=e2x+1
B y=e2x-1
C y=1+ e 2 x
D y=1+e-x

The correct answer is:
y=1+ e 2 x

Solution:
Given equation is a linear differential equation
Integrating factor=e∫2x2dx=e−2x
∴ The solution is y e − 2 x =∫ 2 e − 2 x x 2 dx
or y e − 2 x =∫ e t dt (Putting -2x=t)
⇒y e − 2 x = e t +c
⇒y e − 2 x = e − 2 x +c
⇒y=1+c e 2 x
At x=12,y=e4+1,
⇒c=1
∴ The equation of the curve is y=1+ e 2 x

Topic: Mathematics, Differential Equations, JEE Main
The equation of the curve satisfying the differential equation dydx+yx2=1x2 and passing through 12,e2+1 is

Options:
A y=ex+1
B y=e1x-1
C y=1+e1x
D y=1+e-x

The correct answer is:
y=1+e1x

Solution:
Given equation is linear with I.F. =e∫1x2dx=e-1x
∴ the solution is ye-1x=∫e-1xx2dx

or ye-1x=∫etdt put -1x=t
⇒ye-1x=et+c
⇒y=1+ce1x
At x=12, y=e2+1⇒c=1
∴ the equation of the curve is y=1+e1x

Topic: Mathematics, Differential Equations, JEE Main
The equation of the curve satisfying the differential equation x2dy=2-ydx and passing through P1,4 is

Options:
A y=x2+3
B y=2+2e1x-1
C y=sin⁡x-1+4
D y=2ex-1+2

The correct answer is:
y=2+2e1x-1

Solution:
The given equation is dydx=2-yx2
⇒dydx+y1x2=2x2
Integrating factor =e∫dxx2=e-1x
∴ the solution is ye-1x=∫2x2e-1xdx
Let -1x=t
dxx2=dt
i.e. y⋅e-1x=2∫etdt
⇒ye-1x=2et+c⇒ye-1x=2e-1x+c
⇒y=2+c⋅e1x
Since, the curve is passing through P1,4

⇒2e=c

⇒y=2+2e⋅e1x

Topic: Mathematics, Differential Equations, JEE Main
The equation of the curve satisfying the differential equation xexsinâ¡ydy-x+1excosâ¡ydx=ydy and passing through the origin is

Options:
A xex=y2cosy
B 2xex=ycos⁡y
C 2xexcos⁡y+y2=0
D 2xexcos⁡y=y2

The correct answer is:
2xexcos⁡y+y2=0

Solution:
The given equation is d-xexcos⁡y=ydy
On integrating, we get –xexcos⁡y=y22+c
As it passes through the origin, c=0
∴ the equation of the curve is 2xexcos⁡y+y2=0

Topic: Mathematics, Differential Equations, JEE Main
The equation of the curve satisfying the equation \( \left(x y-x^{2}\right) \frac{d y}{d x}=y^{2} \) and passing through the point \( (-1,1) \) is

Options:
A \( y=(\log \mathrm{y}-1) x \)
B \( y=(\log \mathrm{y}+1) x \)
C \( x=(\log \mathrm{x}-1) y \)
D \( x=(\log \mathrm{x}+1) y \)

The correct answer is:
\( y=(\log \mathrm{y}-1) x \)

Solution:
dy dx = y 2 xy - x 2
Put, y = vx
â dy dx = v + x dv dx
â v + x dv dx = v 2 v - 1
â x dv dx = v 2 v - 1 - v = v v - 1
â â« v - 1 v dv = â« dx x
â â« 1 - 1 v dv = â« dx x
â v - â n v = â n x + c
â y x = â n y + c
Alternate Solution :
We have, x y - x 2 dâ¡ y dâ¡ x = y 2
â y 2 dâ¡ x dâ¡ y = x y - x 2
â 1 x 2 dâ¡ x dâ¡ y - 1 x Â· 1 y = - 1 y 2
Put 1 x = v â - 1 x 2 dâ¡ x dâ¡ y = dâ¡ v dâ¡ y
â´ dâ¡ v dâ¡ y + v y = 1 y 2 , which is linear
â´ IF = eâ¡ â« 1 y dâ¡ y = eâ¡ log y = y
â´ The solution is vy = â« 1 y 2 Â· y dâ¡ y + c
â y x = log |y| + c
â y = x log |y| + c
This posses through the point (â1, 1).
â´ 1 = - 1 log 1 + c
ie, c = - 1
Thus, the equation of the curve is
y = x ( log |y| â 1)

Topic: Mathematics, Differential Equations, JEE Main
The equation of the curve that passes through the point $(1,2)$ and satisfies the differential equation $\frac{d y}{d x}=\frac{-2 x y}{\left(x^2+1\right)}$ is

Options:
A $y\left(x^2+1\right)=4$
B $y\left(x^2+1\right)+4=0$
C $y\left(x^2-1\right)=4$
D None of these

The correct answer is:
$y\left(x^2+1\right)=4$

Solution:
$\frac{d y}{d x}=\frac{-2 x y}{\left(x^2+1\right)} \Rightarrow \frac{d y}{y}=-\frac{2 x}{x^2+1} d x$
On integrating, we get
$\log y=-\log \left(1+x^2\right)+\log c \Rightarrow y\left(1+x^2\right)=c$
$c=2\left(1+1^2\right) \Rightarrow c=4$
Hence solution is $y\left(x^2+1\right)=4$

Topic: Mathematics, Differential Equations, JEE Main
The equation of the curve through the point $(1,0)$ and whose slope is $\frac{y-1}{x^2+x}$ is

Options:
A $(y-1)(x+1)+2 x=0$
B $2 x(y-1)+x+1=0$
C $x(y-1)(x+1)+2=0$
D None of these

The correct answer is:
$(y-1)(x+1)+2 x=0$

Solution:
Slope $=\frac{d y}{d x}$
$\begin{aligned} \Rightarrow & \frac{d y}{d x}=\frac{y-1}{x^2+x} \Rightarrow \frac{d y}{y-1}=\frac{d x}{x^2+x} \\ \Rightarrow & \int \frac{1}{y-1} d y=\int\left(\frac{1}{x}-\frac{1}{x+1}\right) d x+c \Rightarrow \frac{(y-1)(x+1)}{x}=k\end{aligned}$
Putting $x=1, y=0$, we get $k=-2$
Hence the equation is $(y-1)(x+1)+2 x=0$

Topic: Mathematics, Differential Equations, JEE Main
The equation of the curve which passes through the point $(1,1)$ and whose slope is given by $\frac{2 y}{x}$, is

Options:
A $y=x^2$
B $x^2-y^2=0$
C $2 x^2+y^2=3$
D None of these

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

Solution:
Slope $\frac{d y}{d x}=\frac{2 y}{x}$
$\Rightarrow 2 \int \frac{d x}{x}=\int \frac{d y}{y} \Rightarrow 2 \log x=\log y+\log c \Rightarrow x^2=y c$
Since it passes through $(1,1)$, therefore $c=1$
Hence $x^2-y=0 \Rightarrow y=x^2$.

Topic: Mathematics, Differential Equations, JEE Main
The equation of the curve whose slope at any point is equal to y+2x and which passes through the origin is

Options:
A y=2x-1
B y=2ex-x-1
C y=2ex-1
D y=2exx-1

The correct answer is:
y=2ex-x-1

Solution:
We have, dydx=y+2x

⇒dydx-y=2x

Now, IF=e-∫1dx=e-x

y·e-x=∫2xe-xdx+k, k be the constant of integration

=2x∫e-xdx-∫1·-e-xdx+k [using integration by parts]

⇒y·e-x=-2xe-x-2e-x+k ...i

As curve (i) passes through 0,0

∴ 0=0-2+k

⇒k=2

Thus, the curve is

ye-x=-2xe-x-2e-x+2

∴ y=2ex-x-1

Hence, option (b) is correct.

Topic: Mathematics, Differential Equations, JEE Main
The equation of the curve whose vertex is $(0,0)$ and length of latusrectum is $\frac{16}{3}$, is

Options:
A $8 x^{2}+3 y^{2}=72$
B $16 y^{2}=3 x$
C $3 y^{2}=16 x$
D $3 x^{2}+16 y^{2}=48$

The correct answer is:
$3 y^{2}=16 x$

Solution:
For the curve $3 y^{2}=16 x$, the length of latusrectum is $\frac{16}{3}$

Topic: Mathematics, Parabola, JEE Main
The equation of the diameter of the circle $x^2+y^2+2 x-4 y-11=0$ which bisects the chords intercepted on the line $2 x-y+3=0$ is

Options:
A $x+y-7=0$
B $2 x-y-5=0$
C $x+2 y-3=0$
D None of these

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

Solution:
We know that the diameter will be perpendicular to the chord because it bisects the chord. Hence the equation of diameter is $x+2 y+k=0...(i)$
But (i) also passes through the centre of circle $(-1,2)$. Therefore, $-1+4+k=0 \Rightarrow k=-3$
Hence the required equation is $x+2 y-3=0$

Topic: Mathematics, Circle, JEE Main
The equation of the director circle of the ellipse $4 x^2+9 y^2=36$, is given by

Options:
A $x^2+y^2=4$
B $x^2+y^2=9$
C $x^2+y^2=13$
D $x^2+y^2=5$

The correct answer is:
$x^2+y^2=13$

Solution:
Given $4 x^2+9 y^2=36$
Equation ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$
Here $a=3$, and $b=2$
Director circle equation
$\begin{gathered}x^2+y^2=a^2+b^2 \\ x^2+y^2=13\end{gathered}$

Topic: Mathematics, Ellipse, JEE Main
The equation of the director circle of the hyperbola $\frac{x^2}{16}-\frac{y^2}{4}=1$ is given by

Options:
A $x^2+y^2=16$
B $x^2+y^2=4$
C $x^2+y^2=20$
D $x^2+y^2=12$

The correct answer is:
$x^2+y^2=12$

Solution:
Equation of 'director-circle' of hyperbola is $x^2+y^2=a^2-b^2$. Here $a^2=16, b^2=4$
$\therefore x^2+y^2=12$ is the required 'director circle'.

Topic: Mathematics, Hyperbola, JEE Main
The equation of the director circle of the hyperbola $\frac{x^{2}}{16}-\frac{y^{2}}{4}=1$ is given by

Options:
A $x^{2}+y^{2}=16$
B $x^{2}+y^{2}=4$
C $x^{2}+y^{2}=20$
D $x^{2}+y^{2}=12$

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

Solution:
Equation of director circle of hyperbola,
$\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ is $x^{2}+y^{2}=a^{2}-b^{2}$
Hence, director circle of $\frac{x^{2}}{16}-\frac{y^{2}}{4}=1$ is
$$
x^{2}+y^{2}=16-4 \Rightarrow x^{2}+y^{2}=12
$$

Topic: Mathematics, Hyperbola, JEE Main
The equation of the directrices of the conic $x^2+2 x-y^2+5=0$ are

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

The correct answer is:
$y= \pm \sqrt{2}$

Solution:
$(x+1)^2-y^2-1+5=0 \Rightarrow-\frac{(x+1)^2}{4}+\frac{y^2}{4}=1$
Equation of directrices of $\frac{y^2}{b^2}-\frac{x^2}{a^2}=1$ are $y= \pm \frac{b}{e}$
Here $b=2, e=\sqrt{1+1}=\sqrt{2}$
Hence $\quad y= \pm \frac{2}{\sqrt{2}} \Rightarrow y= \pm \sqrt{2}$.

Topic: Mathematics, Hyperbola, JEE Main
The equation of the directrices of the hyperbola $3 x^{2}-3 y^{2}-18 x+12 y+2=0$ is

Options:
A $x=3 \pm \sqrt{\frac{13}{6}}$
B $x=3 \pm \sqrt{\frac{6}{13}}$
C $x=6 \pm \sqrt{\frac{13}{3}}$
D $x=6 \pm \sqrt{\frac{3}{13}}$

The correct answer is:
$x=3 \pm \sqrt{\frac{13}{6}}$

Solution:
Given equation is
$$
3 x^{2}-3 y^{2}-18 x+12 y+2=0
$$
It can be written as
$\frac{(x-3)^{2}}{\left(\sqrt{\frac{13}{3}}\right)^{2}}-\frac{(y-2)^{2}}{\left(\sqrt{\frac{13}{3}}\right)^{2}}=1$
$\begin{aligned} \text { here, } a &=b=\sqrt{\frac{13}{3}} \\ \therefore \quad e &=\sqrt{1+\frac{b^{2}}{a^{2}}} \\ &=\sqrt{1+1} \\ &=\sqrt{2} \end{aligned}$
$\therefore$ Equation of the directrix are
$$
x=3 \pm \sqrt{\frac{13}{6}}
$$

Topic: Mathematics, Hyperbola, JEE Main
The equation of the ellipse whose centre is at origin, major
axis is along $x$ -axis with eccentricity $\frac{3}{4}$ and latus rectum 4
units is

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

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

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

Topic: Mathematics, Ellipse, JEE Main
The equation of the ellipse whose centre is at the origin and the $X$-axis is major axis, which passes through the points $(-3,1)$ and $(2,-2)$ is

Options:
A $5 x^{2}+3 y^{2}=32$
B $3 x^{2}+5 y^{2}=32$
C $5 x^{2}-3 y^{2}=32$
D $3 x^{2}+5 y^{2}+32=0$

The correct answer is:
$3 x^{2}+5 y^{2}=32$

Solution:
Let equation of ellipse be $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$
Since it passes through the points $(-3,1)$ and $(2-2)$
$$
\begin{aligned}
&\therefore \quad \frac{(-3)^{2}}{a^{2}}+\frac{(1)^{2}}{b^{2}}=1 \text { and } \frac{(2)^{2}}{a^{2}}+\frac{(-2)^{2}}{b^{2}}=1 \\
&\Rightarrow \quad \frac{9}{a^{2}}+\frac{1}{b^{2}}=1 \quad \text{...(i)} \\
&\Rightarrow \quad \frac{4}{a^{2}}+\frac{4}{b^{2}}=1 \quad \text{...(ii)}
\end{aligned}
$$
Multiply by 4 in Eq. (i) and subtracting Eq. (i) from Eq. (ii),
$$
3 a^{2}=32 \Rightarrow a^{2}=\frac{32}{3}
$$
Substituting in Eq. (i), gives
$$
\begin{aligned}
& \frac{9 \times 3}{32}+\frac{1}{b^{2}}=1 \\
\Rightarrow \quad & \frac{1}{b^{2}}=1-\frac{27}{32}=\frac{5}{32} \Rightarrow b^{2}=\frac{32}{5}
\end{aligned}
$$
$\therefore$ Required equation is
$$
\frac{3 x^{2}}{32}+\frac{5 y^{2}}{32}=1 \text { or } 3 x^{2}+5 y^{2}=32
$$

Topic: Mathematics, Ellipse, JEE Main
The equation of the ellipse whose foci are $( \pm 5,0)$ and one of its directrix is $5 x=36$, is

Options:
A $\frac{x^2}{36}+\frac{y^2}{11}=1$
B $\frac{x^2}{6}+\frac{y^2}{\sqrt{11}}=1$
C $\frac{x^2}{6}+\frac{y^2}{11}=1$
D None of these

The correct answer is:
$\frac{x^2}{36}+\frac{y^2}{11}=1$

Solution:
Foci $( \pm 5,0) \equiv( \pm a e, 0)$. Directrix $\left(x=\frac{36}{5}\right) \equiv x=\frac{a}{e}$
So, $\frac{a}{e}=\frac{36}{5}, a e=5 \Rightarrow a=6$ and $e=\frac{5}{6}$
Therefore, $b=6 \sqrt{1-\frac{25}{36}}=6 \frac{\sqrt{11}}{6}=\sqrt{11}$
Hence equation is $\frac{x^2}{36}+\frac{y^2}{11}=1$.

Topic: Mathematics, Ellipse, JEE Main
The equation of the ellipse whose one of the vertices is $(0,7)$ and the corresponding directrix is $y=12$, is

Options:
A $95 x^2+144 y^2=4655$
B $144 x^2+95 y^2=4655$
C $95 x^2+144 y^2=13680$
D None of these

The correct answer is:
$144 x^2+95 y^2=4655$

Solution:
Vertex $(0,7)$, directrix $y=12, \therefore b=7$
Also $\frac{b}{e}=12 \Rightarrow e=\frac{7}{12}, a=7 \sqrt{\frac{95}{144}}$
Hence equation of ellipse is $144 x^2+95 y^2=4655$.

Topic: Mathematics, Ellipse, JEE Main
The equation of the ellipse with focus at \((\pm 5,0)\) and \(x=\frac{36}{5}\) as one directrix is

Options:
A \(\frac{x^2}{36}+\frac{y^2}{25}=1\)
B \(\frac{x^2}{36}+\frac{y^2}{11}=1\)
C \(\frac{\mathrm{x}^2}{25}+\frac{\mathrm{y}^2}{11}=1\)
D None of these

The correct answer is:
\(\frac{x^2}{36}+\frac{y^2}{11}=1\)

Solution:
We have \(a e=5 \quad\) [Since focus is \((\pm \mathrm{ae}, 0)]\) and \(\frac{\mathrm{a}}{\mathrm{e}}=\frac{36}{5}\left[\right.\) since directrix is \(\left.\mathrm{x}= \pm \frac{\mathrm{a}}{\mathrm{e}}\right]\)
On solving we get \(\mathrm{a}=6\) and \(\mathrm{e}=\frac{5}{6}\)
\(\Rightarrow \mathrm{b}^2=\mathrm{a}^2\left(1-\mathrm{e}^2\right)=36\left(1-\frac{25}{36}\right)=11\)
Thus, the required equation of the ellipse is \(\frac{x^2}{36}+\frac{y^2}{11}=1\)

Topic: Mathematics, Ellipse, JEE Main
The equation of the ellipse with focus at \(( \pm 5,0)\) and \(x=\frac{36}{5}\) as one directrix is

Options:
A \(\frac{x^2}{36}+\frac{y^2}{25}=1\)
B \(\frac{x^2}{36}+\frac{y^2}{11}=1\)
C \(\frac{x^2}{25}+\frac{y^2}{11}=1\)
D None of these

The correct answer is:
\(\frac{x^2}{36}+\frac{y^2}{11}=1\)

Solution:
We have ae \(=5\) [Since focus is \(( \pm \mathrm{ae}, 0)\)]
and \(\frac{\mathrm{a}}{\mathrm{e}}=\frac{36}{5} \quad\left[\right.\) since directrix is \(\left.\mathrm{x}= \pm \frac{\mathrm{a}}{\mathrm{e}}\right]\)
On solving we get \(\mathrm{a}=6\)
\(\text { and } \mathrm{e}=\frac{5}{6} \Rightarrow \mathrm{b}^2=\mathrm{a}^2\left(1-\mathrm{e}^2\right)=36\left(1-\frac{25}{36}\right)=11\)
Thus, the required equation of the ellipse is
\(\frac{x^2}{36}+\frac{y^2}{11}=1\)

Topic: Mathematics, Ellipse, JEE Main
The equation of the external bisector of ∠BAC of ΔABC with vertices A5,2, B2,3 and C6,5 is

Options:
A 2x+y+12=0
B x+2y-12=0
C 2x+y-12=0
D x-2y-1=0

The correct answer is:
x-2y-1=0

Solution:
AB=9+1=10
AC=1+9=10
⇒AB=AC⇒ΔABC is an isosceles triangle
⇒ Median through A is also an internal angle bisector.
Now, mid-point of B & C is 4,4=D
The slope of AD is 4-24-5=-2
So, the slope of external angle bisector of ∠BAC is 12
⇒ Required equation is y-2x-5=12⇒x-2y-1=0

Topic: Mathematics, Straight Lines, JEE Main