datavorous/entrance-exam-dataset huggingface dataset:
The light beams of intensities in the ratio of $9: 1$ are allowed to interfere. What will be the ratio of the intensities of maxima and minima?

Options:
A $3: 1$
B $4: 1$
C $25: 9$
D $81: 1$

The correct answer is:
$4: 1$

Solution:
Hints: $\frac{\mathrm{A}_1}{\mathrm{~A}_2}=\frac{3}{1}$
$$
\frac{I_{\max }}{I_{\min }}=\frac{16}{4}=\frac{4}{1}
$$

Topic: Physics, Wave Optics, WBJEE, WBJEE 2010
The $\lim _{x \rightarrow \infty}\left(\frac{3 x-1}{3 x+1}\right)^{4 x}$ equals

Options:
A 1
B 0
C $\mathrm{e}^{-8 / 3}$
D $\mathrm{e}^{-4 / 9}$

The correct answer is:
$\mathrm{e}^{-8 / 3}$

Solution:
$\lim _{x \rightarrow \infty}\left(\frac{3 x-1}{3 x+1}\right)^{4 x}=1^{\infty}=e^{L}=e^{-\frac{8}{3}}$
$L=\lim _{x \rightarrow \infty} 4 x\left(\frac{3 x-1}{3 x+1}-1\right)=\lim _{x \rightarrow \infty} 4 x\left(\frac{-2}{3 x+1}\right)=\frac{-8}{3}$

Topic: Mathematics, Limits, WBJEE, WBJEE 2021
The limit of $\sum_{n=1}^{1000}(-1)^{n} x^{n}$ as $x \rightarrow \infty$

Options:
A does not exist
B exists and equals to 0
C exists and approaches to $+\infty$
D exists and approaches $-\infty$

The correct answer is:
exists and approaches to $+\infty$

Solution:
$\lim _{x \rightarrow-} \sum_{n=1}^{1000}(-1)^{n} x^{n}$
$=\lim _{x \rightarrow-}\left\{-x+x^{2}-x^{3}+x^{4}+\ldots+x^{1000}\right\}$
$=\lim _{x \rightarrow \infty}(-x) \cdot\left\{\frac{(-x)^{1000}-1}{(-x-1)}\right\}=\lim _{x \rightarrow-\infty} \frac{x^{1001}-x}{x+1}$
$=\lim _{x \rightarrow-} \frac{x^{1000}-1}{1+\left(\frac{1}{x}\right)}=+\infty$

Topic: Mathematics, Limits, WBJEE, WBJEE 2013
The limit of the interior angle of a regular polygon of $n$ sides as $n \rightarrow \infty$ is

Options:
A $\pi$
B $\frac{\pi}{3}$
C $\frac{3 \pi}{2}$
D $\frac{2 \pi}{3}$

The correct answer is:
$\pi$

Solution:
It is clear that, the limit of the interior angle of a regular polygon of $n$ sides as $n \rightarrow \infty$ is $\pi$ or $180^{\circ}$

Topic: Mathematics, Limits, WBJEE, WBJEE 2019
The limit of $\left\{\frac{1}{x} \sqrt{1+x}-\sqrt{1+\frac{1}{x^{2}}}\right\}$ as $x \rightarrow 0$

Options:
A does not exist
B is equal to $\frac{1}{2}$
C is equal to 0
D is equal to 1

The correct answer is:
is equal to $\frac{1}{2}$

Solution:
$\lim _{x \rightarrow 0}\left\{\frac{\sqrt{1+x}}{x}-\sqrt{1+\frac{1}{x^{2}}}\right\}$
$=\lim _{x \rightarrow 0}\left\{\frac{\sqrt{1+x}-\sqrt{1+x^{2}}}{x}\right\}\left(\frac{0}{0}\right.$ form $)$
$=\lim _{x \rightarrow 0} \frac{\frac{1}{2 \sqrt{1+x}}-\frac{x}{\sqrt{1+x^{2}}}}{1}$ (Use L-Hospital Rule)
$=\frac{1}{2 \sqrt{1+0}}-0=\frac{1}{2}$

Topic: Mathematics, Limits, WBJEE, WBJEE 2013
The limit of $\left[\frac{1}{x^{2}}+\frac{(2013)^{x}}{e^{x}-1}-\frac{1}{e^{x}-1}\right]$ as $x \rightarrow 0$

Options:
A approaches $+\infty$
B approaches $-\infty$
C is equal to $\log_{e}$, (2013)
D does not exist

The correct answer is:
approaches $+\infty$

Solution:
$\lim _{x \rightarrow 0}\left\{\frac{1}{x^{2}}+\frac{(2013)^{x}}{e^{x}-1}-\frac{1}{e^{x}-1}\right\}$
$=\lim _{x \rightarrow 0}\left\{\frac{1}{x^{2}}+\frac{(2013)^{x}-1}{e^{x}-1}\right\}$
$=\lim _{x \rightarrow 0}\left\{\frac{1}{x^{2}}+\frac{(2013)^{x}-1}{x} \cdot \frac{x}{e^{x}-1}\right\}$
$=\lim _{x \rightarrow 0} \frac{1}{x^{2}}+\lim _{x \rightarrow 0} \frac{(2013)^{x}-1}{x} \cdot \lim _{x \rightarrow 0} \frac{x}{e^{x}-1}$
$=+\infty+\log (2013) \cdot 1$
$=+\infty$

Topic: Mathematics, Limits, WBJEE, WBJEE 2013
The limit of $x \sin \left(e^{\frac{1}{x}}\right)$ as $x \rightarrow 0$

Options:
A is equal to 0
B is equal to 1
C is equal to $\frac{e}{2}$
D does not exist

The correct answer is:
is equal to 0

Solution:
$\lim _{x \rightarrow 0} x \sin e^{(1 / x)}$
$\begin{aligned} L H L &=f(0-0)=\lim _{h \rightarrow 0}(-h) \sin e^{(-1 / h)} \\ &=-0 \times \sin \left(e^{-\infty}\right)=-0 \times \sin (0)=0 \end{aligned}$
$=0 \times(\mathrm{a}$ finite number between $-1 \text{ to}+1)$
$$
=0 \quad(\because-1 \leq \sin x \leq 1)
$$
$\because \mathrm{LHL}=\mathrm{RHL}$
$\therefore \lim _{x \rightarrow 0} x \sin \left(e^{1 / x}\right)$ exist and equal to 0

Topic: Mathematics, Limits, WBJEE, WBJEE 2013
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, WBJEE, WBJEE 2015
The line $A B$ cuts off equal intercepts $2 a$ from the axes. From any point $P$ on the line $A B$ perpendiculars $P R$ and $P S$ are drawn on the axes. Locus of mid-point of $R S$ is

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

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

Solution:
$\Rightarrow$
$$
\begin{array}{l}
\frac{x}{2 a}+\frac{y}{2 a}=1 \\
x+y=2 a
\end{array}
$$

Let the coordinates of the mid-point of $\mathrm{RS}$ be $(\mathrm{h}, \mathrm{k})$
So. $R$ and $S$ are $(2 h, 0)$ and $(0,2 k)$ Hence, the mid-point of $R S$ is $(h, k)$
P lies on line AB
Then. from Eq we have
$$
2 h+2 k=2 a
$$
$\Rightarrow \quad h+k=a$
So, the locus of $(h, k)$ is $x+y=a$.

Topic: Mathematics, Straight Lines, WBJEE, WBJEE 2016
The line joining $A(b \cos \alpha, b \sin \alpha)$
and
$B(a \cos \beta, a \sin \beta),$ where $a \neq b,$ is produced to the
point $M(x, y)$ so that $A M: M B=b: a$. Then, $x \cos \frac{\alpha+\beta}{2}+y \sin \frac{\alpha+\beta}{2}$ is equal to

Options:
A 0
B 1
C -1
D $a^{2}+b^{2}$

The correct answer is:
0

Solution:
Since, $A M: B M=b: a$ $\therefore M$ divides $A B$ extemally in the ratio $b: a$...(i)
$\therefore \quad x=\frac{b \cdot a \cos \beta-a b \cos \alpha}{b-a}$
$y=\frac{b a \sin \beta-a b \sin \alpha}{b-a}$
(ii)
Divide Eq. (i) by Eq, (ii), we get
$\frac{x}{y}=\frac{\cos \beta-\cos \alpha}{\sin \beta-\sin \alpha}=\frac{2 \sin \frac{\alpha+\beta}{2} \sin \frac{\alpha-\beta}{2}}{-2 \cos \frac{\alpha+\beta}{2} \sin \frac{\alpha-\beta}{2}}$
$\therefore \quad x \cos \frac{\alpha+\beta}{2}+y \sin \frac{\alpha+\beta}{2}=0$

Topic: Mathematics, Straight Lines, WBJEE, WBJEE 2012
The line segment joining the foci of the hyperbola $x^{2}-y^{2}+1=0$ is one of the diameters of a circle. The equation of the circle is

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

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

Solution:
Given, equation of hyperbola is
$$
x^{2}-y^{2}+1=0
$$
$\Rightarrow \quad y^{2}-x^{2}=1$
On comparing it with $\frac{y^{2}}{b^{2}}-\frac{x^{2}}{a^{2}}=1,$ we get
$$
a=b=1
$$
Now, $e=\sqrt{1+\frac{a^{2}}{b^{2}}}=\sqrt{1+\frac{1}{1}}=\sqrt{2}$
$\therefore$ Foci $=(0, \pm b e)=(0, \pm \sqrt{2})$
since, line joining foci of hyperbola is diameter of circle.
$\therefore$ Centre of circle $=\left(\frac{0+0}{2}, \frac{\sqrt{2}-\sqrt{2}}{2}\right)=(0,0)$
$\begin{aligned} \text { and radius } &=\frac{1}{2} \sqrt{(0-0)^{2}+(\sqrt{2}-(-\sqrt{2}))^{2}} \\ &=\frac{1}{2} \sqrt{(2 \sqrt{2})^{2}}=\sqrt{2} \end{aligned}$
$\therefore$ Equation of circle will be
$$
(x-0)^{2}+(y-0)^{2}=(\sqrt{2})^{2}
$$
$\Rightarrow \quad x^{2}+y^{2}=2$

Topic: Mathematics, Hyperbola, WBJEE, WBJEE 2017
The line through the points $(a, b)$ and $(-a,-b)$
passes through the point

Options:
A (1,1)
B $(3 a,-2 b)$
C $\left(a^{2}, a b\right)$
D $(a, b)$

The correct answer is:
$\left(a^{2}, a b\right)$

Solution:
Let the given points be $A(a, b)$ and $B(-a,-b)$. The equation of line passing through the points A and $B$ is
$$
\begin{array}{ll}
y-y_{1}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\left(x-x_{1}\right) \\
\Rightarrow \quad y-b=\frac{-b-b}{-a-a}(x-a) \\
\Rightarrow \quad y-b=\frac{2 b}{2 a}(x-a)
\end{array}
$$
$\Rightarrow \quad y-b=\frac{b}{a}(x-a)$
$\Rightarrow \quad a y-a b=b x-a b$
$$
\Rightarrow \quad b x=a y ...(i)
$$
Since, from the given points $\left(a^{2}, a b\right)$ and $(a, b)$ are satisty the Eq. (i), but ( $a^{2}$, ab) is the required answer because $(a, b)$ is already given in question.

Topic: Mathematics, Straight Lines, WBJEE, WBJEE 2016
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, WBJEE, WBJEE 2022
The line $x=2 y$ intersects the ellipse $\frac{x^{2}}{4}+y^{2}=1$ at the points $P$ and $Q .$ The equation of the circle with $P Q$ as diameter is

Options:
A $x^{2}+y^{2}=\frac{1}{2}$
B $x^{2}+y^{2}=1$
C $x^{2}+y^{2}=2$
D $x^{2}+y^{2}=\frac{5}{2}$

The correct answer is:
$x^{2}+y^{2}=\frac{5}{2}$

Solution:
Solving $x=2 y$
and $\frac{x^{2}}{4}+y^{2}=1$

Put $x=2 y$ in Eq. (ii), we get $\frac{(2 y)^{2}}{4}+y^{2}=1 \Rightarrow \frac{4 y^{2}}{4}+y^{2}=1$
$=$
$2 y^{2}=1 \Rightarrow y=\pm \frac{1}{\sqrt{2}}$
$\therefore$ From $\mathrm{E} q$ (i), $x=\pm \sqrt{2}$
$\therefore \quad P\left(\sqrt{2} \frac{1}{\sqrt{2}}\right)$ and $Q\left(-\sqrt{2},-\frac{1}{\sqrt{2}}\right)$
Equation of circle with $P Q$ as diameter is
$$
\begin{array}{l}
(x-\sqrt{2})(x+\sqrt{2})+\left(y-\frac{1}{\sqrt{2}}\right)\left(y+\frac{1}{\sqrt{2}}\right) \\
\Rightarrow \quad x^{2}-2+y^{2}-\frac{1}{2}=0 \Rightarrow x^{2}+y^{2}=\frac{5}{2}
\end{array}
$$

Topic: Mathematics, Ellipse, WBJEE, WBJEE 2012
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, WBJEE, WBJEE 2009
The line $y=x+5$ touches

Options:
A the parabola $y^2=20 x$
B the ellipse $9 x^2+16 y^2=144$
C the hyperbola $\frac{x^2}{29}-\frac{y^2}{4}=1$
D the circle $x^2+y^2=25$

The correct answers are:
the parabola $y^2=20 x$, the ellipse $9 x^2+16 y^2=144$, the hyperbola $\frac{x^2}{29}-\frac{y^2}{4}=1$

Solution:
(A) $y^2=20 x=4(5) x \quad \therefore$ Tangent $: y=m x+\frac{a}{m}$ for $m=1, a=5$
$\therefore \mathrm{y}=\mathrm{x}+5$ is a tangent to $\mathrm{y}^2=20 \mathrm{x}$
(B) $9 x^2+16 y^2=144$
$c^2=a^2 m^2+b^2 \quad y=x+5 \Rightarrow m=1 ; c=5$
$\Rightarrow \frac{x^2}{16}+\frac{y^2}{9}=1$
$\therefore 5^2=16(1)^2+9 \Rightarrow$ True
$\therefore y=x+5$ is a tangent to $9 x^2+16 y^2=144$
(C) $c^2=a^2 m^2-b^2$
$\frac{x^2}{29}-\frac{y^2}{4}=1 \rightarrow a^2=29 ; b^2=4$
$5^2=29(1)^2-4 \rightarrow$ True
$\therefore \mathrm{y}=\mathrm{x}+5$ is a tangent
(D)
$\therefore \mathrm{y}=\mathrm{x}+5$ is NOT tangent to $\mathrm{x}^2+\mathrm{y}^2=25$
(A), (B), (C) are correct.

Topic: Mathematics, Parabola, WBJEE, WBJEE 2022
The line $y=x$ intersects the hyperbola $\frac{x^{2}}{9}-\frac{y^{2}}{25}=1$ at the points $P$ and $Q .$ The eccentricity of ellipse with $P Q$ as major axis and minor axis of length $\frac{5}{\sqrt{2}}$ is

Options:
A $\frac{\sqrt{5}}{3}$
B $\frac{5}{\sqrt{3}}$
C $\frac{5}{9}$
D $\frac{2 \sqrt{2}}{3}$

The correct answer is:
$\frac{2 \sqrt{2}}{3}$

Solution:
Given equation of hyperbola and line are $\frac{x^{2}}{9}-\frac{y^{2}}{25}=1$ and $y=x$ respectively.
For intersection point of both curve put $y=x$, we get
$$
\begin{array}{l}
\frac{x^{2}}{9}-\frac{x^{2}}{25}=1 \\
\Rightarrow \quad x^{2}=\frac{9 \times 25}{16}=\left(\frac{15}{4}\right)^{2} \\
\Rightarrow \quad x=\pm \frac{15}{4} \text { and } y=\pm \frac{15}{4}
\end{array}
$$
$\therefore$ Intersetion points $P\left(\frac{15}{4}, \frac{15}{4}\right)$
and $\quad Q\left(\frac{-15}{4}, \frac{-15}{4}\right)$
Since, $P Q$ is major axis, then its length
$$
=2 \sqrt{2} \cdot \frac{15}{4}=\frac{15}{\sqrt{2}}
$$
and length of minor axis is $\frac{5}{\sqrt{2}}$ (given)
i.e., Major axis, $2 a=\frac{15}{\sqrt{2}} \Rightarrow a=\frac{15}{2 \sqrt{2}}$
and minor axis, $2 b=\frac{5}{\sqrt{2}} \Rightarrow b=\frac{5}{2 \sqrt{2}}$
$\therefore$ Eccentricity of an ellipse
$$
=\sqrt{\frac{a^{2}-b^{2}}{a^{2}}}=\sqrt{1-\left(\frac{b}{a}\right)^{2}}
$$
$=\sqrt{1-\left(\frac{1}{3}\right)^{2}}=\sqrt{\frac{8}{9}}=\frac{2 \sqrt{2}}{3}$

Topic: Mathematics, Ellipse, WBJEE, WBJEE 2013
The line $y=x+\lambda$ is tangent to the ellipse $2 x^{2}+3 y^{2}=1 .$ Then $, \lambda$ is

Options:
A -2
B 1
C $\sqrt{\frac{5}{6}}$
D $\sqrt{\frac{2}{3}}$

The correct answer is:
$\sqrt{\frac{5}{6}}$

Solution:
The equation of the line is $y=x+\lambda$
On comparing it with $y=m x+c,$ we get
$$
m=1 \text { and } c=\lambda
$$
The equation of the ellipse is
$$
2 x^{2}+3 y^{2}=1
$$
or
$$
\frac{x^{2}}{\left(\frac{1}{\sqrt{2}}\right)^{2}}+\frac{y^{2}}{\left(\frac{1}{\sqrt{3}}\right)^{2}}=1
$$
On comparing it with $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1,$ we get
$$
a^{2}=\frac{1}{2} \text { and } b^{2}=\frac{1}{3}
$$
If the line touches the ellipse, then
$$
c^{2}=a^{2} m^{2}+b^{2} \Rightarrow \lambda^{2}=\frac{1}{2} \cdot 1+\frac{1}{3}=\frac{5}{6}
$$
$\Rightarrow \quad \lambda^{2}=\frac{5}{6} \Rightarrow \lambda=\sqrt{\frac{5}{6}}$

Topic: Mathematics, Ellipse, WBJEE, WBJEE 2016
The linear system of equations
$\left.\begin{array}{l}8 x-3 y-5 z=0 \\ 5 x-8 y+3 z=0 \\ 3 x+5 y-8 z=0\end{array}\right\}$ has

Options:
A only zero solution
B only finite number of non-zero solutions
C no non-zero solution
D infinitely many non-zero solutions

The correct answer is:
infinitely many non-zero solutions

Solution:
We have,
$$
\begin{aligned}
& 8x-3 y-5 z=0 \\
& 5 x-8 y+3 z=0 \\
& 3 x+5 y-8 z=0 \\
&D=\left|\begin{array}{ccc}
8 & -3 & -5 \\
5 & -8 & 3 \\
3 & 5 & -8
\end{array}\right| \\
&=8(64-15)+3(-40-9)-5(25+24) \\
=& 8 \times 49+3 \times(-49)-5 \times 49 \\
=& 0
\end{aligned}
$$
$\therefore$ The system has infinitely many non-zero solutions.

Topic: Mathematics, Determinants, WBJEE, WBJEE 2017
The locus of points \((x, y)\) in the plane satisfying \(\sin ^2 x+\sin ^2 y=1\) consists of

Options:
A a circle centered at origin
B infinitely many circles that are all centered at the origin
C infinitely many lines with slope \(\pm 1\)
D finitely many lines with slope \(\pm 1\)

The correct answer is:
infinitely many lines with slope \(\pm 1\)

Solution:
\(\begin{array}{r}\text {Hint: }
\sin ^2 y=\cos ^2 x \\
\sin y= \pm \cos x
\end{array}\)
\(\begin{aligned}
& \text { If } \sin y=\cos x=\sin \left(\frac{\pi}{2}-x\right) \\
& \Rightarrow y=n \pi+(-1)^n\left(\frac{\pi}{2}-x\right)
\end{aligned}\)
\(\text { If } \sin y=-\cos x=\sin \left(x-\frac{\pi}{2}\right)\)
\(\Rightarrow \mathrm{y}=\mathrm{n} \pi+(-1)^{\mathrm{n}}\left(\mathrm{x}-\frac{\pi}{2}\right)\)

Topic: Mathematics, Straight Lines, WBJEE, WBJEE 2023
The locus of the center of a variable circle which always touches two given circles externally is

Options:
A an ellipse
B a hyperbola
C a parabola
D a circle

The correct answer is:
a hyperbola

Solution:
$\begin{array}{l}
\mathrm{oo}_{1}=\mathrm{r}+\mathrm{r}_{1} \\
\mathrm{oo}_{2}=\mathrm{r}+\mathrm{r}_{2} \\
\therefore \mathrm{oo}_{2}-\mathrm{oo}_{1}=\mathrm{r}_{2}-\mathrm{r}_{1}
\end{array}$
hence, locus is hyperbola

Topic: Mathematics, Hyperbola, WBJEE, WBJEE 2021
The locus of the centre of a circle which passes through two variable points \((a, 0),(-a, 0)\) is

Options:
A \(\mathrm{x}=1\)
B \(x+y=a\)
C \(x+y=2 a\)
D \(x=0\)

The correct answer is:
\(x=0\)

Solution:
Hint:
Centre lies on \(y\)-axis locus \(x=0\)

Topic: Mathematics, Circle, WBJEE, WBJEE 2011
The locus of the centre of the circles which touch both the circles $x^{2}+y^{2}=a^{2}$ and $x^{2}+y^{2}=4$ax externally is

Options:
A a circle
B a parabola
C an ellipse
D a hyperbola

The correct answer is:
a hyperbola

Solution:
Hint:
Let, centre $\equiv(\mathrm{h}, \mathrm{k})$ and radius $=\mathrm{r}$ for the variable circle So, using $\mathrm{C}_{1} \mathrm{C}_{2}=\mathrm{r}_{1}+\mathrm{r}_{2}$ for both cases we have:
$\begin{array}{l}
\mathrm{h}^{2}+\mathrm{k}^{2}=(\mathrm{r}+\mathrm{a})^{2} \rightarrow(1) \text { and }(\mathrm{h}-2 \mathrm{a})^{2}+\mathrm{k}^{2}=(\mathrm{r}+2 \mathrm{a})^{2} \rightarrow(2) \\
\text { Eq. }(2)-\text { Eq. }(1), \text { gives }: \mathrm{r}=\frac{\mathrm{a}-4 \mathrm{~h}}{2} \rightarrow(3)
\end{array}$
Substitute $(3)$ in $(1)$ to get:
$12 h^{2}-4 k^{2}-24 a h+9 a^{2}=0$
$\therefore$ locus $: 12 x^{2}-4 y^{2}-24 a x+9 a^{2}=0$ i.e. a hyperbola

Topic: Mathematics, Hyperbola, WBJEE, WBJEE 2020
The locus of the middle points of all chords of the parabola \(y^2=4\) ax passing through the vertex is

Options:
A a straight line
B an ellipse
C a parabola
D a circle

The correct answer is:
a parabola

Solution:
\(\text {Hints: } 2 \mathrm{~h}=\mathrm{x}, 2 \mathrm{k}=\mathrm{y}\)
\(\begin{aligned}
& \mathrm{y}^2=4 \mathrm{ax} \\
& \mathrm{k}^2=2 \mathrm{ah} \\
& \mathrm{y}^2=2 \mathrm{ax}
\end{aligned}\)

Topic: Mathematics, Parabola, WBJEE, WBJEE 2011
The locus of the mid-points of all chords of the parabola $y^{2}=4 a x$ through its vertex is another parabola with directrix

Options:
A $x=-a$
B $x=a$
C $x=0$
D $x=-\frac{a}{2}$

The correct answer is:
$x=0$

Solution:
Let $P\left(a t^{2}, 2 a t\right)$ be a moving point on the parabola $y^{2}=4 a x$ and $S(a, 0)$ be its focus. Let $Q(h, k)$ be the mid-point of SP.
Then, $\quad h=\frac{a t^{2}+a}{2}$
and $\quad k=\frac{2 a t+0}{2}$
$\Rightarrow \quad \frac{2 h}{a}=t^{2}+1$
and $\quad t=\frac{k}{a}$
$\Rightarrow \quad \frac{2 h}{a}=\frac{k^{2}}{a^{2}}+1$
$\Rightarrow \quad k^{2}=2 a h-a^{2} \quad[$ on eliminating t $\mid$
Thus, the locus of $(h, k)$ is $y^{2}=2 a x-a^{2}$
Now. $\quad y^{2}=2 a x-a^{2}$
$\Rightarrow \quad(y-0)^{2}=2 a\left(x-\frac{a}{2}\right)$
The equation of the directrix of this parabola is
$$
x-\frac{a}{2}=-\frac{a}{2}, i \cdot e, x=0
$$

Topic: Mathematics, Parabola, WBJEE, WBJEE 2016
The locus of the mid-points of chords of the circle $x^{2}+y^{2}=1$, which subtends a right angle at the origin, is

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

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

Solution:
Let $(h, k)$ be the coordinates of the mid-point of a chord which subtends a right angle at the origin. Then, equation of the chord is
$$
\begin{array}{l}
\quad h x+h y-1=h^{2}+k^{2}-1 \quad[u \operatorname{sing} T=S] \\
\Rightarrow \quad h x+k y=h^{2}+k^{2}
\end{array}
$$
The combined equation of the pair of lines joining the origin to the points of intersection of $x^{2}+y^{2}=1$ and $h x+k y=h^{2}+k^{2}$ is
$$
x^{2}+y^{2}-1\left(\frac{h x+h y}{h^{2}+k^{2}}\right)^{2}=0
$$
Lines given by the above equation are at right angle. Therefore, coefficient of $x^{2}+$ coefficient of $y^{2}=0$
ie. $\quad h^{2}+k^{2}=\frac{1}{2}$
$\therefore x^{2}+y^{2}=\frac{1}{2}$

Topic: Mathematics, Circle, WBJEE, WBJEE 2016
The locus of the mid-points of the chords of an ellipse $x^{2}+4 y^{2}=4$ that are drawn from the positive end of the minor axis, is

Options:
A a circle with centre $\left(\frac{1}{2}, 0\right)$ and radius 1
B a parabola with focus $\left(\frac{1}{2}, 0\right)$ and directrix $x=-1$
C an ellipse with centre $\left(0, \frac{1}{2}\right),$ major axis 1 and minor axis $\frac{1}{2}$
D a hyperbola with centre $\left(0, \frac{1}{2}\right),$ transverse axis 1 and conjugate axis $\frac{1}{2}$

The correct answer is:
an ellipse with centre $\left(0, \frac{1}{2}\right),$ major axis 1 and minor axis $\frac{1}{2}$

Solution:
Given equation of an ellipse is
$$
x^{2}+4 y^{2}=4
$$
$\Rightarrow \quad \frac{x^{2}}{4}+\frac{y^{2}}{1}=1$
$\therefore$ Coordinate of positive end of minor axis is $B(0,1)$ Let mid-point of the chord $B P$ is $M(h, k)$
Then, $\quad(h, k)=\left(\frac{0+x}{2}, \frac{1+y}{2}\right)$
$\Rightarrow \quad h=\frac{x}{2} \quad \Rightarrow x=2 h$
and $\quad k=\frac{1+y}{2} \Rightarrow y=2 k-1$
$\therefore$
$P(x, y)=\{2 h,(2 k-1)\}$
Since, the point '$P$' lies ellipses so form Eq. (i). we get $(2 h)_{1}^{2}+4(2 k-1)^{2}=4$
$\Rightarrow \quad 4 h^{2}+4(2 k-1)^{2}=4$
$\Rightarrow \quad h^{2}+4 k^{2}+1-4 k=1$
$\Rightarrow \quad h^{2}+4 k^{2}-4 k=0$
Thus, required locus is an ellipse whose equation is $x^{2}+4 y^{2}-4 y=0$
$\Rightarrow \quad \frac{(x-0)^{2}}{1}+\frac{\left(y-\frac{1}{2}\right)^{2}}{\left(\frac{1}{4}\right)}=1$
whose centre $\left(0, \frac{1}{2}\right)$ and major and minor axis $\frac{1}{2}$

Topic: Mathematics, Ellipse, WBJEE, WBJEE 2013
The locus of the mid-points of the chords of the circle $x^{2}+y^{2}+2 x-2 y-2=0,$ which make an angle of $90^{\circ}$ at the centre is

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

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

Solution:
Given, equation of circle is $x^{2}+y^{2}+2 x-2 y-2=0$
$\Rightarrow \quad(x+1)^{2}+(y-1)^{2}=4$
$\therefore$ Centre (-1,1) and radius $=2$
Let $(h, k)$ be the mid-point of chord.

From figure,
$$
O P=\sqrt{(h+1)^{2}+(k-1)^{2}}
$$
$\ln \Delta O A P$
$$
\begin{aligned}
\sin 45^{\circ}=& \frac{O P}{O A} \\
\Rightarrow \quad \frac{1}{\sqrt{2}}=\frac{\sqrt{(h+1)^{2}+(k-1)^{2}}}{2}
\end{aligned}
$$
On squaring both sides, we get
$$
(h+1)^{2}+(k-1)^{2}=2
$$
$\Rightarrow \quad h^{2}+k^{2}+2 h-2 k=0$
$\therefore$ Locus of $P$ will be
$$
x^{2}+y^{2}+2 x-2 y=0
$$

Topic: Mathematics, Circle, WBJEE, WBJEE 2017
The locus of the point of intersection of the straight lines $\frac{x}{a}+\frac{y}{b}=K$ and $\frac{x}{a}-\frac{y}{b}=\frac{1}{K}$,
where $K$ is a non-zero real variable, is given by

Options:
A a straight line
B an ellipse
C a parabola
D a hyperbola

The correct answer is:
a hyperbola

Solution:
Given equations of straight lines
$$
\frac{x}{a}+\frac{y}{b}=K
$$
and $\quad \frac{x}{a}-\frac{y}{b}=\frac{1}{K}$
Let the point of intersection be $(\alpha, \beta)$.
So, from Eqs. we get
$$
\frac{\alpha}{a}+\frac{\beta}{b}=K
$$
and
$$
\frac{\alpha}{a}-\frac{\beta}{b}=\frac{1}{K}
$$
$\Rightarrow$
$\left(\frac{\alpha}{a}\right)^{2}-\left(\frac{\beta}{b}\right)^{2}=1 \Rightarrow \frac{\alpha^{2}}{a^{2}}-\frac{\beta^{2}}{b^{2}}=1$
$\therefore$ Locus $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ which is the equation of a hyperbola.

Topic: Mathematics, Straight Lines, WBJEE, WBJEE 2016