datavorous/entrance-exam-dataset huggingface dataset:
The locus of a point so that sum of its distance from two given perpendicular lines is equal to 2 unit in first quadrant, is

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

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

Solution:
We take the coordinate axes as two perpendicular lines. Let $P\left(x_1, y_1\right)$ be the required point. From $P\left(x_1, y_1\right)$, we draw $P M$ and $P N$ perpendicular to $O X$ and $O Y$ respectively.
Given, $P M+P N=2$
But, $\quad P M=y_1, P N=x_1$
Hence from (i), $y_1+x_1=2$
Thus locus of $\left(x_1, y_1\right)$ is $x+y=2$ which is a straight line.

Topic: Mathematics, Straight Lines, JEE Main
The locus of a point which divides the line segment joining the point 0,-1 and a point on the parabola x2=4y internally in the ratio 1:2 is:

Options:
A 9x2-12y=8
B 9x2-3y=2
C x2-3y=2
D 4x2-3y=2

The correct answer is:
9x2-12y=8

Solution:
Let point P be 2t, t2 and Q be h, k .
h=2t3, k=-2+t23
Hence, locus is 3k+2=3h2s2⇒9x2=12y+8

Topic: Mathematics, Parabola, JEE Main
The locus of a point which divides the line segment joining the point 0,-1 and a point on the parabola x2=4y internally in the ratio 1:2 is:

Options:
A 9x2-12y=8
B 9x2-3y=2
C x2-3y=2
D 4x2-3y=2

The correct answer is:
9x2-12y=8

Solution:
Let point P be 2t, t2 and Q be h, k .
h=2t3, k=-2+t23
Hence, locus is 3k+2=3h2s2⇒9x2=12y+8

Topic: Mathematics, Parabola, JEE Main, JEE Main 2020 (08 Jan Shift 1)
The locus of a point which moves such that the difference of its distances from the points 5,0 and -5,0 is 6 units is a conic, whose length of the latus rectum (in units) is equal to

Options:
A 4
B 163
C 8
D 323

The correct answer is:
323

Solution:
Because 6<5--5
Hence, the locus is a hyperbola by definition.
Let's assume the eccentricity, semi-transverse axis and the semi-conjugate axis of the hyperbola is e,a, and b respectively, then

2ae=10 {Distance between focii -5,0 & 5,0=10=2ae }
& 2a=6⇒e=106=53
⇒1+b2a2=e2=259⇒b2a2=259-1=169
⇒2b2a=2×169×a=323= length of the latus rectum

Topic: Mathematics, Hyperbola, JEE Main
The locus of a point which moves such that the difference of its distances from two fixed points is always a constant is

Options:
A A straight line
B A circle
C A circle
D A hyperbola

The correct answer is:
A hyperbola

Solution:
Let $S$ and $S^{\prime}$ be the two fixed points and $P$ be some arbitrary point.
Given that $\left|S P-S^{\prime} \mathrm{P}\right|=$ constant
The above equation satisfy for any equation of hyperbola with $\mathrm{S}$ and $\mathrm{S}^{\prime}$ as its foci

Topic: Mathematics, Hyperbola, JEE Main
The locus of a point, which moves such that the sum of squares of its distances from the points (0,0),(1,0),(0,1)(1,1) is 18 units, is a circle of diameter d. Then d2 is equal to

Options:

The correct answer is:
16

Solution:
Let Px,y be the required point. Then,

x2+y2+(x−0)2+(y−1)2+(x−1)2+(y−0)2+(x−1)2+(y−1)2=18

⇒4x2+4y2−4x−4y+4=18

⇒x2+y2−x−y−144=0

Centre≡12,12 and

Radius,r=14+14+144=2 units

So,

d=2r=2×2=4

⇒d2=16

Topic: Mathematics, Circle, JEE Main, JEE Main 2021 (26 Aug Shift 1)
The locus of a point z represented by the equation z-1=|z-i| on the argand plane is (where, zâC, i=-1)

Options:
A a circle of radius 1
B an ellipse with foci at 1 and –i
C a line passing through the origin
D a circle on the line joining 1 and âi as diameter

The correct answer is:
a line passing through the origin

Solution:
Given, z-1=z-i
â z lies on the perpendicular bisector of the line joining (1, 0) and (0, 1) and it is a straight line passing through the origin.

Topic: Mathematics, Complex Number, JEE Main
The locus of a variable point whose chord of contact w.r.t. the hyperbola x2a2-y2b2=1 subtends a right angle at the origin is

Options:
A x24a2-y24b2=1
B x2a2-y2b2=x2a4+y2b4
C xa-yb=1a2+1b2
D x2a4+y2b4=1a2-1b2

The correct answer is:
x2a4+y2b4=1a2-1b2

Solution:
Given,

The equation of the given hyperbola is x2a2-y2b2=1 ⋯1

Now let h,k be the pole of a chord PQ of the hyperbola

Then the straight line PQ is the polar of the point h,k w.r.t. the hyperbola and so its equation is

xha2-ykb2=1 ⋯2

Now the center of the hyperbola is the origin, let it be point C,

Now making 1 homogeneous with the help of 2, the combined equation of CP and CQ is

x2a2-y2b2=xha2-ykb22 ⋯3

Now the chord PQ subtends a right angle at 0,0

Therefore the lines CP and CQ given by 3 are perpendicular and so in the equation 3, we have the coefficient of x2+coefficient of y2=0

⇒1a2-h2a4+-1b2-k2b4=0

⇒h2a4+k2b4=1a2-1b2

Thus locus is

x2a4+y2b4=1a2-1b2

Topic: Mathematics, Hyperbola, JEE Main
The locus of all points that are at a distance greater than 2 units from -3, 0, is

Options:
A (x,y)∣x2+y2+6x+5>0
B (x,y)∣x2+y2+6x+7>0
C (x,y)∣x2+y2-6x+5>0
D (x,y)∣x2+y2-6x+5≤0

The correct answer is:
(x,y)∣x2+y2+6x+7>0

Solution:
Let, h,k be the point whose distance from -3,0 is greater than 2.

⇒h+32+k2>2

⇒h2+k2+6h+7>0

Replace h by x and k by y

⇒x2+y2+6x+7>0

Topic: Mathematics, Straight Lines, JEE Main
The locus of mid-points of all chords of parabola y2=4x, for which all circles drawn taking them as diameters passes through the vertex of the parabola is a conic whose length of the smallest focal chord is equal to

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

The correct answer is:
2 units

Solution:
Circle drawn by taking AB as diameter passes through vertex hence AV is perpendicular to BV

Let h,k be the mid-point of the chord
⇒t1t2=-4, h=t12+t222, k=2t1+t22
k2=t12+t22+2t1t2
⇒y2=2x-8

Hence, the locus of the conic is y2=2x-4
⇒ Length of the latus rectum (smallest focal chord) =2 units

Topic: Mathematics, Parabola, JEE Main
The locus of mid-points of the line segments joining -3,-5 and the points on the ellipse x24+y29=1 is :

Options:
A 36x2+16y2+90x+56y+145=0
B 36x2+16y2+108x+80y+145=0
C 9x2+4y2+18x+8y+145=0
D 36x2+16y2+72x+32y+145=0

The correct answer is:
36x2+16y2+108x+80y+145=0

Solution:
Parametric point on the given ellipse is 2sinθ, 3cosθ

Let, the mid-point of line segments joining -3,-5 and 2sinθ,3cosθ is h, k

Then, by using mid-point formula, we get

2sinθ-32=h, 3cosθ-52=k

⇒2sinθ=2h+3, 3cosθ=2k+5

⇒sinθ=2h+32, cosθ=2k+53

We know, sin2θ+cos2θ=1

⇒2h+322+2k+532=1

⇒144h2+9+12h+194k2+25+20k=1

⇒36h2+16k2+108h+80k+145=0

Hence, the locus of x,y is 36x2+16y2+108x+80y+145=0

Topic: Mathematics, Ellipse, JEE Main, JEE Main 2021 (31 Aug Shift 2)
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, JEE Main
The locus of the centre of a circle which cuts the circle x2- 20x+y2+4=0 orthogonally and also touches the line x=2 is

Options:
A y2=16x+4
B x2=16y
C x2 =16 y+4
D y2=16x

The correct answer is:
y2=16x

Solution:
Let the general equation of circle be
x2+y2+2gx+2fy+c=0 ....... (i)
It cuts the circle x2+y2- 20x+4=0 orthogonally
â´2 -10g+0 Ãf = c+4 â- 20 g=c+4 .......(ii)
âµ circle (i) touches x=2
therefore, perpendicular distance from centre to the tangent to the circle=radius
â-g+0-212+02=g2+f2-c
âg+22= g2+f2-c
âg2+4+4g=g2+f2-câ 4g+4=f2-c ... (iii)
on eliminating c from (ii) and (iii) we get
-16g+4=f2+4âf2 +16g=0
Hence, locus of -g, -f is,
y2- 16x=0 (replacing -f & -g by x & y )

Topic: Mathematics, Circle, JEE Main
The locus of the centre of a circle, which touches externally the given two circles, is

Options:
A Circle
B Parabola
C Hyperbola
D Ellipse

The correct answer is:
Hyperbola

Solution:
We know that when a circle touches externally to the two given circles, then the locus of the circle will be hyperbola.

Topic: Mathematics, Hyperbola, JEE Main
The locus of the centre of a circle which touches the circle $\left|z-z_1\right|=a$ and $\left|z-z_2\right|=b$ externally ( $z, z_1$ and $z_2$ are complex numbers) will be

Options:
A
an ellipse

B
a hyperbola

C
a circle

D
none of these

The correct answer is:
a hyperbola

Solution:
No Solution Available

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2002
The locus of the centre of the circle described on any focal chord of the parabola y2=4ax as the diameter is

Options:
A y2=2ax+a
B y2=ax+a
C y2=2ax-a
D y2=4ax-a

The correct answer is:
y2=2ax-a

Solution:
Let, Pat12,2at1 and Qat22,2at2 be the extremities of a focal chord PQ of the parabola y2=4ax. Then, t1t2=-1.
Let, h,k be the coordinates of the centre of the circle described on PQ as diameter. Then,
h=a2t12+t22 and k=at1+t2
⇒2ha=t12+t22 and ka2=t1+t22
⇒2ha=t12+t22 and k2a2=t12+t22+2t1t2
⇒k2a2=2ha-2 ∵t1t2=-1
⇒k2=2ah-a
Hence, the locus of h,k is
y2=2ax-a.

Topic: Mathematics, Parabola, JEE Main
The locus of the centre of the circle which makes equal intercepts on the lines x+y=1 and x+y=5 is

Options:
A x-y=2
B x+y=6
C x+y=3
D x-y=0

The correct answer is:
x+y=3

Solution:
Let center be h,k
Since it makes equal intercepts on the two given lines
Hence, the center is equidistant from both the lines
i.e. h+k-12=h+k-52âh+k=3

Topic: Mathematics, Circle, JEE Main
The locus of the centres of the circles, which touch the circle, x2+y2=1 externally, also touch the y-axis and lie in the first quadrant, is:

Options:
A y=1+2x, x≥0
B y=1+4x, x≥0
C x=1+2y, y≥0
D x=1+4y, y≥0

The correct answer is:
y=1+2x, x≥0

Solution:
Let, the centre of the circle whose locus is to be determined is P(h, k) and it touches y-axis in the first quadrant and we know that if a circle touches the y-axis, then its radius is equal to the absolute value of the x-coordinate of the centre.

Hence, the radius of the circle is =h.

Also, we know that if two circles touches externally, then the distance between their centres is equal to the sum of their radii.

And, the centre and radius of a circle x2+y2=r2 is 0, 0 and 1 respectively.

Now, since the circle with centre h, k and radius h touches x2+y2=1externally, then

⇒h+1=h2+k2

Squaring both the sides, we get

⇒h2+2h+1=h2+k2

⇒k2=2h+1

Now, to get the locus, replace h, k by x, y, to get

y2=2x+1.

Hence, the required locus is y2=2x+1

⇒y=1+2x, x≥0.

Topic: Mathematics, Circle, JEE Main, JEE Main 2019 (10 Apr Shift 2)
The locus of the centres of the circles, which touch the circle, \( x^{2}+y^{2}=1 \) externally, also touch the \( y \)-axis and lie in the first quadrant, is:

Options:
A \( y=\sqrt{1+2 x}, \quad x \geq 0 \)
B \( \mathrm{y}=\sqrt{1+4 \mathrm{x}}, \quad \mathrm{x} \geq 0 \)
C \( \mathrm{x}=\sqrt{1+2 \mathrm{y}}, \quad \mathrm{y} \geq 0 \)
D \( \mathrm{x}=\sqrt{1+4 \mathrm{y}}, \quad \mathrm{y} \geq 0 \)

The correct answer is:
\( y=\sqrt{1+2 x}, \quad x \geq 0 \)

Solution:
Let I =∫dxx3+3x2+2x

Factorizing the denominator of the integral I, we get

⇒ I=∫dxxx+1x+2

⇒ I=∫1x+1-1x+21xdx

⇒ I=∫1xx+1dx-∫1xx+2dx

⇒ I=∫ 1x-1x+1dx-12∫1x-1x+2dx

Using ∫1xdx=logx

⇒ I=logx-logx+1-12logx+12logx+2+c

⇒ I=12logx+logx+2-2logx+1+c

Using logmn=nlogm

⇒ I=12logx+logx+2-logx+12+c

Using logm+logn=logmn & logm-logn=logmn

The required solution is I=12logx2+2xx+12+c.

Let the centre of one such circle be P(h,k) since it touches y-axis in the first quadrant
⇒ its radius =h
Now, since it touches x2+y2=1 externally
⇒h+1=h2+k2
⇒h2+2h+1=h2k2
⇒k2=2h+1
Hence the required locus is
y2=2x+1
⇒y= 1+2x, x≥0

Topic: Mathematics, Circle, JEE Main
The locus of the centroid of the triangle formed by any point P on the hyperbola 16x2-9y2+32x+36y-164=0 and its foci is

Options:
A 16x2-9y2+32x+36y-36=0
B 9x2-16y2+36x+32y-144=0
C 16x2-9y2+32x+36y-144=0
D 9x2-16y2+36x+32y-36=0

The correct answer is:
16x2-9y2+32x+36y-36=0

Solution:
Given hyperbola is

16x2-9y2+32x+36y-164=0

⇒16x+12-9y-22=164+16-36

⇒16x+12-9y-22=144....1

⇒(x+1)29-(y-2)216=1

Compare the above equation with ⇒(x-h)2a2-(y-k)2b2=1

We get, h=-1, k= 2, a2=9 and b2=16

∴ Eccentricity, e=1+169=53

We know that, focii are h+ae, k and h-ae, k

Hence, focii are (4,2) and (-6,2)

Let the centroid be (h,k) & A(α,β) be point on hyperbola

So h=α-6+43, k=β+2+23

⇒α=3h+2, β=3k-4

Put the values of α,β in the equation1,

163h+2+12-93k-4-22=144

⇒144(h+1)2-81(k-2)2=144

⇒16h2+2h+1-9k2-4k+4=16

⇒16h2-9k2+32h+36k-36=0

So, the locus of the centroid is

16x2-9y2+32x+36y-36=0

Topic: Mathematics, Hyperbola, JEE Main
The locus of the centroid of the triangle formed by any point P on the hyperbola 16x2-9y2+32x+36y-164=0 and its foci is

Options:
A 16x2-9y2+32x+36y-36=0
B 9x2-16y2+36x+32y-144=0
C 16x2-9y2+32x+36y-144=0
D 9x2-16y2+36x+32y-36=0

The correct answer is:
16x2-9y2+32x+36y-36=0

Solution:
Given hyperbola is

16x2-9y2+32x+36y-164=0

⇒16x+12-9y-22=164+16-36

⇒16x+12-9y-22=144....1

⇒(x+1)29-(y-2)216=1

Compare the above equation with ⇒(x-h)2a2-(y-k)2b2=1

We get, h=-1, k= 2, a2=9 and b2=16

∴ Eccentricity, e=1+169=53

We know that, focii are h+ae, k and h-ae, k

Hence, focii are (4,2) and (-6,2)

Let the centroid be (h,k) & A(α,β) be point on hyperbola

So h=α-6+43, k=β+2+23

⇒α=3h+2, β=3k-4

Put the values of α,β in the equation1,

163h+2+12-93k-4-22=144

⇒144(h+1)2-81(k-2)2=144

⇒16h2+2h+1-9k2-4k+4=16

⇒16h2-9k2+32h+36k-36=0

So, the locus of the centroid is

16x2-9y2+32x+36y-36=0

Topic: Mathematics, Hyperbola, JEE Main, JEE Main 2021 (25 Jul Shift 1)
The locus of the centroid of the triangle whose vertices are \( (3 \cos \alpha, 3 \sin \alpha),(9 \sin \alpha,-9 \cos \alpha) \) and \( (1,0) \) is a circle of radius \( R \), then the value of \( \frac{9 R^{2}}{2} \) is equal to

Options:

The correct answer is:
15

Solution:
Let centroid of ∆ABC is h,k

⇒h=3cosα+9sinα+13 & k=0+3sinα-9cosα3

⇒cosα+3sinα=3h-1 & sinα-3cosα=3k

⇒cosα+3sinα2=3h-12 & sinα-3cosα2=9k2

⇒3h-12+9k2=cosα+3sinα2+sinα-3cosα2

Replace h by x and k by y

⇒3x-12+9y2=10cos2α+sin2α

⇒x-132+y2=109

⇒R=103

⇒9R22=5

Topic: Mathematics, Straight Lines, JEE Main
The locus of the foot of perpendicular drawn from the centre of the ellipse x2+3y2=6 on any tangent to it is

Options:
A x2 + y22=6x2+2y2
B x2+ y22=6x2-2y2
C x2-y22=6x2+2y2
D x2-y22=6x2-2y2

The correct answer is:
x2 + y22=6x2+2y2

Solution:
The centre of the given ellipse 0,0
The general equation of a tangent to the ellipse

x 2 a 2 + y 2 b 2 = 1 is y=mx±a2m2+b2 ...1

Given ellipse : x26+y22=1

A perpendicular line from the centre is y=-xm ...2

Eliminating m, from 1 and 2

y=-xyx± a2x2y2+b2

y2=-x2±y a2x2y2+b2

∴ x2+y2=±a2x2+b2y2

Squaring both sides,

x2+y22=a2x2+b2y2=6x2+2y2

Topic: Mathematics, Parabola, JEE Main, JEE Main 2014 (06 Apr)
The locus of the mid point of the line joining the focus and any point on the parabola $y^{2}=4 a x$ is a parabola with the equation of directrix as

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

The correct answer is:
$\mathrm{x}=0$

Solution:
Let the coordinates of focus be $S(a, 0)$.
Let any point on the parabola be $P$ (at ${ }^{2}, 2$ at). Let the coordinates of mid point of $P$ and $S$ be $\left(x_{1}, y_{1}\right)$.
$$
\begin{aligned}
& & \mathrm{x}_{1} &=\frac{\mathrm{a}+\mathrm{at}^{2}}{2}, \mathrm{y}_{1} \\
\Rightarrow & & \mathrm{at}^{2} &=2 \mathrm{x}_{1}-\mathrm{a}, \quad \mathrm{y}_{1} \\
\Rightarrow & \mathrm{a}\left(\frac{\mathrm{y}_{1}}{\mathrm{a}}\right)^{2} &=2 \mathrm{x}_{1}-\mathrm{a} \\
\Rightarrow & \mathrm{y}_{1}^{2} &=2 \mathrm{x}_{1} \mathrm{a}-\mathrm{a}^{2}
\end{aligned}
$$
Hence, the locus of the mid point is
$$
\mathrm{y}^{2}=2 \mathrm{a}\left(\mathrm{x}-\frac{\mathrm{a}}{2}\right)
$$
$\therefore$ Equation of directrix is $x-\frac{a}{2}=-\frac{a}{2}$
$$
x=0
$$

Topic: Mathematics, Parabola, JEE Main
The locus of the mid points of the intercepted portion of the tangents by the coordinate axes, which are drawn to the ellipse $x^2+2 y^2=2$ is

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

The correct answer is:
$\frac{1}{2 x^2}+\frac{1}{4 y^2}=1$

Solution:
$$
\frac{x^2}{2}+\frac{y^2}{1}=1 \Rightarrow \frac{x^2}{a^2}+\frac{y^2}{b^2}=1
$$
Equation of tangent at $P(a \cos \theta, b \sin \theta)$
$\begin{aligned} & \frac{x}{\sqrt{2}} \cos \theta+y \sin \theta=1, \text { For } A, y=0 \\ & x=\sqrt{2} \sec \theta, \text { For } B, x=0, y=\operatorname{cosec} \theta \\ & \therefore \quad \text { Midpoint of } A B=\left(\frac{x}{2}, \frac{y}{2}\right) \\ & M \equiv\left(\frac{\sec \theta}{\sqrt{2}}, \frac{\operatorname{cosec} \theta}{2}\right) \equiv\left(\left(\frac{x}{2}\right),\left(\frac{y}{2}\right)\right) \equiv(h, k) \\ & \because \quad \sin ^2 \theta+\cos ^2 \theta=-1 \\ & \Rightarrow \quad\left(\frac{1}{2 k}\right)^2+\left(\frac{1}{\sqrt{2} h}\right)^2=1 \Rightarrow \frac{1}{4 k^2}+\frac{1}{2 h^2}=1 \\ & \therefore \quad \frac{1}{2 x^2}+\frac{1}{4 y^2}=1\end{aligned}$

Topic: Mathematics, Ellipse, JEE Main
The locus of the middle points of the chords of the parabola $y^2=4 a x$ which passes through the origin

Options:
A $y^2=a x$
B $y^2=2 a x$
C $y^2=4 a x$
D $x^2=4 a y$

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

Solution:
Any line through origin $(0,0)$ is $y=m x$. It intersects $y^2=4 a x$ in $\left(\frac{4 a}{m^2}, \frac{4 a}{m}\right)$.
Mid point of the chord is $\left(\frac{2 a}{m^2}, \frac{2 a}{m}\right)$
$x=\frac{2 a}{m^2}, y=\frac{2 a}{m} \Rightarrow \frac{2 a}{x}=\frac{4 a^2}{y^2}$ or $y^2=2 a x$, which is a parabola.

Topic: Mathematics, Parabola, JEE Main
The locus of the middle points of the chords of the parabola y2=4ax, which passes through the origin is

Options:
A y2=ax
B y2=2ax
C y2=4ax
D x2=4ay

The correct answer is:
y2=2ax

Solution:
Let mid point be(x1,y1)
â´ Equation of chord is
T=S1
yy1-2ax+x1=y12-4ax1
Since, it passes through origin
â´ -2ax1=y12-4ax1
â y12=2ax1
â´ Locus is y2=2ax

Topic: Mathematics, Parabola, JEE Main
The locus of the middle points of the chords of the parabola $y^2=4 a x$ which passes through the origin

Options:
A $y^2=a x$
B $y^2=2 a x$
C $y^2=4 a x$
D $x^2=4 a y$

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

Solution:
Any line through origin $(0,0)$ is $y=m x$. It intersects $y^2=4 a x$ in $\left(\frac{4 a}{m^2}, \frac{4 a}{m}\right)$
Mid point of the chord is $\left(\frac{2 a}{m^2}, \frac{2 a}{m}\right)$
$x=\frac{2 a}{m^2}$, $y=\frac{2 a}{m}$ $\Rightarrow \frac{2 a}{x}=\frac{4 a^2}{y^2}$ or $y^2=2 a x$, which is a parabola.

Topic: Mathematics, Parabola, JEE Main
The locus of the mid-point of the chord if contact of tangents drawn from points lying on the straight line $4 x-5 y=20$ to the circle $x^2+y^2=9$ is

Options:
A
$20\left(x^2+y^2\right)-36 x+45 y=0$

B
$20\left(x^2+y^2\right)+36 x-45 y=0$

C
$36\left(x^2+y^2\right)-20 x+45 y=0$

D
$36\left(x^2+y^2\right)+20 x-45 y=0$

The correct answer is:

$20\left(x^2+y^2\right)-36 x+45 y=0$

Solution:
Let $P\left(t, \frac{4 t-20}{5}\right)$ be a point on the line
$4 x-5 y=20$. Then the chord of contact of tangents drawn from $P$ to the circle $x^2+y^2=9$ is
$t x+\left(\frac{4 t-20}{5}\right) y=9\quad \quad \dots(i)$
Let $\theta(h, k)$ be the mid-point of this chord of contact, then its equation is also $h x+k y=h^2+k^2\quad \quad \dots(ii)$
$\left[\right.$ using $\left.T=S^{\prime}\right]$
Clearly Eqs. (i) and (ii) represent the same line
$\begin{aligned}
& \therefore \frac{t}{h}=\frac{4 t-20}{5 k}=\frac{9}{h^2+k^2} \\
& \Rightarrow \frac{t}{h}=\frac{9}{h^2+k^2} \text { and } \frac{t}{h}=\frac{4 t-20}{5 k} \\
& \Rightarrow t=\frac{9 h}{h^2+k^2} \text { and } t=\frac{20 h}{4 h-5 k} \\
& \Rightarrow \frac{9 h}{h^2+k^2}=\frac{20 h}{4 h-5 k} \\
& \Rightarrow h\left\{20\left(h^2+k^2\right)-36 h+45 k\right\}=0 \\
& x=0, \text { or }\left[20\left(x^2+y^2\right)-36 x+45 y\right]=0
\end{aligned}$

Topic: Mathematics, Circle, JEE Main
The locus of the midpoint of the chords of the hyperbola x225-y236=1 which passes through the point 2,4 is a hyperbola, whose transverse axis length (in units) is equal to

Options:
A 165
B 43
C 85
D 6125

The correct answer is:
165

Solution:
The equation of the chord whose mid-point is (h,k) is T=S1
⇒xh25-yk36=h225-k236
Since it passes through 2,4
2h25-4k36=h225-k236⇒h2-2h+125-k2-4k+436=125-436
Hence, the locus of h,k is
x-12169-y-226425=-1
This is the equation of a hyperbola whose transverse axis length is 165 units

Topic: Mathematics, Hyperbola, JEE Main
The locus of the mid-point of the chords of the hyperbola x2-y2=4, that touches the parabola y2=8x is

Options:
A x2x-2=y3
B y2x-2=x3
C x3x-2=y2
D y3x-2=x2

The correct answer is:
y2x-2=x3

Solution:
Let midpoint of the chord be Ph,k.

Then, the equation of the chord will be xh-yk=h2-k2
⇒y=hkx+k2-h2k … 1
Now, the line y=mx+c is tangent to y2=4ax if c=am
Hence, equation 1 is tangent to y2=8x if
k2-h2k=2kh⇒y2x-x3=2y2
⇒y2x-2=x3

Topic: Mathematics, Hyperbola, JEE Main
The locus of the mid-point of the line segment joining the focus of the parabola y2=4ax to a moving point of the parabola, is another parabola whose directrix is:

Options:
A x=a
B x=0
C x=-a2
D x=a2

The correct answer is:
x=0

Solution:
h=at2+a2, k=2at+02

⇒ t2=2 h-aa and t=ka

⇒ k2a2=2 h-aa

⇒ Locus of h, k is y2=a2x-a

⇒ y2=2ax-a2

Its directrix is x-a2=-a2⇒x=0

Topic: Mathematics, Parabola, JEE Main
The locus of the mid-point of the line segment joining the focus of the parabola y2=4ax to a moving point of the parabola, is another parabola whose directrix is:

Options:
A x=a
B x=0
C x=-a2
D x=a2

The correct answer is:
x=0

Solution:
h=at2+a2, k=2at+02

⇒ t2=2 h-aa and t=ka

⇒ k2a2=2 h-aa

⇒ Locus of h, k is y2=a2x-a

⇒ y2=2ax-a2

Its directrix is x-a2=-a2⇒x=0

Topic: Mathematics, Parabola, JEE Main, JEE Main 2021 (24 Feb Shift 1)
The locus of the midpoints of the chord of the circle, x2+y2=25 which is tangent to the hyperbola, x29-y216=1 is :

Options:
A x2+y22-16x2+9y2=0
B x2+y22-9x2+144y2=0
C x2+y22-9x2-16y2=0
D x2+y22-9x2+16y2=0

The correct answer is:
x2+y22-9x2+16y2=0

Solution:
Equation of chord

y-k=-hkx-h

ky-k2=-hx+h2

hx+ky=h2+k2

y=-hxk+h2+k2k

tangent to x29-y216=1

c2=a2m2-b2

h2+k2k2=9-hk2-16

x2+y22=9x2-16y2

Topic: Mathematics, Hyperbola, JEE Main, JEE Main 2021 (16 Mar Shift 1)
The locus of the mid-points of the chords of the circle $x^2+y^2=16$ which are the tangents to the hyperbola $9 x^2-16 y^2=144$ is

Options:
A $3 x^2-4 y^2=16\left(x^2+y^2\right)$
B $4 x^2-3 y^2=9\left(x^2+y^2\right)$
C $16 x^2-9 y^2=\left(x^2+y^2\right)^2$
D $16 x^2-9 y^2=4\left(x^2+y^2\right)$

The correct answer is:
$16 x^2-9 y^2=\left(x^2+y^2\right)^2$

Solution:
Equation of chord with mid-point $(h, k)$ to the circle $x^2+y^2=16$ is $T=S_1$
$$
\begin{gathered}
h x+k y-16=h^2+k^2-16 \\
y=-\frac{h}{k} x+\left(\frac{h^2+k^2}{k}\right)
\end{gathered}
$$
For hyperbola $9 x^2-16 y^2=144$
$$
\Rightarrow \quad \frac{x^2}{16}-\frac{y^2}{9}=1
$$
As we know that, condition of tangency of a line $y=m x+c$ is $a^2 m^2-b^2=c^2$
So,
$$
\begin{aligned}
\text { So, } & & 16\left(-\frac{h}{k}\right)^2-9 & =\left(\frac{h^2+k^2}{k}\right)^2 \\
\Rightarrow & & 16 h^2-9 k^2 & =\left(h^2+k^2\right)^2
\end{aligned}
$$
Replace $h, k$ by $x, y$
$$
\therefore \quad 16 x^2-9 y^2=\left(x^2+y^2\right)^2
$$

Topic: Mathematics, Hyperbola, JEE Main
The locus of the mid-points of the chords of the hyperbola 3x2-2y2+4x-6y=0 which are parallel to the line y=2x+4 is

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

The correct answer is:
3x-4y=4

Solution:
Let mid-point of the chord is h,k
⇒ equation of the chord is
3hx-2ky+4x+h2-6y+k2=3h2-2k2+4h-6k
⇒3h+2x+-2k-3y=3h2-2k2+2h-3k
⇒ Slope of the chord is equal to 3h+22k+3=2
⇒ locus of the mid-points is 3x+2=4y+6
⇒3x-4y-4=0

Topic: Mathematics, Hyperbola, JEE Main
The locus of the mid-points of the parallel chords with slope m of the rectangular hyperbola xy=c2 is

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

The correct answer is:
y+mx=0

Solution:
Let, the mid-point is h,k, then
Equation of AB is xh+yk=2
m=-kh∴y=-mx

Topic: Mathematics, Hyperbola, JEE Main
The locus of the mid-points of the perpendiculars drawn from points on the line x=2y, to the line x=y, is.

Options:
A 2x-3y=0
B 5x-7y=0
C 3x-2y=0
D 7x-5y=0

The correct answer is:
5x-7y=0

Solution:
Slope of PQ=k-αh-2α=-1
k-α=-h+2α ⇒α=h+k3 …1
Also, 2h=2α+β and 2k=α+β
2h=α+2k⇒α=2h-2k …2
From 1 & 2,
h+k3=2h-k
So, locus is 6x-6y=x+y⇒5x=7y.

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2020 (07 Jan Shift 2)