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Co-Ordinate system

88460 A point moves in such a way that the difference of its distance from two points $(8, 0)$ and $(- 8, 0)$ always remains 4 . Then, the locus of the point is

1 a circle

2 a parabola

3 an ellipse

4 a hyperbola

Explanation:

(D) : Given that- A point moves in such away that the difference of its distance from two point $(8, 0)$ and $(- 8, 0)$ always remains 4 .
We have to find that the locus of the point = ?
Solve-
We know by the definition of hyperbola-
" A hyperbola is the locus of a point which moves in such away that the difference between two fixed points remains constant,
$i.e, | PS - {PS}^{'} | = 2 a$
So, locus of the point $=$ a hyperbola

WB JEE-2012

Co-Ordinate system

88461 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

1 A straight line

2 An ellipse

3 A parabola

4 A hyperbola

Explanation:

(D) : Given equation of straight line-
$\frac{x}{a} + \frac{y}{b} = k$
$\frac{x}{a} - \frac{y}{b} = \frac{1}{k}$
Let the point of intersection be $(α, β)$
So, from equation (i) and (ii), we get-
$\frac{α}{a} + \frac{β}{b} = k$
and $\frac{α}{a} - \frac{β}{b} = \frac{1}{k}$
${(\frac{α}{a})}^{2} - {(\frac{β}{b})}^{2} = 1$
$\frac{α^{2}}{a^{2}} - \frac{β^{2}}{b^{2}} = 1$
$∴$ Locus $\frac{x^{2}}{a^{2}} - {\frac{y}{}}^{2}^{2} = 1$ ,
${\frac{x}{a^{2}}}^{2} - {\frac{y}{b^{2}}}^{2} = 1$
Which is the equation of a hyperbola.

WB JEE-2016

Co-Ordinate system

88462 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 RS is

1

x - y = \frac{a}{2}

2

x + y = a

3

x^{2} + y^{2} = 4 a^{2}

4

x^{2} - y^{2} = 2 a^{2}

Explanation:

(B) : Let the equation of line $AB$ is-
$\frac{x}{2 a} + \frac{y}{2 a} = 1$
$x + y = 2 a$

Let the coordinate of the midpoint of RS be (h, k)
$∴ R$ and $S$ are $(2 h, 0)$ and $(0, 2 k)$
So the mid point of Rs is $(h, k)$
$∵ P$ lies on line $A B$
The from equation(i), we have
$2 h + 2 k = 2 a$
$h + k = a$
So, the locus of $(h, k)$ is $x + y = a$ $x + y = a$

WB JEE-2016

Co-Ordinate system

88463 The angle between a pair of tangents drawn from a point $P$ to the circle
$x^{2} + y^{2} + 4 x - 6 y + 9 \sin^{2} α + 13 \cos^{2} α = 0$ is $2 α$ . The equation of the locus of the point $P$ is

1

x^{2} + y^{2} + 4 x + 6 y + 9 = 0

2

x^{2} + y^{2} - 4 x + 6 y + 9 = 0

3

x^{2} + y^{2} - 4 x - 6 y + 9 = 0

4

x^{2} + y^{2} + 4 x - 6 y + 9 = 0

Explanation:

(D) : Given equation of circle-
$x^{2} + y^{2} + 4 x - 6 y + 9 \sin^{2} α + 13 \cos^{2} α = 0$
Here, $C = (- 2, 3)$
$Radius = \sqrt{(- 2)^{2} + (3)^{2} - (9 \sin^{2} α + 13 \cos^{2} α)}$
$= \sqrt{4 + 9 - 9 \sin^{2} α - 13 \cos^{2} α}$
$= \sqrt{13 \sin^{2} α - 9 \sin^{2} α}$
$= \sqrt{4 \sin^{2} α} = 2 \sin α$

Here, $\sin α = \frac{AC}{PC}$
$PC \sin α = AC$
${PC}^{2} \sin^{2} α = {AC}^{2} = (2 \sin α)^{2}$
$[(h + 2)^{2} + (k - 3)^{2}] \sin^{2} α = 4 \sin^{2} α$
$(h + 2)^{2} + (k - 3)^{2} = 4$
$h^{2} + k^{2} + 4 h + k^{2} + 9 - 6 k = 4$
$h^{2} + k^{2} + 4 h - 6 k + 9 = 0$
So, locus of a point is $x^{2} + y^{2} + 4 x - 6 y + 9 = 0$

Co-Ordinate system

88471 Let $A$ be the fixed point $(0, 4)$ and $B$ be a moving point on $x$ -axis. Let $M$ be the midpoint of $A B$ and let the perpendicular bisector of $A B$ meets the $y$ -axis at $R$ , The locus of the midpoint $P$ of MR is

1

y + x^{2} = 2

2

x^{2} + (y - 2)^{2} = \frac{1}{4}

3

(y - 2)^{2} - x^{2} = \frac{1}{4}

4

x^{2} + y^{2} = 16

Explanation:

(A): Let $B$ is $(2 p, 0)$
Therefore co-ordinate of $M$ are $(b, 2)$
Now, slope of $A B = \frac{4 - 0}{0 - 2 p} = \frac{- 2}{p}$
Therefore,
Slope of MR $= \frac{p}{2}$
Therefore of MR $= y - 2 = \frac{p}{2} (x - p)^{0, 00}, p$
On putting $x = 0$ ,

$y - 2 = \frac{- p^{2}}{2}$
$y = 2 - \frac{p^{2}}{2}$
Therefore co-ordinate of $R$ is $(0, 2 - \frac{p^{2}}{2})$
Therefore of $p$ is $[\frac{p}{2}, \frac{2 - \frac{p^{2}}{2} + 2}{2}]$
i.e $[\frac{p}{2}, 2 - \frac{p^{2}}{4}]$
$∴ y = 2 - x^{2}$
$y + x^{2} = 2$

WB JEE-2021

Co-Ordinate system

88460 A point moves in such a way that the difference of its distance from two points $(8, 0)$ and $(- 8, 0)$ always remains 4 . Then, the locus of the point is

1 a circle

2 a parabola

3 an ellipse

4 a hyperbola

Explanation:

(D) : Given that- A point moves in such away that the difference of its distance from two point $(8, 0)$ and $(- 8, 0)$ always remains 4 .
We have to find that the locus of the point = ?
Solve-
We know by the definition of hyperbola-
" A hyperbola is the locus of a point which moves in such away that the difference between two fixed points remains constant,
$i.e, | PS - {PS}^{'} | = 2 a$
So, locus of the point $=$ a hyperbola

WB JEE-2012

Co-Ordinate system

88461 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

1 A straight line

2 An ellipse

3 A parabola

4 A hyperbola

Explanation:

(D) : Given equation of straight line-
$\frac{x}{a} + \frac{y}{b} = k$
$\frac{x}{a} - \frac{y}{b} = \frac{1}{k}$
Let the point of intersection be $(α, β)$
So, from equation (i) and (ii), we get-
$\frac{α}{a} + \frac{β}{b} = k$
and $\frac{α}{a} - \frac{β}{b} = \frac{1}{k}$
${(\frac{α}{a})}^{2} - {(\frac{β}{b})}^{2} = 1$
$\frac{α^{2}}{a^{2}} - \frac{β^{2}}{b^{2}} = 1$
$∴$ Locus $\frac{x^{2}}{a^{2}} - {\frac{y}{}}^{2}^{2} = 1$ ,
${\frac{x}{a^{2}}}^{2} - {\frac{y}{b^{2}}}^{2} = 1$
Which is the equation of a hyperbola.

WB JEE-2016

Co-Ordinate system

88462 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 RS is

1

x - y = \frac{a}{2}

2

x + y = a

3

x^{2} + y^{2} = 4 a^{2}

4

x^{2} - y^{2} = 2 a^{2}

Explanation:

(B) : Let the equation of line $AB$ is-
$\frac{x}{2 a} + \frac{y}{2 a} = 1$
$x + y = 2 a$

Let the coordinate of the midpoint of RS be (h, k)
$∴ R$ and $S$ are $(2 h, 0)$ and $(0, 2 k)$
So the mid point of Rs is $(h, k)$
$∵ P$ lies on line $A B$
The from equation(i), we have
$2 h + 2 k = 2 a$
$h + k = a$
So, the locus of $(h, k)$ is $x + y = a$ $x + y = a$

WB JEE-2016

Co-Ordinate system

88463 The angle between a pair of tangents drawn from a point $P$ to the circle
$x^{2} + y^{2} + 4 x - 6 y + 9 \sin^{2} α + 13 \cos^{2} α = 0$ is $2 α$ . The equation of the locus of the point $P$ is

1

x^{2} + y^{2} + 4 x + 6 y + 9 = 0

2

x^{2} + y^{2} - 4 x + 6 y + 9 = 0

3

x^{2} + y^{2} - 4 x - 6 y + 9 = 0

4

x^{2} + y^{2} + 4 x - 6 y + 9 = 0

Explanation:

(D) : Given equation of circle-
$x^{2} + y^{2} + 4 x - 6 y + 9 \sin^{2} α + 13 \cos^{2} α = 0$
Here, $C = (- 2, 3)$
$Radius = \sqrt{(- 2)^{2} + (3)^{2} - (9 \sin^{2} α + 13 \cos^{2} α)}$
$= \sqrt{4 + 9 - 9 \sin^{2} α - 13 \cos^{2} α}$
$= \sqrt{13 \sin^{2} α - 9 \sin^{2} α}$
$= \sqrt{4 \sin^{2} α} = 2 \sin α$

Here, $\sin α = \frac{AC}{PC}$
$PC \sin α = AC$
${PC}^{2} \sin^{2} α = {AC}^{2} = (2 \sin α)^{2}$
$[(h + 2)^{2} + (k - 3)^{2}] \sin^{2} α = 4 \sin^{2} α$
$(h + 2)^{2} + (k - 3)^{2} = 4$
$h^{2} + k^{2} + 4 h + k^{2} + 9 - 6 k = 4$
$h^{2} + k^{2} + 4 h - 6 k + 9 = 0$
So, locus of a point is $x^{2} + y^{2} + 4 x - 6 y + 9 = 0$

Co-Ordinate system

88471 Let $A$ be the fixed point $(0, 4)$ and $B$ be a moving point on $x$ -axis. Let $M$ be the midpoint of $A B$ and let the perpendicular bisector of $A B$ meets the $y$ -axis at $R$ , The locus of the midpoint $P$ of MR is

1

y + x^{2} = 2

2

x^{2} + (y - 2)^{2} = \frac{1}{4}

3

(y - 2)^{2} - x^{2} = \frac{1}{4}

4

x^{2} + y^{2} = 16

Explanation:

(A): Let $B$ is $(2 p, 0)$
Therefore co-ordinate of $M$ are $(b, 2)$
Now, slope of $A B = \frac{4 - 0}{0 - 2 p} = \frac{- 2}{p}$
Therefore,
Slope of MR $= \frac{p}{2}$
Therefore of MR $= y - 2 = \frac{p}{2} (x - p)^{0, 00}, p$
On putting $x = 0$ ,

$y - 2 = \frac{- p^{2}}{2}$
$y = 2 - \frac{p^{2}}{2}$
Therefore co-ordinate of $R$ is $(0, 2 - \frac{p^{2}}{2})$
Therefore of $p$ is $[\frac{p}{2}, \frac{2 - \frac{p^{2}}{2} + 2}{2}]$
i.e $[\frac{p}{2}, 2 - \frac{p^{2}}{4}]$
$∴ y = 2 - x^{2}$
$y + x^{2} = 2$

WB JEE-2021

Co-Ordinate system

88460 A point moves in such a way that the difference of its distance from two points $(8, 0)$ and $(- 8, 0)$ always remains 4 . Then, the locus of the point is

1 a circle

2 a parabola

3 an ellipse

4 a hyperbola

Explanation:

(D) : Given that- A point moves in such away that the difference of its distance from two point $(8, 0)$ and $(- 8, 0)$ always remains 4 .
We have to find that the locus of the point = ?
Solve-
We know by the definition of hyperbola-
" A hyperbola is the locus of a point which moves in such away that the difference between two fixed points remains constant,
$i.e, | PS - {PS}^{'} | = 2 a$
So, locus of the point $=$ a hyperbola

WB JEE-2012

Co-Ordinate system

88461 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

1 A straight line

2 An ellipse

3 A parabola

4 A hyperbola

Explanation:

(D) : Given equation of straight line-
$\frac{x}{a} + \frac{y}{b} = k$
$\frac{x}{a} - \frac{y}{b} = \frac{1}{k}$
Let the point of intersection be $(α, β)$
So, from equation (i) and (ii), we get-
$\frac{α}{a} + \frac{β}{b} = k$
and $\frac{α}{a} - \frac{β}{b} = \frac{1}{k}$
${(\frac{α}{a})}^{2} - {(\frac{β}{b})}^{2} = 1$
$\frac{α^{2}}{a^{2}} - \frac{β^{2}}{b^{2}} = 1$
$∴$ Locus $\frac{x^{2}}{a^{2}} - {\frac{y}{}}^{2}^{2} = 1$ ,
${\frac{x}{a^{2}}}^{2} - {\frac{y}{b^{2}}}^{2} = 1$
Which is the equation of a hyperbola.

WB JEE-2016

Co-Ordinate system

88462 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 RS is

1

x - y = \frac{a}{2}

2

x + y = a

3

x^{2} + y^{2} = 4 a^{2}

4

x^{2} - y^{2} = 2 a^{2}

Explanation:

(B) : Let the equation of line $AB$ is-
$\frac{x}{2 a} + \frac{y}{2 a} = 1$
$x + y = 2 a$

Let the coordinate of the midpoint of RS be (h, k)
$∴ R$ and $S$ are $(2 h, 0)$ and $(0, 2 k)$
So the mid point of Rs is $(h, k)$
$∵ P$ lies on line $A B$
The from equation(i), we have
$2 h + 2 k = 2 a$
$h + k = a$
So, the locus of $(h, k)$ is $x + y = a$ $x + y = a$

WB JEE-2016

Co-Ordinate system

88463 The angle between a pair of tangents drawn from a point $P$ to the circle
$x^{2} + y^{2} + 4 x - 6 y + 9 \sin^{2} α + 13 \cos^{2} α = 0$ is $2 α$ . The equation of the locus of the point $P$ is

1

x^{2} + y^{2} + 4 x + 6 y + 9 = 0

2

x^{2} + y^{2} - 4 x + 6 y + 9 = 0

3

x^{2} + y^{2} - 4 x - 6 y + 9 = 0

4

x^{2} + y^{2} + 4 x - 6 y + 9 = 0

Explanation:

(D) : Given equation of circle-
$x^{2} + y^{2} + 4 x - 6 y + 9 \sin^{2} α + 13 \cos^{2} α = 0$
Here, $C = (- 2, 3)$
$Radius = \sqrt{(- 2)^{2} + (3)^{2} - (9 \sin^{2} α + 13 \cos^{2} α)}$
$= \sqrt{4 + 9 - 9 \sin^{2} α - 13 \cos^{2} α}$
$= \sqrt{13 \sin^{2} α - 9 \sin^{2} α}$
$= \sqrt{4 \sin^{2} α} = 2 \sin α$

Here, $\sin α = \frac{AC}{PC}$
$PC \sin α = AC$
${PC}^{2} \sin^{2} α = {AC}^{2} = (2 \sin α)^{2}$
$[(h + 2)^{2} + (k - 3)^{2}] \sin^{2} α = 4 \sin^{2} α$
$(h + 2)^{2} + (k - 3)^{2} = 4$
$h^{2} + k^{2} + 4 h + k^{2} + 9 - 6 k = 4$
$h^{2} + k^{2} + 4 h - 6 k + 9 = 0$
So, locus of a point is $x^{2} + y^{2} + 4 x - 6 y + 9 = 0$

Co-Ordinate system

88471 Let $A$ be the fixed point $(0, 4)$ and $B$ be a moving point on $x$ -axis. Let $M$ be the midpoint of $A B$ and let the perpendicular bisector of $A B$ meets the $y$ -axis at $R$ , The locus of the midpoint $P$ of MR is

1

y + x^{2} = 2

2

x^{2} + (y - 2)^{2} = \frac{1}{4}

3

(y - 2)^{2} - x^{2} = \frac{1}{4}

4

x^{2} + y^{2} = 16

Explanation:

(A): Let $B$ is $(2 p, 0)$
Therefore co-ordinate of $M$ are $(b, 2)$
Now, slope of $A B = \frac{4 - 0}{0 - 2 p} = \frac{- 2}{p}$
Therefore,
Slope of MR $= \frac{p}{2}$
Therefore of MR $= y - 2 = \frac{p}{2} (x - p)^{0, 00}, p$
On putting $x = 0$ ,

$y - 2 = \frac{- p^{2}}{2}$
$y = 2 - \frac{p^{2}}{2}$
Therefore co-ordinate of $R$ is $(0, 2 - \frac{p^{2}}{2})$
Therefore of $p$ is $[\frac{p}{2}, \frac{2 - \frac{p^{2}}{2} + 2}{2}]$
i.e $[\frac{p}{2}, 2 - \frac{p^{2}}{4}]$
$∴ y = 2 - x^{2}$
$y + x^{2} = 2$

WB JEE-2021

Co-Ordinate system

88460 A point moves in such a way that the difference of its distance from two points $(8, 0)$ and $(- 8, 0)$ always remains 4 . Then, the locus of the point is

1 a circle

2 a parabola

3 an ellipse

4 a hyperbola

Explanation:

(D) : Given that- A point moves in such away that the difference of its distance from two point $(8, 0)$ and $(- 8, 0)$ always remains 4 .
We have to find that the locus of the point = ?
Solve-
We know by the definition of hyperbola-
" A hyperbola is the locus of a point which moves in such away that the difference between two fixed points remains constant,
$i.e, | PS - {PS}^{'} | = 2 a$
So, locus of the point $=$ a hyperbola

WB JEE-2012

Co-Ordinate system

88461 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

1 A straight line

2 An ellipse

3 A parabola

4 A hyperbola

Explanation:

(D) : Given equation of straight line-
$\frac{x}{a} + \frac{y}{b} = k$
$\frac{x}{a} - \frac{y}{b} = \frac{1}{k}$
Let the point of intersection be $(α, β)$
So, from equation (i) and (ii), we get-
$\frac{α}{a} + \frac{β}{b} = k$
and $\frac{α}{a} - \frac{β}{b} = \frac{1}{k}$
${(\frac{α}{a})}^{2} - {(\frac{β}{b})}^{2} = 1$
$\frac{α^{2}}{a^{2}} - \frac{β^{2}}{b^{2}} = 1$
$∴$ Locus $\frac{x^{2}}{a^{2}} - {\frac{y}{}}^{2}^{2} = 1$ ,
${\frac{x}{a^{2}}}^{2} - {\frac{y}{b^{2}}}^{2} = 1$
Which is the equation of a hyperbola.

WB JEE-2016

Co-Ordinate system

88462 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 RS is

1

x - y = \frac{a}{2}

2

x + y = a

3

x^{2} + y^{2} = 4 a^{2}

4

x^{2} - y^{2} = 2 a^{2}

Explanation:

(B) : Let the equation of line $AB$ is-
$\frac{x}{2 a} + \frac{y}{2 a} = 1$
$x + y = 2 a$

Let the coordinate of the midpoint of RS be (h, k)
$∴ R$ and $S$ are $(2 h, 0)$ and $(0, 2 k)$
So the mid point of Rs is $(h, k)$
$∵ P$ lies on line $A B$
The from equation(i), we have
$2 h + 2 k = 2 a$
$h + k = a$
So, the locus of $(h, k)$ is $x + y = a$ $x + y = a$

WB JEE-2016

Co-Ordinate system

88463 The angle between a pair of tangents drawn from a point $P$ to the circle
$x^{2} + y^{2} + 4 x - 6 y + 9 \sin^{2} α + 13 \cos^{2} α = 0$ is $2 α$ . The equation of the locus of the point $P$ is

1

x^{2} + y^{2} + 4 x + 6 y + 9 = 0

2

x^{2} + y^{2} - 4 x + 6 y + 9 = 0

3

x^{2} + y^{2} - 4 x - 6 y + 9 = 0

4

x^{2} + y^{2} + 4 x - 6 y + 9 = 0

Explanation:

(D) : Given equation of circle-
$x^{2} + y^{2} + 4 x - 6 y + 9 \sin^{2} α + 13 \cos^{2} α = 0$
Here, $C = (- 2, 3)$
$Radius = \sqrt{(- 2)^{2} + (3)^{2} - (9 \sin^{2} α + 13 \cos^{2} α)}$
$= \sqrt{4 + 9 - 9 \sin^{2} α - 13 \cos^{2} α}$
$= \sqrt{13 \sin^{2} α - 9 \sin^{2} α}$
$= \sqrt{4 \sin^{2} α} = 2 \sin α$

Here, $\sin α = \frac{AC}{PC}$
$PC \sin α = AC$
${PC}^{2} \sin^{2} α = {AC}^{2} = (2 \sin α)^{2}$
$[(h + 2)^{2} + (k - 3)^{2}] \sin^{2} α = 4 \sin^{2} α$
$(h + 2)^{2} + (k - 3)^{2} = 4$
$h^{2} + k^{2} + 4 h + k^{2} + 9 - 6 k = 4$
$h^{2} + k^{2} + 4 h - 6 k + 9 = 0$
So, locus of a point is $x^{2} + y^{2} + 4 x - 6 y + 9 = 0$

Co-Ordinate system

88471 Let $A$ be the fixed point $(0, 4)$ and $B$ be a moving point on $x$ -axis. Let $M$ be the midpoint of $A B$ and let the perpendicular bisector of $A B$ meets the $y$ -axis at $R$ , The locus of the midpoint $P$ of MR is

1

y + x^{2} = 2

2

x^{2} + (y - 2)^{2} = \frac{1}{4}

3

(y - 2)^{2} - x^{2} = \frac{1}{4}

4

x^{2} + y^{2} = 16

Explanation:

(A): Let $B$ is $(2 p, 0)$
Therefore co-ordinate of $M$ are $(b, 2)$
Now, slope of $A B = \frac{4 - 0}{0 - 2 p} = \frac{- 2}{p}$
Therefore,
Slope of MR $= \frac{p}{2}$
Therefore of MR $= y - 2 = \frac{p}{2} (x - p)^{0, 00}, p$
On putting $x = 0$ ,

$y - 2 = \frac{- p^{2}}{2}$
$y = 2 - \frac{p^{2}}{2}$
Therefore co-ordinate of $R$ is $(0, 2 - \frac{p^{2}}{2})$
Therefore of $p$ is $[\frac{p}{2}, \frac{2 - \frac{p^{2}}{2} + 2}{2}]$
i.e $[\frac{p}{2}, 2 - \frac{p^{2}}{4}]$
$∴ y = 2 - x^{2}$
$y + x^{2} = 2$

WB JEE-2021

Co-Ordinate system

88460 A point moves in such a way that the difference of its distance from two points $(8, 0)$ and $(- 8, 0)$ always remains 4 . Then, the locus of the point is

1 a circle

2 a parabola

3 an ellipse

4 a hyperbola

Explanation:

(D) : Given that- A point moves in such away that the difference of its distance from two point $(8, 0)$ and $(- 8, 0)$ always remains 4 .
We have to find that the locus of the point = ?
Solve-
We know by the definition of hyperbola-
" A hyperbola is the locus of a point which moves in such away that the difference between two fixed points remains constant,
$i.e, | PS - {PS}^{'} | = 2 a$
So, locus of the point $=$ a hyperbola

WB JEE-2012

Co-Ordinate system

88461 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

1 A straight line

2 An ellipse

3 A parabola

4 A hyperbola

Explanation:

(D) : Given equation of straight line-
$\frac{x}{a} + \frac{y}{b} = k$
$\frac{x}{a} - \frac{y}{b} = \frac{1}{k}$
Let the point of intersection be $(α, β)$
So, from equation (i) and (ii), we get-
$\frac{α}{a} + \frac{β}{b} = k$
and $\frac{α}{a} - \frac{β}{b} = \frac{1}{k}$
${(\frac{α}{a})}^{2} - {(\frac{β}{b})}^{2} = 1$
$\frac{α^{2}}{a^{2}} - \frac{β^{2}}{b^{2}} = 1$
$∴$ Locus $\frac{x^{2}}{a^{2}} - {\frac{y}{}}^{2}^{2} = 1$ ,
${\frac{x}{a^{2}}}^{2} - {\frac{y}{b^{2}}}^{2} = 1$
Which is the equation of a hyperbola.

WB JEE-2016

Co-Ordinate system

88462 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 RS is

1

x - y = \frac{a}{2}

2

x + y = a

3

x^{2} + y^{2} = 4 a^{2}

4

x^{2} - y^{2} = 2 a^{2}

Explanation:

(B) : Let the equation of line $AB$ is-
$\frac{x}{2 a} + \frac{y}{2 a} = 1$
$x + y = 2 a$

Let the coordinate of the midpoint of RS be (h, k)
$∴ R$ and $S$ are $(2 h, 0)$ and $(0, 2 k)$
So the mid point of Rs is $(h, k)$
$∵ P$ lies on line $A B$
The from equation(i), we have
$2 h + 2 k = 2 a$
$h + k = a$
So, the locus of $(h, k)$ is $x + y = a$ $x + y = a$

WB JEE-2016

Co-Ordinate system

88463 The angle between a pair of tangents drawn from a point $P$ to the circle
$x^{2} + y^{2} + 4 x - 6 y + 9 \sin^{2} α + 13 \cos^{2} α = 0$ is $2 α$ . The equation of the locus of the point $P$ is

1

x^{2} + y^{2} + 4 x + 6 y + 9 = 0

2

x^{2} + y^{2} - 4 x + 6 y + 9 = 0

3

x^{2} + y^{2} - 4 x - 6 y + 9 = 0

4

x^{2} + y^{2} + 4 x - 6 y + 9 = 0

Explanation:

(D) : Given equation of circle-
$x^{2} + y^{2} + 4 x - 6 y + 9 \sin^{2} α + 13 \cos^{2} α = 0$
Here, $C = (- 2, 3)$
$Radius = \sqrt{(- 2)^{2} + (3)^{2} - (9 \sin^{2} α + 13 \cos^{2} α)}$
$= \sqrt{4 + 9 - 9 \sin^{2} α - 13 \cos^{2} α}$
$= \sqrt{13 \sin^{2} α - 9 \sin^{2} α}$
$= \sqrt{4 \sin^{2} α} = 2 \sin α$

Here, $\sin α = \frac{AC}{PC}$
$PC \sin α = AC$
${PC}^{2} \sin^{2} α = {AC}^{2} = (2 \sin α)^{2}$
$[(h + 2)^{2} + (k - 3)^{2}] \sin^{2} α = 4 \sin^{2} α$
$(h + 2)^{2} + (k - 3)^{2} = 4$
$h^{2} + k^{2} + 4 h + k^{2} + 9 - 6 k = 4$
$h^{2} + k^{2} + 4 h - 6 k + 9 = 0$
So, locus of a point is $x^{2} + y^{2} + 4 x - 6 y + 9 = 0$

Co-Ordinate system

88471 Let $A$ be the fixed point $(0, 4)$ and $B$ be a moving point on $x$ -axis. Let $M$ be the midpoint of $A B$ and let the perpendicular bisector of $A B$ meets the $y$ -axis at $R$ , The locus of the midpoint $P$ of MR is

1

y + x^{2} = 2

2

x^{2} + (y - 2)^{2} = \frac{1}{4}

3

(y - 2)^{2} - x^{2} = \frac{1}{4}

4

x^{2} + y^{2} = 16

Explanation:

(A): Let $B$ is $(2 p, 0)$
Therefore co-ordinate of $M$ are $(b, 2)$
Now, slope of $A B = \frac{4 - 0}{0 - 2 p} = \frac{- 2}{p}$
Therefore,
Slope of MR $= \frac{p}{2}$
Therefore of MR $= y - 2 = \frac{p}{2} (x - p)^{0, 00}, p$
On putting $x = 0$ ,

$y - 2 = \frac{- p^{2}}{2}$
$y = 2 - \frac{p^{2}}{2}$
Therefore co-ordinate of $R$ is $(0, 2 - \frac{p^{2}}{2})$
Therefore of $p$ is $[\frac{p}{2}, \frac{2 - \frac{p^{2}}{2} + 2}{2}]$
i.e $[\frac{p}{2}, 2 - \frac{p^{2}}{4}]$
$∴ y = 2 - x^{2}$
$y + x^{2} = 2$

WB JEE-2021