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, \(\text { i.e, }\left|\mathrm{PS}-\mathrm{PS}^{\prime}\right|=2 \mathrm{a}\) 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- \(\frac{\mathrm{x}}{\mathrm{a}}+\frac{\mathrm{y}}{\mathrm{b}}=\mathrm{k} \tag{i}\) \(\frac{\mathrm{x}}{\mathrm{a}}-\frac{\mathrm{y}}{\mathrm{b}}=\frac{1}{\mathrm{k}} \tag{ii}\) Let the point of intersection be \((\alpha, \beta)\) So, from equation (i) and (ii), we get- \(\frac{\alpha}{\mathrm{a}}+\frac{\beta}{\mathrm{b}}=\mathrm{k}\) and \(\quad \frac{\alpha}{\mathrm{a}}-\frac{\beta}{\mathrm{b}}=\frac{1}{\mathrm{k}}\) \(\left(\frac{\alpha}{a}\right)^{2}-\left(\frac{\beta}{b}\right)^{2}=1\) \(\frac{\alpha^{2}}{\mathrm{a}^{2}}-\frac{\beta^{2}}{\mathrm{~b}^{2}}=1\) \(\therefore \quad\) Locus \(\frac{\mathrm{x}^{2}}{\mathrm{a}^{2}}-\frac{\mathrm{y}}{}^{2}{ }^{2}=1\), \({\frac{x}{a^{2}}}^{2}-{\frac{y}{b^{2}}}^{2}=1\) 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
(B) : Let the equation of line \(\mathrm{AB}\) is- \(\frac{x}{2 a}+\frac{y}{2 a}=1\) \(x+y=2 a \tag{i}\) Let the coordinate of the midpoint of RS be (h, k) \(\therefore \quad \mathrm{R}\) and \(\mathrm{S}\) are \((2 \mathrm{~h}, 0)\) and \((0,2 \mathrm{k})\) So the mid point of Rs is \((\mathrm{h}, \mathrm{k})\) \(\because \quad P\) lies on line \(A B\) The from equation(i), we have \(2 \mathrm{~h}+2 \mathrm{k}=2 \mathrm{a}\) \(\mathrm{h}+\mathrm{k}=\mathrm{a}\) So, the locus of \((h, k)\) is \(x+y=a\) \(\mathrm{x}+\mathrm{y}=\mathrm{a}\)
WB JEE-2016
Co-Ordinate system
88463
The angle between a pair of tangents drawn from a point \(P\) to the circle \(x^{2}+y^{2}+4 x-6 y+9 \sin ^{2} \alpha+13 \cos ^{2} \alpha=0\) is \(2 \alpha\). The equation of the locus of the point \(P\) is
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 \(\mathrm{x}^{2}+\mathrm{y}^{2}=16\)
Explanation:
(A): Let \(B\) is \((2 \mathrm{p}, 0)\) Therefore co-ordinate of \(\mathrm{M}\) are \((\mathrm{b}, 2)\) Now, slope of \(A B=\frac{4-0}{0-2 p}=\frac{-2}{p}\) Therefore, Slope of MR \(=\frac{\mathrm{p}}{2}\) 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 \(\mathrm{R}\) is \(\left(0,2-\frac{\mathrm{p}^{2}}{2}\right)\) Therefore of \(p\) is \(\left[\frac{p}{2}, \frac{2-\frac{p^{2}}{2}+2}{2}\right]\) i.e \(\quad\left[\frac{\mathrm{p}}{2}, 2-\frac{\mathrm{p}^{2}}{4}\right]\) \(\therefore \quad \mathrm{y}=2-\mathrm{x}^{2}\) \(y+x^{2}=2\)
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, \(\text { i.e, }\left|\mathrm{PS}-\mathrm{PS}^{\prime}\right|=2 \mathrm{a}\) 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- \(\frac{\mathrm{x}}{\mathrm{a}}+\frac{\mathrm{y}}{\mathrm{b}}=\mathrm{k} \tag{i}\) \(\frac{\mathrm{x}}{\mathrm{a}}-\frac{\mathrm{y}}{\mathrm{b}}=\frac{1}{\mathrm{k}} \tag{ii}\) Let the point of intersection be \((\alpha, \beta)\) So, from equation (i) and (ii), we get- \(\frac{\alpha}{\mathrm{a}}+\frac{\beta}{\mathrm{b}}=\mathrm{k}\) and \(\quad \frac{\alpha}{\mathrm{a}}-\frac{\beta}{\mathrm{b}}=\frac{1}{\mathrm{k}}\) \(\left(\frac{\alpha}{a}\right)^{2}-\left(\frac{\beta}{b}\right)^{2}=1\) \(\frac{\alpha^{2}}{\mathrm{a}^{2}}-\frac{\beta^{2}}{\mathrm{~b}^{2}}=1\) \(\therefore \quad\) Locus \(\frac{\mathrm{x}^{2}}{\mathrm{a}^{2}}-\frac{\mathrm{y}}{}^{2}{ }^{2}=1\), \({\frac{x}{a^{2}}}^{2}-{\frac{y}{b^{2}}}^{2}=1\) 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
(B) : Let the equation of line \(\mathrm{AB}\) is- \(\frac{x}{2 a}+\frac{y}{2 a}=1\) \(x+y=2 a \tag{i}\) Let the coordinate of the midpoint of RS be (h, k) \(\therefore \quad \mathrm{R}\) and \(\mathrm{S}\) are \((2 \mathrm{~h}, 0)\) and \((0,2 \mathrm{k})\) So the mid point of Rs is \((\mathrm{h}, \mathrm{k})\) \(\because \quad P\) lies on line \(A B\) The from equation(i), we have \(2 \mathrm{~h}+2 \mathrm{k}=2 \mathrm{a}\) \(\mathrm{h}+\mathrm{k}=\mathrm{a}\) So, the locus of \((h, k)\) is \(x+y=a\) \(\mathrm{x}+\mathrm{y}=\mathrm{a}\)
WB JEE-2016
Co-Ordinate system
88463
The angle between a pair of tangents drawn from a point \(P\) to the circle \(x^{2}+y^{2}+4 x-6 y+9 \sin ^{2} \alpha+13 \cos ^{2} \alpha=0\) is \(2 \alpha\). The equation of the locus of the point \(P\) is
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 \(\mathrm{x}^{2}+\mathrm{y}^{2}=16\)
Explanation:
(A): Let \(B\) is \((2 \mathrm{p}, 0)\) Therefore co-ordinate of \(\mathrm{M}\) are \((\mathrm{b}, 2)\) Now, slope of \(A B=\frac{4-0}{0-2 p}=\frac{-2}{p}\) Therefore, Slope of MR \(=\frac{\mathrm{p}}{2}\) 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 \(\mathrm{R}\) is \(\left(0,2-\frac{\mathrm{p}^{2}}{2}\right)\) Therefore of \(p\) is \(\left[\frac{p}{2}, \frac{2-\frac{p^{2}}{2}+2}{2}\right]\) i.e \(\quad\left[\frac{\mathrm{p}}{2}, 2-\frac{\mathrm{p}^{2}}{4}\right]\) \(\therefore \quad \mathrm{y}=2-\mathrm{x}^{2}\) \(y+x^{2}=2\)
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, \(\text { i.e, }\left|\mathrm{PS}-\mathrm{PS}^{\prime}\right|=2 \mathrm{a}\) 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- \(\frac{\mathrm{x}}{\mathrm{a}}+\frac{\mathrm{y}}{\mathrm{b}}=\mathrm{k} \tag{i}\) \(\frac{\mathrm{x}}{\mathrm{a}}-\frac{\mathrm{y}}{\mathrm{b}}=\frac{1}{\mathrm{k}} \tag{ii}\) Let the point of intersection be \((\alpha, \beta)\) So, from equation (i) and (ii), we get- \(\frac{\alpha}{\mathrm{a}}+\frac{\beta}{\mathrm{b}}=\mathrm{k}\) and \(\quad \frac{\alpha}{\mathrm{a}}-\frac{\beta}{\mathrm{b}}=\frac{1}{\mathrm{k}}\) \(\left(\frac{\alpha}{a}\right)^{2}-\left(\frac{\beta}{b}\right)^{2}=1\) \(\frac{\alpha^{2}}{\mathrm{a}^{2}}-\frac{\beta^{2}}{\mathrm{~b}^{2}}=1\) \(\therefore \quad\) Locus \(\frac{\mathrm{x}^{2}}{\mathrm{a}^{2}}-\frac{\mathrm{y}}{}^{2}{ }^{2}=1\), \({\frac{x}{a^{2}}}^{2}-{\frac{y}{b^{2}}}^{2}=1\) 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
(B) : Let the equation of line \(\mathrm{AB}\) is- \(\frac{x}{2 a}+\frac{y}{2 a}=1\) \(x+y=2 a \tag{i}\) Let the coordinate of the midpoint of RS be (h, k) \(\therefore \quad \mathrm{R}\) and \(\mathrm{S}\) are \((2 \mathrm{~h}, 0)\) and \((0,2 \mathrm{k})\) So the mid point of Rs is \((\mathrm{h}, \mathrm{k})\) \(\because \quad P\) lies on line \(A B\) The from equation(i), we have \(2 \mathrm{~h}+2 \mathrm{k}=2 \mathrm{a}\) \(\mathrm{h}+\mathrm{k}=\mathrm{a}\) So, the locus of \((h, k)\) is \(x+y=a\) \(\mathrm{x}+\mathrm{y}=\mathrm{a}\)
WB JEE-2016
Co-Ordinate system
88463
The angle between a pair of tangents drawn from a point \(P\) to the circle \(x^{2}+y^{2}+4 x-6 y+9 \sin ^{2} \alpha+13 \cos ^{2} \alpha=0\) is \(2 \alpha\). The equation of the locus of the point \(P\) is
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 \(\mathrm{x}^{2}+\mathrm{y}^{2}=16\)
Explanation:
(A): Let \(B\) is \((2 \mathrm{p}, 0)\) Therefore co-ordinate of \(\mathrm{M}\) are \((\mathrm{b}, 2)\) Now, slope of \(A B=\frac{4-0}{0-2 p}=\frac{-2}{p}\) Therefore, Slope of MR \(=\frac{\mathrm{p}}{2}\) 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 \(\mathrm{R}\) is \(\left(0,2-\frac{\mathrm{p}^{2}}{2}\right)\) Therefore of \(p\) is \(\left[\frac{p}{2}, \frac{2-\frac{p^{2}}{2}+2}{2}\right]\) i.e \(\quad\left[\frac{\mathrm{p}}{2}, 2-\frac{\mathrm{p}^{2}}{4}\right]\) \(\therefore \quad \mathrm{y}=2-\mathrm{x}^{2}\) \(y+x^{2}=2\)
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, \(\text { i.e, }\left|\mathrm{PS}-\mathrm{PS}^{\prime}\right|=2 \mathrm{a}\) 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- \(\frac{\mathrm{x}}{\mathrm{a}}+\frac{\mathrm{y}}{\mathrm{b}}=\mathrm{k} \tag{i}\) \(\frac{\mathrm{x}}{\mathrm{a}}-\frac{\mathrm{y}}{\mathrm{b}}=\frac{1}{\mathrm{k}} \tag{ii}\) Let the point of intersection be \((\alpha, \beta)\) So, from equation (i) and (ii), we get- \(\frac{\alpha}{\mathrm{a}}+\frac{\beta}{\mathrm{b}}=\mathrm{k}\) and \(\quad \frac{\alpha}{\mathrm{a}}-\frac{\beta}{\mathrm{b}}=\frac{1}{\mathrm{k}}\) \(\left(\frac{\alpha}{a}\right)^{2}-\left(\frac{\beta}{b}\right)^{2}=1\) \(\frac{\alpha^{2}}{\mathrm{a}^{2}}-\frac{\beta^{2}}{\mathrm{~b}^{2}}=1\) \(\therefore \quad\) Locus \(\frac{\mathrm{x}^{2}}{\mathrm{a}^{2}}-\frac{\mathrm{y}}{}^{2}{ }^{2}=1\), \({\frac{x}{a^{2}}}^{2}-{\frac{y}{b^{2}}}^{2}=1\) 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
(B) : Let the equation of line \(\mathrm{AB}\) is- \(\frac{x}{2 a}+\frac{y}{2 a}=1\) \(x+y=2 a \tag{i}\) Let the coordinate of the midpoint of RS be (h, k) \(\therefore \quad \mathrm{R}\) and \(\mathrm{S}\) are \((2 \mathrm{~h}, 0)\) and \((0,2 \mathrm{k})\) So the mid point of Rs is \((\mathrm{h}, \mathrm{k})\) \(\because \quad P\) lies on line \(A B\) The from equation(i), we have \(2 \mathrm{~h}+2 \mathrm{k}=2 \mathrm{a}\) \(\mathrm{h}+\mathrm{k}=\mathrm{a}\) So, the locus of \((h, k)\) is \(x+y=a\) \(\mathrm{x}+\mathrm{y}=\mathrm{a}\)
WB JEE-2016
Co-Ordinate system
88463
The angle between a pair of tangents drawn from a point \(P\) to the circle \(x^{2}+y^{2}+4 x-6 y+9 \sin ^{2} \alpha+13 \cos ^{2} \alpha=0\) is \(2 \alpha\). The equation of the locus of the point \(P\) is
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 \(\mathrm{x}^{2}+\mathrm{y}^{2}=16\)
Explanation:
(A): Let \(B\) is \((2 \mathrm{p}, 0)\) Therefore co-ordinate of \(\mathrm{M}\) are \((\mathrm{b}, 2)\) Now, slope of \(A B=\frac{4-0}{0-2 p}=\frac{-2}{p}\) Therefore, Slope of MR \(=\frac{\mathrm{p}}{2}\) 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 \(\mathrm{R}\) is \(\left(0,2-\frac{\mathrm{p}^{2}}{2}\right)\) Therefore of \(p\) is \(\left[\frac{p}{2}, \frac{2-\frac{p^{2}}{2}+2}{2}\right]\) i.e \(\quad\left[\frac{\mathrm{p}}{2}, 2-\frac{\mathrm{p}^{2}}{4}\right]\) \(\therefore \quad \mathrm{y}=2-\mathrm{x}^{2}\) \(y+x^{2}=2\)
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, \(\text { i.e, }\left|\mathrm{PS}-\mathrm{PS}^{\prime}\right|=2 \mathrm{a}\) 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- \(\frac{\mathrm{x}}{\mathrm{a}}+\frac{\mathrm{y}}{\mathrm{b}}=\mathrm{k} \tag{i}\) \(\frac{\mathrm{x}}{\mathrm{a}}-\frac{\mathrm{y}}{\mathrm{b}}=\frac{1}{\mathrm{k}} \tag{ii}\) Let the point of intersection be \((\alpha, \beta)\) So, from equation (i) and (ii), we get- \(\frac{\alpha}{\mathrm{a}}+\frac{\beta}{\mathrm{b}}=\mathrm{k}\) and \(\quad \frac{\alpha}{\mathrm{a}}-\frac{\beta}{\mathrm{b}}=\frac{1}{\mathrm{k}}\) \(\left(\frac{\alpha}{a}\right)^{2}-\left(\frac{\beta}{b}\right)^{2}=1\) \(\frac{\alpha^{2}}{\mathrm{a}^{2}}-\frac{\beta^{2}}{\mathrm{~b}^{2}}=1\) \(\therefore \quad\) Locus \(\frac{\mathrm{x}^{2}}{\mathrm{a}^{2}}-\frac{\mathrm{y}}{}^{2}{ }^{2}=1\), \({\frac{x}{a^{2}}}^{2}-{\frac{y}{b^{2}}}^{2}=1\) 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
(B) : Let the equation of line \(\mathrm{AB}\) is- \(\frac{x}{2 a}+\frac{y}{2 a}=1\) \(x+y=2 a \tag{i}\) Let the coordinate of the midpoint of RS be (h, k) \(\therefore \quad \mathrm{R}\) and \(\mathrm{S}\) are \((2 \mathrm{~h}, 0)\) and \((0,2 \mathrm{k})\) So the mid point of Rs is \((\mathrm{h}, \mathrm{k})\) \(\because \quad P\) lies on line \(A B\) The from equation(i), we have \(2 \mathrm{~h}+2 \mathrm{k}=2 \mathrm{a}\) \(\mathrm{h}+\mathrm{k}=\mathrm{a}\) So, the locus of \((h, k)\) is \(x+y=a\) \(\mathrm{x}+\mathrm{y}=\mathrm{a}\)
WB JEE-2016
Co-Ordinate system
88463
The angle between a pair of tangents drawn from a point \(P\) to the circle \(x^{2}+y^{2}+4 x-6 y+9 \sin ^{2} \alpha+13 \cos ^{2} \alpha=0\) is \(2 \alpha\). The equation of the locus of the point \(P\) is
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 \(\mathrm{x}^{2}+\mathrm{y}^{2}=16\)
Explanation:
(A): Let \(B\) is \((2 \mathrm{p}, 0)\) Therefore co-ordinate of \(\mathrm{M}\) are \((\mathrm{b}, 2)\) Now, slope of \(A B=\frac{4-0}{0-2 p}=\frac{-2}{p}\) Therefore, Slope of MR \(=\frac{\mathrm{p}}{2}\) 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 \(\mathrm{R}\) is \(\left(0,2-\frac{\mathrm{p}^{2}}{2}\right)\) Therefore of \(p\) is \(\left[\frac{p}{2}, \frac{2-\frac{p^{2}}{2}+2}{2}\right]\) i.e \(\quad\left[\frac{\mathrm{p}}{2}, 2-\frac{\mathrm{p}^{2}}{4}\right]\) \(\therefore \quad \mathrm{y}=2-\mathrm{x}^{2}\) \(y+x^{2}=2\)
