88429
A straight rod of length 9 units slides with its ends \(A, B\) always on the \(X\) and \(Y\)-axis respectively. Then the locus of the centroid of \(\triangle O A B\) is :
1 \(x^{2}+y^{2}=3\)
2 \(x^{2}+y^{2}=9\)
3 \(x^{2}+y^{2}=1\)
4 \(\mathrm{x}^{2}+\mathrm{y}^{2}=81\)
Explanation:
(B) : Let the point be, \(\mathrm{A}(\mathrm{a}, 0)\) and \(\mathrm{B}(0, \mathrm{~b})\) then, \(a^{2}+b^{2}=81\) If centroid \((\mathrm{G})=\left(\frac{\mathrm{a}}{3}, \frac{\mathrm{b}}{3}\right)\) \(\mathrm{a}=3 \mathrm{x}\) and \(\mathrm{b}=3 \mathrm{y}\) Hence, \(9 x^{2}+9 y^{2}=81\) \(x^{2}+y^{2}=9\)
BITSAT-2020
Co-Ordinate system
88430
The locus of a point of intersection of two lines \(x \sqrt{3}-y=k \sqrt{3}\) and \(\sqrt{3} k x+k y=\sqrt{3}, k \in R\), describes
1 an ellipse
2 a hyperbola
3 a pair of lines
4 a parabola
Explanation:
(B) : Given, \(x \sqrt{3}-y= k \sqrt{3} \tag{1}\) \(\sqrt{3} k x+k y=\sqrt{3} \tag{2}\) We will equate value of \(\mathrm{k}\) from eq. (1) and (2) \(\frac{x \sqrt{3}-y}{\sqrt{3}}=\frac{\sqrt{3}}{x \sqrt{3}+y}\) \(\therefore \quad(x \sqrt{3}-y)(x \sqrt{3}+y)=(\sqrt{3})(\sqrt{3})\) \(\therefore \quad 3 x^{2}-y^{2}=3 \Rightarrow \frac{3 x^{2}}{3}-\frac{y^{2}}{3}=\frac{3}{3}\) i.e. \(\frac{x^{2}}{1}-\frac{y^{2}}{(\sqrt{3})^{2}}=1\), which is an equation of a hyperbola.
MHT CET-2020
Co-Ordinate system
88431
A variable plane remains at constant distance \(p\) from the origin. If it meets the coordinate axes at the points \(A, B, C\) then the locus of the Centroid of \(\triangle \mathrm{ABC}\) is
(A) : Let \(\mathrm{A} \equiv(\mathrm{a}, 0,0), \mathrm{B} \equiv(0, \mathrm{~b}, 0), \mathrm{C} \equiv(0,0, \mathrm{c})\), then equation of the plane is \(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\) Its distance from the origin, \(\frac{1}{\mathrm{a}^{2}}+\frac{1}{\mathrm{~b}^{2}}+\frac{1}{\mathrm{c}^{2}}=\frac{1}{\mathrm{p}^{2}}\) If \((x, y, z)\) be centroid of \(\triangle A B C\), then \(\mathrm{x}=\frac{\mathrm{a}}{3}, \mathrm{y}=\frac{\mathrm{b}}{3}, \mathrm{z}=\frac{\mathrm{c}}{3}\) Eliminating a,b,c from (i) and (ii) required Locus is \(\mathrm{x}^{-2}+\mathrm{y}^{-2}+\mathrm{z}^{-2}=9 \mathrm{p}^{-2}\)
VITEEE-2016
Co-Ordinate system
88432
The centres of a set of circles, each of radius 3, lie on the circle \(x^{2}+y^{2}=25\). The locus of any point in the set is
1 \(4 \leq x^{2}+y^{2} \leq 64\)
2 \(x^{2}+y^{2} \leq 25\)
3 \(\mathrm{x}^{2}+\mathrm{y}^{2} \geq 25\)
4 \(3 \leq x^{2}+y^{2} \geq 9\)
Explanation:
(A) : Let (x, y) be any point in the set. \(C\) lying on the circle \(x^{2}+y^{2}=25\) and radius 3 from the origin lies between \((\mathrm{OA})=5-3\) \(=2\) and \(\mathrm{OB}=5+3=8\) Hence \(4 \leq \mathrm{x}^{2}+\mathrm{y}^{2} \leq 64\)
VITEEE-2013
Co-Ordinate system
88433
The line joining \((5,0)\) to \((10 \cos \theta, 10 \sin \theta)\) is divided internally in the ratio \(2: 3\) at \(P\). If \(\theta\) varies, then the locus of \(P\) is
1 a straight line
2 a pair of straight lines
3 a circle
4 None of the above
Explanation:
(C) : Let coordinates of \(\mathrm{P}\) be \((\mathrm{h}, \mathrm{k})\), then \(\mathrm{h}=\frac{2(10 \cos \theta)+3(5)}{2+3}=4 \cos \theta+3\) \(\text { and } \mathrm{k}=\frac{2(10 \sin \theta)+3(0)}{2+3}=4 \sin \theta\) [Using the internal section formula] \(\frac{\mathrm{h}-3}{4}=\cos \theta \text { and } \frac{\mathrm{k}}{4}=\sin \theta\) Squaring and adding both of these equations, \(\frac{(\mathrm{h}-3)^{2}}{16}+\frac{\mathrm{k}^{2}}{16}=\cos ^{2} \theta+\sin ^{2} \theta\) \(\Rightarrow(\mathrm{h}-3)^{2}+\mathrm{k}^{2}=16\) Therefore, locus of point \(\mathrm{P}\) is \((x-3)^{2}+y^{2}=16 \text { which is a circle. }\)
88429
A straight rod of length 9 units slides with its ends \(A, B\) always on the \(X\) and \(Y\)-axis respectively. Then the locus of the centroid of \(\triangle O A B\) is :
1 \(x^{2}+y^{2}=3\)
2 \(x^{2}+y^{2}=9\)
3 \(x^{2}+y^{2}=1\)
4 \(\mathrm{x}^{2}+\mathrm{y}^{2}=81\)
Explanation:
(B) : Let the point be, \(\mathrm{A}(\mathrm{a}, 0)\) and \(\mathrm{B}(0, \mathrm{~b})\) then, \(a^{2}+b^{2}=81\) If centroid \((\mathrm{G})=\left(\frac{\mathrm{a}}{3}, \frac{\mathrm{b}}{3}\right)\) \(\mathrm{a}=3 \mathrm{x}\) and \(\mathrm{b}=3 \mathrm{y}\) Hence, \(9 x^{2}+9 y^{2}=81\) \(x^{2}+y^{2}=9\)
BITSAT-2020
Co-Ordinate system
88430
The locus of a point of intersection of two lines \(x \sqrt{3}-y=k \sqrt{3}\) and \(\sqrt{3} k x+k y=\sqrt{3}, k \in R\), describes
1 an ellipse
2 a hyperbola
3 a pair of lines
4 a parabola
Explanation:
(B) : Given, \(x \sqrt{3}-y= k \sqrt{3} \tag{1}\) \(\sqrt{3} k x+k y=\sqrt{3} \tag{2}\) We will equate value of \(\mathrm{k}\) from eq. (1) and (2) \(\frac{x \sqrt{3}-y}{\sqrt{3}}=\frac{\sqrt{3}}{x \sqrt{3}+y}\) \(\therefore \quad(x \sqrt{3}-y)(x \sqrt{3}+y)=(\sqrt{3})(\sqrt{3})\) \(\therefore \quad 3 x^{2}-y^{2}=3 \Rightarrow \frac{3 x^{2}}{3}-\frac{y^{2}}{3}=\frac{3}{3}\) i.e. \(\frac{x^{2}}{1}-\frac{y^{2}}{(\sqrt{3})^{2}}=1\), which is an equation of a hyperbola.
MHT CET-2020
Co-Ordinate system
88431
A variable plane remains at constant distance \(p\) from the origin. If it meets the coordinate axes at the points \(A, B, C\) then the locus of the Centroid of \(\triangle \mathrm{ABC}\) is
(A) : Let \(\mathrm{A} \equiv(\mathrm{a}, 0,0), \mathrm{B} \equiv(0, \mathrm{~b}, 0), \mathrm{C} \equiv(0,0, \mathrm{c})\), then equation of the plane is \(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\) Its distance from the origin, \(\frac{1}{\mathrm{a}^{2}}+\frac{1}{\mathrm{~b}^{2}}+\frac{1}{\mathrm{c}^{2}}=\frac{1}{\mathrm{p}^{2}}\) If \((x, y, z)\) be centroid of \(\triangle A B C\), then \(\mathrm{x}=\frac{\mathrm{a}}{3}, \mathrm{y}=\frac{\mathrm{b}}{3}, \mathrm{z}=\frac{\mathrm{c}}{3}\) Eliminating a,b,c from (i) and (ii) required Locus is \(\mathrm{x}^{-2}+\mathrm{y}^{-2}+\mathrm{z}^{-2}=9 \mathrm{p}^{-2}\)
VITEEE-2016
Co-Ordinate system
88432
The centres of a set of circles, each of radius 3, lie on the circle \(x^{2}+y^{2}=25\). The locus of any point in the set is
1 \(4 \leq x^{2}+y^{2} \leq 64\)
2 \(x^{2}+y^{2} \leq 25\)
3 \(\mathrm{x}^{2}+\mathrm{y}^{2} \geq 25\)
4 \(3 \leq x^{2}+y^{2} \geq 9\)
Explanation:
(A) : Let (x, y) be any point in the set. \(C\) lying on the circle \(x^{2}+y^{2}=25\) and radius 3 from the origin lies between \((\mathrm{OA})=5-3\) \(=2\) and \(\mathrm{OB}=5+3=8\) Hence \(4 \leq \mathrm{x}^{2}+\mathrm{y}^{2} \leq 64\)
VITEEE-2013
Co-Ordinate system
88433
The line joining \((5,0)\) to \((10 \cos \theta, 10 \sin \theta)\) is divided internally in the ratio \(2: 3\) at \(P\). If \(\theta\) varies, then the locus of \(P\) is
1 a straight line
2 a pair of straight lines
3 a circle
4 None of the above
Explanation:
(C) : Let coordinates of \(\mathrm{P}\) be \((\mathrm{h}, \mathrm{k})\), then \(\mathrm{h}=\frac{2(10 \cos \theta)+3(5)}{2+3}=4 \cos \theta+3\) \(\text { and } \mathrm{k}=\frac{2(10 \sin \theta)+3(0)}{2+3}=4 \sin \theta\) [Using the internal section formula] \(\frac{\mathrm{h}-3}{4}=\cos \theta \text { and } \frac{\mathrm{k}}{4}=\sin \theta\) Squaring and adding both of these equations, \(\frac{(\mathrm{h}-3)^{2}}{16}+\frac{\mathrm{k}^{2}}{16}=\cos ^{2} \theta+\sin ^{2} \theta\) \(\Rightarrow(\mathrm{h}-3)^{2}+\mathrm{k}^{2}=16\) Therefore, locus of point \(\mathrm{P}\) is \((x-3)^{2}+y^{2}=16 \text { which is a circle. }\)
88429
A straight rod of length 9 units slides with its ends \(A, B\) always on the \(X\) and \(Y\)-axis respectively. Then the locus of the centroid of \(\triangle O A B\) is :
1 \(x^{2}+y^{2}=3\)
2 \(x^{2}+y^{2}=9\)
3 \(x^{2}+y^{2}=1\)
4 \(\mathrm{x}^{2}+\mathrm{y}^{2}=81\)
Explanation:
(B) : Let the point be, \(\mathrm{A}(\mathrm{a}, 0)\) and \(\mathrm{B}(0, \mathrm{~b})\) then, \(a^{2}+b^{2}=81\) If centroid \((\mathrm{G})=\left(\frac{\mathrm{a}}{3}, \frac{\mathrm{b}}{3}\right)\) \(\mathrm{a}=3 \mathrm{x}\) and \(\mathrm{b}=3 \mathrm{y}\) Hence, \(9 x^{2}+9 y^{2}=81\) \(x^{2}+y^{2}=9\)
BITSAT-2020
Co-Ordinate system
88430
The locus of a point of intersection of two lines \(x \sqrt{3}-y=k \sqrt{3}\) and \(\sqrt{3} k x+k y=\sqrt{3}, k \in R\), describes
1 an ellipse
2 a hyperbola
3 a pair of lines
4 a parabola
Explanation:
(B) : Given, \(x \sqrt{3}-y= k \sqrt{3} \tag{1}\) \(\sqrt{3} k x+k y=\sqrt{3} \tag{2}\) We will equate value of \(\mathrm{k}\) from eq. (1) and (2) \(\frac{x \sqrt{3}-y}{\sqrt{3}}=\frac{\sqrt{3}}{x \sqrt{3}+y}\) \(\therefore \quad(x \sqrt{3}-y)(x \sqrt{3}+y)=(\sqrt{3})(\sqrt{3})\) \(\therefore \quad 3 x^{2}-y^{2}=3 \Rightarrow \frac{3 x^{2}}{3}-\frac{y^{2}}{3}=\frac{3}{3}\) i.e. \(\frac{x^{2}}{1}-\frac{y^{2}}{(\sqrt{3})^{2}}=1\), which is an equation of a hyperbola.
MHT CET-2020
Co-Ordinate system
88431
A variable plane remains at constant distance \(p\) from the origin. If it meets the coordinate axes at the points \(A, B, C\) then the locus of the Centroid of \(\triangle \mathrm{ABC}\) is
(A) : Let \(\mathrm{A} \equiv(\mathrm{a}, 0,0), \mathrm{B} \equiv(0, \mathrm{~b}, 0), \mathrm{C} \equiv(0,0, \mathrm{c})\), then equation of the plane is \(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\) Its distance from the origin, \(\frac{1}{\mathrm{a}^{2}}+\frac{1}{\mathrm{~b}^{2}}+\frac{1}{\mathrm{c}^{2}}=\frac{1}{\mathrm{p}^{2}}\) If \((x, y, z)\) be centroid of \(\triangle A B C\), then \(\mathrm{x}=\frac{\mathrm{a}}{3}, \mathrm{y}=\frac{\mathrm{b}}{3}, \mathrm{z}=\frac{\mathrm{c}}{3}\) Eliminating a,b,c from (i) and (ii) required Locus is \(\mathrm{x}^{-2}+\mathrm{y}^{-2}+\mathrm{z}^{-2}=9 \mathrm{p}^{-2}\)
VITEEE-2016
Co-Ordinate system
88432
The centres of a set of circles, each of radius 3, lie on the circle \(x^{2}+y^{2}=25\). The locus of any point in the set is
1 \(4 \leq x^{2}+y^{2} \leq 64\)
2 \(x^{2}+y^{2} \leq 25\)
3 \(\mathrm{x}^{2}+\mathrm{y}^{2} \geq 25\)
4 \(3 \leq x^{2}+y^{2} \geq 9\)
Explanation:
(A) : Let (x, y) be any point in the set. \(C\) lying on the circle \(x^{2}+y^{2}=25\) and radius 3 from the origin lies between \((\mathrm{OA})=5-3\) \(=2\) and \(\mathrm{OB}=5+3=8\) Hence \(4 \leq \mathrm{x}^{2}+\mathrm{y}^{2} \leq 64\)
VITEEE-2013
Co-Ordinate system
88433
The line joining \((5,0)\) to \((10 \cos \theta, 10 \sin \theta)\) is divided internally in the ratio \(2: 3\) at \(P\). If \(\theta\) varies, then the locus of \(P\) is
1 a straight line
2 a pair of straight lines
3 a circle
4 None of the above
Explanation:
(C) : Let coordinates of \(\mathrm{P}\) be \((\mathrm{h}, \mathrm{k})\), then \(\mathrm{h}=\frac{2(10 \cos \theta)+3(5)}{2+3}=4 \cos \theta+3\) \(\text { and } \mathrm{k}=\frac{2(10 \sin \theta)+3(0)}{2+3}=4 \sin \theta\) [Using the internal section formula] \(\frac{\mathrm{h}-3}{4}=\cos \theta \text { and } \frac{\mathrm{k}}{4}=\sin \theta\) Squaring and adding both of these equations, \(\frac{(\mathrm{h}-3)^{2}}{16}+\frac{\mathrm{k}^{2}}{16}=\cos ^{2} \theta+\sin ^{2} \theta\) \(\Rightarrow(\mathrm{h}-3)^{2}+\mathrm{k}^{2}=16\) Therefore, locus of point \(\mathrm{P}\) is \((x-3)^{2}+y^{2}=16 \text { which is a circle. }\)
88429
A straight rod of length 9 units slides with its ends \(A, B\) always on the \(X\) and \(Y\)-axis respectively. Then the locus of the centroid of \(\triangle O A B\) is :
1 \(x^{2}+y^{2}=3\)
2 \(x^{2}+y^{2}=9\)
3 \(x^{2}+y^{2}=1\)
4 \(\mathrm{x}^{2}+\mathrm{y}^{2}=81\)
Explanation:
(B) : Let the point be, \(\mathrm{A}(\mathrm{a}, 0)\) and \(\mathrm{B}(0, \mathrm{~b})\) then, \(a^{2}+b^{2}=81\) If centroid \((\mathrm{G})=\left(\frac{\mathrm{a}}{3}, \frac{\mathrm{b}}{3}\right)\) \(\mathrm{a}=3 \mathrm{x}\) and \(\mathrm{b}=3 \mathrm{y}\) Hence, \(9 x^{2}+9 y^{2}=81\) \(x^{2}+y^{2}=9\)
BITSAT-2020
Co-Ordinate system
88430
The locus of a point of intersection of two lines \(x \sqrt{3}-y=k \sqrt{3}\) and \(\sqrt{3} k x+k y=\sqrt{3}, k \in R\), describes
1 an ellipse
2 a hyperbola
3 a pair of lines
4 a parabola
Explanation:
(B) : Given, \(x \sqrt{3}-y= k \sqrt{3} \tag{1}\) \(\sqrt{3} k x+k y=\sqrt{3} \tag{2}\) We will equate value of \(\mathrm{k}\) from eq. (1) and (2) \(\frac{x \sqrt{3}-y}{\sqrt{3}}=\frac{\sqrt{3}}{x \sqrt{3}+y}\) \(\therefore \quad(x \sqrt{3}-y)(x \sqrt{3}+y)=(\sqrt{3})(\sqrt{3})\) \(\therefore \quad 3 x^{2}-y^{2}=3 \Rightarrow \frac{3 x^{2}}{3}-\frac{y^{2}}{3}=\frac{3}{3}\) i.e. \(\frac{x^{2}}{1}-\frac{y^{2}}{(\sqrt{3})^{2}}=1\), which is an equation of a hyperbola.
MHT CET-2020
Co-Ordinate system
88431
A variable plane remains at constant distance \(p\) from the origin. If it meets the coordinate axes at the points \(A, B, C\) then the locus of the Centroid of \(\triangle \mathrm{ABC}\) is
(A) : Let \(\mathrm{A} \equiv(\mathrm{a}, 0,0), \mathrm{B} \equiv(0, \mathrm{~b}, 0), \mathrm{C} \equiv(0,0, \mathrm{c})\), then equation of the plane is \(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\) Its distance from the origin, \(\frac{1}{\mathrm{a}^{2}}+\frac{1}{\mathrm{~b}^{2}}+\frac{1}{\mathrm{c}^{2}}=\frac{1}{\mathrm{p}^{2}}\) If \((x, y, z)\) be centroid of \(\triangle A B C\), then \(\mathrm{x}=\frac{\mathrm{a}}{3}, \mathrm{y}=\frac{\mathrm{b}}{3}, \mathrm{z}=\frac{\mathrm{c}}{3}\) Eliminating a,b,c from (i) and (ii) required Locus is \(\mathrm{x}^{-2}+\mathrm{y}^{-2}+\mathrm{z}^{-2}=9 \mathrm{p}^{-2}\)
VITEEE-2016
Co-Ordinate system
88432
The centres of a set of circles, each of radius 3, lie on the circle \(x^{2}+y^{2}=25\). The locus of any point in the set is
1 \(4 \leq x^{2}+y^{2} \leq 64\)
2 \(x^{2}+y^{2} \leq 25\)
3 \(\mathrm{x}^{2}+\mathrm{y}^{2} \geq 25\)
4 \(3 \leq x^{2}+y^{2} \geq 9\)
Explanation:
(A) : Let (x, y) be any point in the set. \(C\) lying on the circle \(x^{2}+y^{2}=25\) and radius 3 from the origin lies between \((\mathrm{OA})=5-3\) \(=2\) and \(\mathrm{OB}=5+3=8\) Hence \(4 \leq \mathrm{x}^{2}+\mathrm{y}^{2} \leq 64\)
VITEEE-2013
Co-Ordinate system
88433
The line joining \((5,0)\) to \((10 \cos \theta, 10 \sin \theta)\) is divided internally in the ratio \(2: 3\) at \(P\). If \(\theta\) varies, then the locus of \(P\) is
1 a straight line
2 a pair of straight lines
3 a circle
4 None of the above
Explanation:
(C) : Let coordinates of \(\mathrm{P}\) be \((\mathrm{h}, \mathrm{k})\), then \(\mathrm{h}=\frac{2(10 \cos \theta)+3(5)}{2+3}=4 \cos \theta+3\) \(\text { and } \mathrm{k}=\frac{2(10 \sin \theta)+3(0)}{2+3}=4 \sin \theta\) [Using the internal section formula] \(\frac{\mathrm{h}-3}{4}=\cos \theta \text { and } \frac{\mathrm{k}}{4}=\sin \theta\) Squaring and adding both of these equations, \(\frac{(\mathrm{h}-3)^{2}}{16}+\frac{\mathrm{k}^{2}}{16}=\cos ^{2} \theta+\sin ^{2} \theta\) \(\Rightarrow(\mathrm{h}-3)^{2}+\mathrm{k}^{2}=16\) Therefore, locus of point \(\mathrm{P}\) is \((x-3)^{2}+y^{2}=16 \text { which is a circle. }\)
88429
A straight rod of length 9 units slides with its ends \(A, B\) always on the \(X\) and \(Y\)-axis respectively. Then the locus of the centroid of \(\triangle O A B\) is :
1 \(x^{2}+y^{2}=3\)
2 \(x^{2}+y^{2}=9\)
3 \(x^{2}+y^{2}=1\)
4 \(\mathrm{x}^{2}+\mathrm{y}^{2}=81\)
Explanation:
(B) : Let the point be, \(\mathrm{A}(\mathrm{a}, 0)\) and \(\mathrm{B}(0, \mathrm{~b})\) then, \(a^{2}+b^{2}=81\) If centroid \((\mathrm{G})=\left(\frac{\mathrm{a}}{3}, \frac{\mathrm{b}}{3}\right)\) \(\mathrm{a}=3 \mathrm{x}\) and \(\mathrm{b}=3 \mathrm{y}\) Hence, \(9 x^{2}+9 y^{2}=81\) \(x^{2}+y^{2}=9\)
BITSAT-2020
Co-Ordinate system
88430
The locus of a point of intersection of two lines \(x \sqrt{3}-y=k \sqrt{3}\) and \(\sqrt{3} k x+k y=\sqrt{3}, k \in R\), describes
1 an ellipse
2 a hyperbola
3 a pair of lines
4 a parabola
Explanation:
(B) : Given, \(x \sqrt{3}-y= k \sqrt{3} \tag{1}\) \(\sqrt{3} k x+k y=\sqrt{3} \tag{2}\) We will equate value of \(\mathrm{k}\) from eq. (1) and (2) \(\frac{x \sqrt{3}-y}{\sqrt{3}}=\frac{\sqrt{3}}{x \sqrt{3}+y}\) \(\therefore \quad(x \sqrt{3}-y)(x \sqrt{3}+y)=(\sqrt{3})(\sqrt{3})\) \(\therefore \quad 3 x^{2}-y^{2}=3 \Rightarrow \frac{3 x^{2}}{3}-\frac{y^{2}}{3}=\frac{3}{3}\) i.e. \(\frac{x^{2}}{1}-\frac{y^{2}}{(\sqrt{3})^{2}}=1\), which is an equation of a hyperbola.
MHT CET-2020
Co-Ordinate system
88431
A variable plane remains at constant distance \(p\) from the origin. If it meets the coordinate axes at the points \(A, B, C\) then the locus of the Centroid of \(\triangle \mathrm{ABC}\) is
(A) : Let \(\mathrm{A} \equiv(\mathrm{a}, 0,0), \mathrm{B} \equiv(0, \mathrm{~b}, 0), \mathrm{C} \equiv(0,0, \mathrm{c})\), then equation of the plane is \(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\) Its distance from the origin, \(\frac{1}{\mathrm{a}^{2}}+\frac{1}{\mathrm{~b}^{2}}+\frac{1}{\mathrm{c}^{2}}=\frac{1}{\mathrm{p}^{2}}\) If \((x, y, z)\) be centroid of \(\triangle A B C\), then \(\mathrm{x}=\frac{\mathrm{a}}{3}, \mathrm{y}=\frac{\mathrm{b}}{3}, \mathrm{z}=\frac{\mathrm{c}}{3}\) Eliminating a,b,c from (i) and (ii) required Locus is \(\mathrm{x}^{-2}+\mathrm{y}^{-2}+\mathrm{z}^{-2}=9 \mathrm{p}^{-2}\)
VITEEE-2016
Co-Ordinate system
88432
The centres of a set of circles, each of radius 3, lie on the circle \(x^{2}+y^{2}=25\). The locus of any point in the set is
1 \(4 \leq x^{2}+y^{2} \leq 64\)
2 \(x^{2}+y^{2} \leq 25\)
3 \(\mathrm{x}^{2}+\mathrm{y}^{2} \geq 25\)
4 \(3 \leq x^{2}+y^{2} \geq 9\)
Explanation:
(A) : Let (x, y) be any point in the set. \(C\) lying on the circle \(x^{2}+y^{2}=25\) and radius 3 from the origin lies between \((\mathrm{OA})=5-3\) \(=2\) and \(\mathrm{OB}=5+3=8\) Hence \(4 \leq \mathrm{x}^{2}+\mathrm{y}^{2} \leq 64\)
VITEEE-2013
Co-Ordinate system
88433
The line joining \((5,0)\) to \((10 \cos \theta, 10 \sin \theta)\) is divided internally in the ratio \(2: 3\) at \(P\). If \(\theta\) varies, then the locus of \(P\) is
1 a straight line
2 a pair of straight lines
3 a circle
4 None of the above
Explanation:
(C) : Let coordinates of \(\mathrm{P}\) be \((\mathrm{h}, \mathrm{k})\), then \(\mathrm{h}=\frac{2(10 \cos \theta)+3(5)}{2+3}=4 \cos \theta+3\) \(\text { and } \mathrm{k}=\frac{2(10 \sin \theta)+3(0)}{2+3}=4 \sin \theta\) [Using the internal section formula] \(\frac{\mathrm{h}-3}{4}=\cos \theta \text { and } \frac{\mathrm{k}}{4}=\sin \theta\) Squaring and adding both of these equations, \(\frac{(\mathrm{h}-3)^{2}}{16}+\frac{\mathrm{k}^{2}}{16}=\cos ^{2} \theta+\sin ^{2} \theta\) \(\Rightarrow(\mathrm{h}-3)^{2}+\mathrm{k}^{2}=16\) Therefore, locus of point \(\mathrm{P}\) is \((x-3)^{2}+y^{2}=16 \text { which is a circle. }\)
