88473
The side \(A B\) of \(\triangle A B C\) is fixed and is of length 2 a unit. The vertex moves in the plane such that the vertical angle is always constant and is \(\alpha\). Let \(x\)-axis be along \(A B\) and the origin be at \(A\). Then the locus of the vertex is
1 \(x^{2}+y^{2}+2 a x \sin \alpha+a^{2} \cos \alpha=0\)
2 \(x^{2}+y^{2}-2 a x-2 a y \cot \alpha=0\)
3 \(x^{2}+y^{2}-2 a x \cos \alpha-a^{2}=0\)
4 \(x^{2}+y^{2}-a x \sin \alpha-a y \cos \alpha=0\)
Explanation:
(B) : Let \(\mathrm{AC}=\mathrm{BC}=\mathrm{r}\) Now, from \(\triangle \mathrm{ACE}\) \(\mathrm{r} \sin \alpha=\mathrm{a}\) Now again \(\triangle \mathrm{ACE}\) CE \(r \cos \alpha(a \operatorname{cosec} \alpha) \cos \alpha=a \cot \alpha\) Now locus of the vertex will be a circle with radius \(\mathrm{r}=\) a \(\operatorname{cosec} \alpha\) and center \(\mathrm{c}(\mathrm{a}, \mathrm{a} \cot \alpha)\) So required locus is \(\left(x^{2}-a\right)^{2}+(y-a \cot \alpha)_{2}=a^{2} \operatorname{cosec}^{2} \alpha\) \(x^{2}+a^{2}-2 a x+y^{2}+a^{2} \cot ^{2} \alpha-2 a y \cot \alpha=a^{2} \operatorname{cosec}^{2} \alpha\) \(\mathrm{x}^{2}+\mathrm{y}^{2}-2 \mathrm{ax}-2 \mathrm{ay} \cot \alpha=0\) So, the locus of the vertex
WB JEE-2022
Co-Ordinate system
88474
If the sum of the distances of a point from two perpendicular lines in a plane is 1 unit, then its locus is
1 a square
2 a circle
3 a straight line
4 two intersecting lines
Explanation:
(C) : Let, the locus of a point in a plane be \(\mathrm{P}(\mathrm{h}, \mathrm{k})\). According to the question, \(|\mathrm{PA}|+|\mathrm{PB}|=1\) \(|\mathrm{X}|+|\mathrm{k}|=1\) So, locus of a point is \(|x|+|y|=1\) Which represent the straight line.
WB JEE-2022
Co-Ordinate system
88475
Locus of the centroid of a triangle whose vertices are \((1,0),(a \cos t, a \sin t)\), \((b \sin t,-b \cos t)\) is \(9 x^{2}+9 y^{2}-6 x=k\). Then, the value of \(k=\)
1 \(a^{2}+b^{2}\)
2 \(\mathrm{a}^{2}+\mathrm{b}^{2}-1\)
3 \(a^{2}+b^{2}+1\)
4 0
Explanation:
(B) : Given that - locus of the centroid of a triangle whose vertices are \((1,0)\) ( a cost,\(a \sin t), b \sin t,-b\) \(\cos \mathrm{t})\) is \(9 \mathrm{x}^{2}+9 \mathrm{y}^{2}-6 \mathrm{x}=\mathrm{k}\) \(x=\frac{x_{1}+x_{2}+x_{3}}{3}\) \(y=\frac{y_{1}+y_{2}+y_{3}}{3}\) So, \(x=\frac{1+a \cos +b \sin t}{3}\) \(y=\frac{0+a \sin t+(-b \cos t)}{3} \tag{ii}\) \(\Rightarrow \quad (3 x-1)=a \cos t+b \sin t \tag{i}\) And \(\quad 3 y=a \sin t-b \cos t\) Squaring and adding equation (i) and (ii) \(9 x^{2}-6 x+1+9 y^{2}=a^{2}+b^{2}\) \(9 x^{2}+9 y^{2}-6 x=a^{2}+b^{2}-1\) \(\begin{array}{ll}9 x+9 y -6 x=a^{2}+b^{2}-1\end{array}\)
AP EAMCET-2020-17.09.2020
Co-Ordinate system
88476
If it is mentioned such that for a curve passing through \((3,4)\), the slope of the curve at any point is the reciprocal of twice the ordinate of that point, then that curve is a ......
1 Ellipse
2 Parabola
3 Hyperbola
4 Circle
Explanation:
(B) : Given, the slope of the curve at any pointy is the reciprocal of twice of ordinate of that point. slope \(=\frac{1}{2 \mathrm{y}}\) \(\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{2 \mathrm{y}} \Rightarrow 2 \mathrm{y}\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)=1\) \(\mathrm{y}^{2}=\mathrm{x}+\mathrm{c} \quad[\mathrm{x}=3, \mathrm{y}=4]\) given that \(16=3+\mathrm{c}\) \(\mathrm{c}=13\) So, \(\mathrm{y}^{2}=\mathrm{x}+\mathrm{c}\) \((4)^{2}=(3)+13\) Which is form of \(\mathrm{y}^{2}=4 \mathrm{ax}\) So that carve is a parabola.
88473
The side \(A B\) of \(\triangle A B C\) is fixed and is of length 2 a unit. The vertex moves in the plane such that the vertical angle is always constant and is \(\alpha\). Let \(x\)-axis be along \(A B\) and the origin be at \(A\). Then the locus of the vertex is
1 \(x^{2}+y^{2}+2 a x \sin \alpha+a^{2} \cos \alpha=0\)
2 \(x^{2}+y^{2}-2 a x-2 a y \cot \alpha=0\)
3 \(x^{2}+y^{2}-2 a x \cos \alpha-a^{2}=0\)
4 \(x^{2}+y^{2}-a x \sin \alpha-a y \cos \alpha=0\)
Explanation:
(B) : Let \(\mathrm{AC}=\mathrm{BC}=\mathrm{r}\) Now, from \(\triangle \mathrm{ACE}\) \(\mathrm{r} \sin \alpha=\mathrm{a}\) Now again \(\triangle \mathrm{ACE}\) CE \(r \cos \alpha(a \operatorname{cosec} \alpha) \cos \alpha=a \cot \alpha\) Now locus of the vertex will be a circle with radius \(\mathrm{r}=\) a \(\operatorname{cosec} \alpha\) and center \(\mathrm{c}(\mathrm{a}, \mathrm{a} \cot \alpha)\) So required locus is \(\left(x^{2}-a\right)^{2}+(y-a \cot \alpha)_{2}=a^{2} \operatorname{cosec}^{2} \alpha\) \(x^{2}+a^{2}-2 a x+y^{2}+a^{2} \cot ^{2} \alpha-2 a y \cot \alpha=a^{2} \operatorname{cosec}^{2} \alpha\) \(\mathrm{x}^{2}+\mathrm{y}^{2}-2 \mathrm{ax}-2 \mathrm{ay} \cot \alpha=0\) So, the locus of the vertex
WB JEE-2022
Co-Ordinate system
88474
If the sum of the distances of a point from two perpendicular lines in a plane is 1 unit, then its locus is
1 a square
2 a circle
3 a straight line
4 two intersecting lines
Explanation:
(C) : Let, the locus of a point in a plane be \(\mathrm{P}(\mathrm{h}, \mathrm{k})\). According to the question, \(|\mathrm{PA}|+|\mathrm{PB}|=1\) \(|\mathrm{X}|+|\mathrm{k}|=1\) So, locus of a point is \(|x|+|y|=1\) Which represent the straight line.
WB JEE-2022
Co-Ordinate system
88475
Locus of the centroid of a triangle whose vertices are \((1,0),(a \cos t, a \sin t)\), \((b \sin t,-b \cos t)\) is \(9 x^{2}+9 y^{2}-6 x=k\). Then, the value of \(k=\)
1 \(a^{2}+b^{2}\)
2 \(\mathrm{a}^{2}+\mathrm{b}^{2}-1\)
3 \(a^{2}+b^{2}+1\)
4 0
Explanation:
(B) : Given that - locus of the centroid of a triangle whose vertices are \((1,0)\) ( a cost,\(a \sin t), b \sin t,-b\) \(\cos \mathrm{t})\) is \(9 \mathrm{x}^{2}+9 \mathrm{y}^{2}-6 \mathrm{x}=\mathrm{k}\) \(x=\frac{x_{1}+x_{2}+x_{3}}{3}\) \(y=\frac{y_{1}+y_{2}+y_{3}}{3}\) So, \(x=\frac{1+a \cos +b \sin t}{3}\) \(y=\frac{0+a \sin t+(-b \cos t)}{3} \tag{ii}\) \(\Rightarrow \quad (3 x-1)=a \cos t+b \sin t \tag{i}\) And \(\quad 3 y=a \sin t-b \cos t\) Squaring and adding equation (i) and (ii) \(9 x^{2}-6 x+1+9 y^{2}=a^{2}+b^{2}\) \(9 x^{2}+9 y^{2}-6 x=a^{2}+b^{2}-1\) \(\begin{array}{ll}9 x+9 y -6 x=a^{2}+b^{2}-1\end{array}\)
AP EAMCET-2020-17.09.2020
Co-Ordinate system
88476
If it is mentioned such that for a curve passing through \((3,4)\), the slope of the curve at any point is the reciprocal of twice the ordinate of that point, then that curve is a ......
1 Ellipse
2 Parabola
3 Hyperbola
4 Circle
Explanation:
(B) : Given, the slope of the curve at any pointy is the reciprocal of twice of ordinate of that point. slope \(=\frac{1}{2 \mathrm{y}}\) \(\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{2 \mathrm{y}} \Rightarrow 2 \mathrm{y}\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)=1\) \(\mathrm{y}^{2}=\mathrm{x}+\mathrm{c} \quad[\mathrm{x}=3, \mathrm{y}=4]\) given that \(16=3+\mathrm{c}\) \(\mathrm{c}=13\) So, \(\mathrm{y}^{2}=\mathrm{x}+\mathrm{c}\) \((4)^{2}=(3)+13\) Which is form of \(\mathrm{y}^{2}=4 \mathrm{ax}\) So that carve is a parabola.
88473
The side \(A B\) of \(\triangle A B C\) is fixed and is of length 2 a unit. The vertex moves in the plane such that the vertical angle is always constant and is \(\alpha\). Let \(x\)-axis be along \(A B\) and the origin be at \(A\). Then the locus of the vertex is
1 \(x^{2}+y^{2}+2 a x \sin \alpha+a^{2} \cos \alpha=0\)
2 \(x^{2}+y^{2}-2 a x-2 a y \cot \alpha=0\)
3 \(x^{2}+y^{2}-2 a x \cos \alpha-a^{2}=0\)
4 \(x^{2}+y^{2}-a x \sin \alpha-a y \cos \alpha=0\)
Explanation:
(B) : Let \(\mathrm{AC}=\mathrm{BC}=\mathrm{r}\) Now, from \(\triangle \mathrm{ACE}\) \(\mathrm{r} \sin \alpha=\mathrm{a}\) Now again \(\triangle \mathrm{ACE}\) CE \(r \cos \alpha(a \operatorname{cosec} \alpha) \cos \alpha=a \cot \alpha\) Now locus of the vertex will be a circle with radius \(\mathrm{r}=\) a \(\operatorname{cosec} \alpha\) and center \(\mathrm{c}(\mathrm{a}, \mathrm{a} \cot \alpha)\) So required locus is \(\left(x^{2}-a\right)^{2}+(y-a \cot \alpha)_{2}=a^{2} \operatorname{cosec}^{2} \alpha\) \(x^{2}+a^{2}-2 a x+y^{2}+a^{2} \cot ^{2} \alpha-2 a y \cot \alpha=a^{2} \operatorname{cosec}^{2} \alpha\) \(\mathrm{x}^{2}+\mathrm{y}^{2}-2 \mathrm{ax}-2 \mathrm{ay} \cot \alpha=0\) So, the locus of the vertex
WB JEE-2022
Co-Ordinate system
88474
If the sum of the distances of a point from two perpendicular lines in a plane is 1 unit, then its locus is
1 a square
2 a circle
3 a straight line
4 two intersecting lines
Explanation:
(C) : Let, the locus of a point in a plane be \(\mathrm{P}(\mathrm{h}, \mathrm{k})\). According to the question, \(|\mathrm{PA}|+|\mathrm{PB}|=1\) \(|\mathrm{X}|+|\mathrm{k}|=1\) So, locus of a point is \(|x|+|y|=1\) Which represent the straight line.
WB JEE-2022
Co-Ordinate system
88475
Locus of the centroid of a triangle whose vertices are \((1,0),(a \cos t, a \sin t)\), \((b \sin t,-b \cos t)\) is \(9 x^{2}+9 y^{2}-6 x=k\). Then, the value of \(k=\)
1 \(a^{2}+b^{2}\)
2 \(\mathrm{a}^{2}+\mathrm{b}^{2}-1\)
3 \(a^{2}+b^{2}+1\)
4 0
Explanation:
(B) : Given that - locus of the centroid of a triangle whose vertices are \((1,0)\) ( a cost,\(a \sin t), b \sin t,-b\) \(\cos \mathrm{t})\) is \(9 \mathrm{x}^{2}+9 \mathrm{y}^{2}-6 \mathrm{x}=\mathrm{k}\) \(x=\frac{x_{1}+x_{2}+x_{3}}{3}\) \(y=\frac{y_{1}+y_{2}+y_{3}}{3}\) So, \(x=\frac{1+a \cos +b \sin t}{3}\) \(y=\frac{0+a \sin t+(-b \cos t)}{3} \tag{ii}\) \(\Rightarrow \quad (3 x-1)=a \cos t+b \sin t \tag{i}\) And \(\quad 3 y=a \sin t-b \cos t\) Squaring and adding equation (i) and (ii) \(9 x^{2}-6 x+1+9 y^{2}=a^{2}+b^{2}\) \(9 x^{2}+9 y^{2}-6 x=a^{2}+b^{2}-1\) \(\begin{array}{ll}9 x+9 y -6 x=a^{2}+b^{2}-1\end{array}\)
AP EAMCET-2020-17.09.2020
Co-Ordinate system
88476
If it is mentioned such that for a curve passing through \((3,4)\), the slope of the curve at any point is the reciprocal of twice the ordinate of that point, then that curve is a ......
1 Ellipse
2 Parabola
3 Hyperbola
4 Circle
Explanation:
(B) : Given, the slope of the curve at any pointy is the reciprocal of twice of ordinate of that point. slope \(=\frac{1}{2 \mathrm{y}}\) \(\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{2 \mathrm{y}} \Rightarrow 2 \mathrm{y}\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)=1\) \(\mathrm{y}^{2}=\mathrm{x}+\mathrm{c} \quad[\mathrm{x}=3, \mathrm{y}=4]\) given that \(16=3+\mathrm{c}\) \(\mathrm{c}=13\) So, \(\mathrm{y}^{2}=\mathrm{x}+\mathrm{c}\) \((4)^{2}=(3)+13\) Which is form of \(\mathrm{y}^{2}=4 \mathrm{ax}\) So that carve is a parabola.
88473
The side \(A B\) of \(\triangle A B C\) is fixed and is of length 2 a unit. The vertex moves in the plane such that the vertical angle is always constant and is \(\alpha\). Let \(x\)-axis be along \(A B\) and the origin be at \(A\). Then the locus of the vertex is
1 \(x^{2}+y^{2}+2 a x \sin \alpha+a^{2} \cos \alpha=0\)
2 \(x^{2}+y^{2}-2 a x-2 a y \cot \alpha=0\)
3 \(x^{2}+y^{2}-2 a x \cos \alpha-a^{2}=0\)
4 \(x^{2}+y^{2}-a x \sin \alpha-a y \cos \alpha=0\)
Explanation:
(B) : Let \(\mathrm{AC}=\mathrm{BC}=\mathrm{r}\) Now, from \(\triangle \mathrm{ACE}\) \(\mathrm{r} \sin \alpha=\mathrm{a}\) Now again \(\triangle \mathrm{ACE}\) CE \(r \cos \alpha(a \operatorname{cosec} \alpha) \cos \alpha=a \cot \alpha\) Now locus of the vertex will be a circle with radius \(\mathrm{r}=\) a \(\operatorname{cosec} \alpha\) and center \(\mathrm{c}(\mathrm{a}, \mathrm{a} \cot \alpha)\) So required locus is \(\left(x^{2}-a\right)^{2}+(y-a \cot \alpha)_{2}=a^{2} \operatorname{cosec}^{2} \alpha\) \(x^{2}+a^{2}-2 a x+y^{2}+a^{2} \cot ^{2} \alpha-2 a y \cot \alpha=a^{2} \operatorname{cosec}^{2} \alpha\) \(\mathrm{x}^{2}+\mathrm{y}^{2}-2 \mathrm{ax}-2 \mathrm{ay} \cot \alpha=0\) So, the locus of the vertex
WB JEE-2022
Co-Ordinate system
88474
If the sum of the distances of a point from two perpendicular lines in a plane is 1 unit, then its locus is
1 a square
2 a circle
3 a straight line
4 two intersecting lines
Explanation:
(C) : Let, the locus of a point in a plane be \(\mathrm{P}(\mathrm{h}, \mathrm{k})\). According to the question, \(|\mathrm{PA}|+|\mathrm{PB}|=1\) \(|\mathrm{X}|+|\mathrm{k}|=1\) So, locus of a point is \(|x|+|y|=1\) Which represent the straight line.
WB JEE-2022
Co-Ordinate system
88475
Locus of the centroid of a triangle whose vertices are \((1,0),(a \cos t, a \sin t)\), \((b \sin t,-b \cos t)\) is \(9 x^{2}+9 y^{2}-6 x=k\). Then, the value of \(k=\)
1 \(a^{2}+b^{2}\)
2 \(\mathrm{a}^{2}+\mathrm{b}^{2}-1\)
3 \(a^{2}+b^{2}+1\)
4 0
Explanation:
(B) : Given that - locus of the centroid of a triangle whose vertices are \((1,0)\) ( a cost,\(a \sin t), b \sin t,-b\) \(\cos \mathrm{t})\) is \(9 \mathrm{x}^{2}+9 \mathrm{y}^{2}-6 \mathrm{x}=\mathrm{k}\) \(x=\frac{x_{1}+x_{2}+x_{3}}{3}\) \(y=\frac{y_{1}+y_{2}+y_{3}}{3}\) So, \(x=\frac{1+a \cos +b \sin t}{3}\) \(y=\frac{0+a \sin t+(-b \cos t)}{3} \tag{ii}\) \(\Rightarrow \quad (3 x-1)=a \cos t+b \sin t \tag{i}\) And \(\quad 3 y=a \sin t-b \cos t\) Squaring and adding equation (i) and (ii) \(9 x^{2}-6 x+1+9 y^{2}=a^{2}+b^{2}\) \(9 x^{2}+9 y^{2}-6 x=a^{2}+b^{2}-1\) \(\begin{array}{ll}9 x+9 y -6 x=a^{2}+b^{2}-1\end{array}\)
AP EAMCET-2020-17.09.2020
Co-Ordinate system
88476
If it is mentioned such that for a curve passing through \((3,4)\), the slope of the curve at any point is the reciprocal of twice the ordinate of that point, then that curve is a ......
1 Ellipse
2 Parabola
3 Hyperbola
4 Circle
Explanation:
(B) : Given, the slope of the curve at any pointy is the reciprocal of twice of ordinate of that point. slope \(=\frac{1}{2 \mathrm{y}}\) \(\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{2 \mathrm{y}} \Rightarrow 2 \mathrm{y}\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)=1\) \(\mathrm{y}^{2}=\mathrm{x}+\mathrm{c} \quad[\mathrm{x}=3, \mathrm{y}=4]\) given that \(16=3+\mathrm{c}\) \(\mathrm{c}=13\) So, \(\mathrm{y}^{2}=\mathrm{x}+\mathrm{c}\) \((4)^{2}=(3)+13\) Which is form of \(\mathrm{y}^{2}=4 \mathrm{ax}\) So that carve is a parabola.
