NEET Test Series from KOTA - 10 Papers In MS WORD
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Co-Ordinate system
88464
A straight line meets the \(X\) and \(Y\) axes at the points \(A, B\) respectively. If \(A B=6\) units, then the locus of the point \(P\) which divides the line segment \(A B\) such that \(A P: P B=2: 1\), is
1 \(3 x^{2}+y^{2}=36\)
2 \(4 x^{2}+y^{2}=36\)
3 \(3 x^{2}+y^{2}=16\)
4 \(4 x^{2}+y^{2}=16\)
Explanation:
(D) : Let \(A(a, o)\) and \(B(o, b)\) and point \(\mathrm{p}(\mathrm{h}, \mathrm{k})\) divides the line \(\mathrm{AB}\). Now, \(p[h, k)=\left(\frac{a \times 1+0}{1+2}, \frac{0 \times 2 b}{1+2}\right)\) \(\mathrm{p}(\mathrm{h}, \mathrm{k})=\left(\frac{\mathrm{a}}{3}, \frac{2 \mathrm{~b}}{3}\right)\) \(\therefore \quad \mathrm{h}=\frac{\mathrm{a}}{3}\) and \(\mathrm{k}=\frac{2 \mathrm{~b}}{3}\) \(\mathrm{a}=3 \mathrm{~h}, \quad \mathrm{~b}=\frac{3 \mathrm{k}}{2}\) \(\mathrm{AB}=6\) \(\sqrt{(\mathrm{a}-0)^{2}+(0-\mathrm{b})^{2}}=6\) (given that) \(\mathrm{a}^{2}+\mathrm{b}^{2}=36\) \(9 \mathrm{~h}^{2}+\frac{9 \mathrm{k}^{2}}{4}=36\) \(\mathrm{h}^{2}+\frac{\mathrm{k}^{2}}{4}=4\) \(4 \mathrm{~h}^{2}+\mathrm{k}^{2}=16\) \(\therefore \quad\) required locus of \(\mathrm{P}\) is \(4 x^{2}+y^{2}=16\)
AP EAMCET-2019-22.04.2019
Co-Ordinate system
88465
If the sum of the distances of a point from two perpendicular lines in a plane is 1 , then its locus is
1 Two intersecting lines
2 Square
3 A straight line
4 Circle
Explanation:
(B) : : Let \(\mathrm{x}, \mathrm{y}\) - axis be two perpendicular lines Let \(\mathrm{P}=(\mathrm{h}, \mathrm{k})\) be any point in the locus. \(\therefore\) Distance from \(\mathrm{p}(\mathrm{h}, \mathrm{k})\) to \((\mathrm{x}\)-axis \(=1), \quad\) (given that) \(|\mathrm{k}|=1\) Distance from \(\phi(\mathrm{h}, \mathrm{k})\) to \(\mathrm{y}\) axis \(=1 \Rightarrow|\mathrm{h}|=1\) According to the question, \(|\mathrm{h}|+|\mathrm{k}|=1\) \(\therefore\) The locus of point \((h, k)\) is \(|x|+|y|=1\) So, It is a square.
AP EAMCET-2020-22.09.2020
Co-Ordinate system
88466
\(A B\) is a line segment moving between the axes such that ' \(A\) ' lies on \(X\)-axis and ' \(B\) ' lies on \(Y\) axis. If \(P\) is a point on \(A B\) such that \(P A=b\) and \(P B=a\), then the equation of locus of \(P\) is
(B) : Let's consider- \(\mathrm{p}(\mathrm{h}, \mathrm{k})\) be any point in the locus \(\mathrm{cm}\). Let \(\quad \mathrm{A}=(\mathrm{x}, 0), \mathrm{B}=(0, \mathrm{y})\) \(\mathrm{PA}=\mathrm{b}, \mathrm{PB}=\mathrm{a}\) (give that) In \(\triangle \mathrm{BNP} \cos \theta=\frac{\mathrm{h}}{\mathrm{a}}\) In \(\triangle \mathrm{PMA}\), \(\sin \theta=\frac{\mathrm{k}}{\mathrm{b}}\) \(\sin ^{2} \theta+\cos ^{2} \theta=1\) (we know) \(\frac{\mathrm{k}^{2}}{\mathrm{~b}^{2}}+\frac{\mathrm{b}^{2}}{\mathrm{a}^{2}}=1\) So, \(\quad \frac{\mathrm{h}^{2}}{\mathrm{a}^{2}}+\frac{\mathrm{k}^{2}}{\mathrm{~b}^{2}}=1\) \(\therefore \quad\) Required locus of point \(\mathrm{p}(\mathrm{h}, \mathrm{k})\) is \(\frac{\mathrm{x}^{2}}{\mathrm{a}^{2}}+\frac{\mathrm{y}^{2}}{\mathrm{~b}^{2}}=1\)
AP EAMCET-2020-22.09.2020
Co-Ordinate system
88467
The locus of midpoints of points of intersection of \(x \cos \theta+y \sin \theta=1\) with the coordinate axes is
(B) : : Given that-locus of mid points of intersection of \(x \cos \theta+y \sin \theta=1\). Let the line cut the axis is \(A\) and \(B\) and if \((h, k)\) be the mid-point of \(\mathrm{AB}\), then- \(2 \mathrm{~h}=\frac{1}{\cos \theta}, 2 \mathrm{k}=\frac{1}{\sin \theta}\) In order to find the locus, eliminate the variable ' \(\theta\) ' by \(\cos ^{2} \theta \sin ^{2} \theta=1\) \(\frac{1}{4 \mathrm{~h}^{2}}+\frac{1}{4 \mathrm{k}^{2}}=1, \frac{1}{\mathrm{x}^{2}}+\frac{1}{\mathrm{y}^{2}}=4\) So, \(\quad\) Coordinate axis \(=\frac{1}{\mathrm{x}^{2}}+\frac{1}{\mathrm{y}^{2}}=4\)
88464
A straight line meets the \(X\) and \(Y\) axes at the points \(A, B\) respectively. If \(A B=6\) units, then the locus of the point \(P\) which divides the line segment \(A B\) such that \(A P: P B=2: 1\), is
1 \(3 x^{2}+y^{2}=36\)
2 \(4 x^{2}+y^{2}=36\)
3 \(3 x^{2}+y^{2}=16\)
4 \(4 x^{2}+y^{2}=16\)
Explanation:
(D) : Let \(A(a, o)\) and \(B(o, b)\) and point \(\mathrm{p}(\mathrm{h}, \mathrm{k})\) divides the line \(\mathrm{AB}\). Now, \(p[h, k)=\left(\frac{a \times 1+0}{1+2}, \frac{0 \times 2 b}{1+2}\right)\) \(\mathrm{p}(\mathrm{h}, \mathrm{k})=\left(\frac{\mathrm{a}}{3}, \frac{2 \mathrm{~b}}{3}\right)\) \(\therefore \quad \mathrm{h}=\frac{\mathrm{a}}{3}\) and \(\mathrm{k}=\frac{2 \mathrm{~b}}{3}\) \(\mathrm{a}=3 \mathrm{~h}, \quad \mathrm{~b}=\frac{3 \mathrm{k}}{2}\) \(\mathrm{AB}=6\) \(\sqrt{(\mathrm{a}-0)^{2}+(0-\mathrm{b})^{2}}=6\) (given that) \(\mathrm{a}^{2}+\mathrm{b}^{2}=36\) \(9 \mathrm{~h}^{2}+\frac{9 \mathrm{k}^{2}}{4}=36\) \(\mathrm{h}^{2}+\frac{\mathrm{k}^{2}}{4}=4\) \(4 \mathrm{~h}^{2}+\mathrm{k}^{2}=16\) \(\therefore \quad\) required locus of \(\mathrm{P}\) is \(4 x^{2}+y^{2}=16\)
AP EAMCET-2019-22.04.2019
Co-Ordinate system
88465
If the sum of the distances of a point from two perpendicular lines in a plane is 1 , then its locus is
1 Two intersecting lines
2 Square
3 A straight line
4 Circle
Explanation:
(B) : : Let \(\mathrm{x}, \mathrm{y}\) - axis be two perpendicular lines Let \(\mathrm{P}=(\mathrm{h}, \mathrm{k})\) be any point in the locus. \(\therefore\) Distance from \(\mathrm{p}(\mathrm{h}, \mathrm{k})\) to \((\mathrm{x}\)-axis \(=1), \quad\) (given that) \(|\mathrm{k}|=1\) Distance from \(\phi(\mathrm{h}, \mathrm{k})\) to \(\mathrm{y}\) axis \(=1 \Rightarrow|\mathrm{h}|=1\) According to the question, \(|\mathrm{h}|+|\mathrm{k}|=1\) \(\therefore\) The locus of point \((h, k)\) is \(|x|+|y|=1\) So, It is a square.
AP EAMCET-2020-22.09.2020
Co-Ordinate system
88466
\(A B\) is a line segment moving between the axes such that ' \(A\) ' lies on \(X\)-axis and ' \(B\) ' lies on \(Y\) axis. If \(P\) is a point on \(A B\) such that \(P A=b\) and \(P B=a\), then the equation of locus of \(P\) is
(B) : Let's consider- \(\mathrm{p}(\mathrm{h}, \mathrm{k})\) be any point in the locus \(\mathrm{cm}\). Let \(\quad \mathrm{A}=(\mathrm{x}, 0), \mathrm{B}=(0, \mathrm{y})\) \(\mathrm{PA}=\mathrm{b}, \mathrm{PB}=\mathrm{a}\) (give that) In \(\triangle \mathrm{BNP} \cos \theta=\frac{\mathrm{h}}{\mathrm{a}}\) In \(\triangle \mathrm{PMA}\), \(\sin \theta=\frac{\mathrm{k}}{\mathrm{b}}\) \(\sin ^{2} \theta+\cos ^{2} \theta=1\) (we know) \(\frac{\mathrm{k}^{2}}{\mathrm{~b}^{2}}+\frac{\mathrm{b}^{2}}{\mathrm{a}^{2}}=1\) So, \(\quad \frac{\mathrm{h}^{2}}{\mathrm{a}^{2}}+\frac{\mathrm{k}^{2}}{\mathrm{~b}^{2}}=1\) \(\therefore \quad\) Required locus of point \(\mathrm{p}(\mathrm{h}, \mathrm{k})\) is \(\frac{\mathrm{x}^{2}}{\mathrm{a}^{2}}+\frac{\mathrm{y}^{2}}{\mathrm{~b}^{2}}=1\)
AP EAMCET-2020-22.09.2020
Co-Ordinate system
88467
The locus of midpoints of points of intersection of \(x \cos \theta+y \sin \theta=1\) with the coordinate axes is
(B) : : Given that-locus of mid points of intersection of \(x \cos \theta+y \sin \theta=1\). Let the line cut the axis is \(A\) and \(B\) and if \((h, k)\) be the mid-point of \(\mathrm{AB}\), then- \(2 \mathrm{~h}=\frac{1}{\cos \theta}, 2 \mathrm{k}=\frac{1}{\sin \theta}\) In order to find the locus, eliminate the variable ' \(\theta\) ' by \(\cos ^{2} \theta \sin ^{2} \theta=1\) \(\frac{1}{4 \mathrm{~h}^{2}}+\frac{1}{4 \mathrm{k}^{2}}=1, \frac{1}{\mathrm{x}^{2}}+\frac{1}{\mathrm{y}^{2}}=4\) So, \(\quad\) Coordinate axis \(=\frac{1}{\mathrm{x}^{2}}+\frac{1}{\mathrm{y}^{2}}=4\)
88464
A straight line meets the \(X\) and \(Y\) axes at the points \(A, B\) respectively. If \(A B=6\) units, then the locus of the point \(P\) which divides the line segment \(A B\) such that \(A P: P B=2: 1\), is
1 \(3 x^{2}+y^{2}=36\)
2 \(4 x^{2}+y^{2}=36\)
3 \(3 x^{2}+y^{2}=16\)
4 \(4 x^{2}+y^{2}=16\)
Explanation:
(D) : Let \(A(a, o)\) and \(B(o, b)\) and point \(\mathrm{p}(\mathrm{h}, \mathrm{k})\) divides the line \(\mathrm{AB}\). Now, \(p[h, k)=\left(\frac{a \times 1+0}{1+2}, \frac{0 \times 2 b}{1+2}\right)\) \(\mathrm{p}(\mathrm{h}, \mathrm{k})=\left(\frac{\mathrm{a}}{3}, \frac{2 \mathrm{~b}}{3}\right)\) \(\therefore \quad \mathrm{h}=\frac{\mathrm{a}}{3}\) and \(\mathrm{k}=\frac{2 \mathrm{~b}}{3}\) \(\mathrm{a}=3 \mathrm{~h}, \quad \mathrm{~b}=\frac{3 \mathrm{k}}{2}\) \(\mathrm{AB}=6\) \(\sqrt{(\mathrm{a}-0)^{2}+(0-\mathrm{b})^{2}}=6\) (given that) \(\mathrm{a}^{2}+\mathrm{b}^{2}=36\) \(9 \mathrm{~h}^{2}+\frac{9 \mathrm{k}^{2}}{4}=36\) \(\mathrm{h}^{2}+\frac{\mathrm{k}^{2}}{4}=4\) \(4 \mathrm{~h}^{2}+\mathrm{k}^{2}=16\) \(\therefore \quad\) required locus of \(\mathrm{P}\) is \(4 x^{2}+y^{2}=16\)
AP EAMCET-2019-22.04.2019
Co-Ordinate system
88465
If the sum of the distances of a point from two perpendicular lines in a plane is 1 , then its locus is
1 Two intersecting lines
2 Square
3 A straight line
4 Circle
Explanation:
(B) : : Let \(\mathrm{x}, \mathrm{y}\) - axis be two perpendicular lines Let \(\mathrm{P}=(\mathrm{h}, \mathrm{k})\) be any point in the locus. \(\therefore\) Distance from \(\mathrm{p}(\mathrm{h}, \mathrm{k})\) to \((\mathrm{x}\)-axis \(=1), \quad\) (given that) \(|\mathrm{k}|=1\) Distance from \(\phi(\mathrm{h}, \mathrm{k})\) to \(\mathrm{y}\) axis \(=1 \Rightarrow|\mathrm{h}|=1\) According to the question, \(|\mathrm{h}|+|\mathrm{k}|=1\) \(\therefore\) The locus of point \((h, k)\) is \(|x|+|y|=1\) So, It is a square.
AP EAMCET-2020-22.09.2020
Co-Ordinate system
88466
\(A B\) is a line segment moving between the axes such that ' \(A\) ' lies on \(X\)-axis and ' \(B\) ' lies on \(Y\) axis. If \(P\) is a point on \(A B\) such that \(P A=b\) and \(P B=a\), then the equation of locus of \(P\) is
(B) : Let's consider- \(\mathrm{p}(\mathrm{h}, \mathrm{k})\) be any point in the locus \(\mathrm{cm}\). Let \(\quad \mathrm{A}=(\mathrm{x}, 0), \mathrm{B}=(0, \mathrm{y})\) \(\mathrm{PA}=\mathrm{b}, \mathrm{PB}=\mathrm{a}\) (give that) In \(\triangle \mathrm{BNP} \cos \theta=\frac{\mathrm{h}}{\mathrm{a}}\) In \(\triangle \mathrm{PMA}\), \(\sin \theta=\frac{\mathrm{k}}{\mathrm{b}}\) \(\sin ^{2} \theta+\cos ^{2} \theta=1\) (we know) \(\frac{\mathrm{k}^{2}}{\mathrm{~b}^{2}}+\frac{\mathrm{b}^{2}}{\mathrm{a}^{2}}=1\) So, \(\quad \frac{\mathrm{h}^{2}}{\mathrm{a}^{2}}+\frac{\mathrm{k}^{2}}{\mathrm{~b}^{2}}=1\) \(\therefore \quad\) Required locus of point \(\mathrm{p}(\mathrm{h}, \mathrm{k})\) is \(\frac{\mathrm{x}^{2}}{\mathrm{a}^{2}}+\frac{\mathrm{y}^{2}}{\mathrm{~b}^{2}}=1\)
AP EAMCET-2020-22.09.2020
Co-Ordinate system
88467
The locus of midpoints of points of intersection of \(x \cos \theta+y \sin \theta=1\) with the coordinate axes is
(B) : : Given that-locus of mid points of intersection of \(x \cos \theta+y \sin \theta=1\). Let the line cut the axis is \(A\) and \(B\) and if \((h, k)\) be the mid-point of \(\mathrm{AB}\), then- \(2 \mathrm{~h}=\frac{1}{\cos \theta}, 2 \mathrm{k}=\frac{1}{\sin \theta}\) In order to find the locus, eliminate the variable ' \(\theta\) ' by \(\cos ^{2} \theta \sin ^{2} \theta=1\) \(\frac{1}{4 \mathrm{~h}^{2}}+\frac{1}{4 \mathrm{k}^{2}}=1, \frac{1}{\mathrm{x}^{2}}+\frac{1}{\mathrm{y}^{2}}=4\) So, \(\quad\) Coordinate axis \(=\frac{1}{\mathrm{x}^{2}}+\frac{1}{\mathrm{y}^{2}}=4\)
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Co-Ordinate system
88464
A straight line meets the \(X\) and \(Y\) axes at the points \(A, B\) respectively. If \(A B=6\) units, then the locus of the point \(P\) which divides the line segment \(A B\) such that \(A P: P B=2: 1\), is
1 \(3 x^{2}+y^{2}=36\)
2 \(4 x^{2}+y^{2}=36\)
3 \(3 x^{2}+y^{2}=16\)
4 \(4 x^{2}+y^{2}=16\)
Explanation:
(D) : Let \(A(a, o)\) and \(B(o, b)\) and point \(\mathrm{p}(\mathrm{h}, \mathrm{k})\) divides the line \(\mathrm{AB}\). Now, \(p[h, k)=\left(\frac{a \times 1+0}{1+2}, \frac{0 \times 2 b}{1+2}\right)\) \(\mathrm{p}(\mathrm{h}, \mathrm{k})=\left(\frac{\mathrm{a}}{3}, \frac{2 \mathrm{~b}}{3}\right)\) \(\therefore \quad \mathrm{h}=\frac{\mathrm{a}}{3}\) and \(\mathrm{k}=\frac{2 \mathrm{~b}}{3}\) \(\mathrm{a}=3 \mathrm{~h}, \quad \mathrm{~b}=\frac{3 \mathrm{k}}{2}\) \(\mathrm{AB}=6\) \(\sqrt{(\mathrm{a}-0)^{2}+(0-\mathrm{b})^{2}}=6\) (given that) \(\mathrm{a}^{2}+\mathrm{b}^{2}=36\) \(9 \mathrm{~h}^{2}+\frac{9 \mathrm{k}^{2}}{4}=36\) \(\mathrm{h}^{2}+\frac{\mathrm{k}^{2}}{4}=4\) \(4 \mathrm{~h}^{2}+\mathrm{k}^{2}=16\) \(\therefore \quad\) required locus of \(\mathrm{P}\) is \(4 x^{2}+y^{2}=16\)
AP EAMCET-2019-22.04.2019
Co-Ordinate system
88465
If the sum of the distances of a point from two perpendicular lines in a plane is 1 , then its locus is
1 Two intersecting lines
2 Square
3 A straight line
4 Circle
Explanation:
(B) : : Let \(\mathrm{x}, \mathrm{y}\) - axis be two perpendicular lines Let \(\mathrm{P}=(\mathrm{h}, \mathrm{k})\) be any point in the locus. \(\therefore\) Distance from \(\mathrm{p}(\mathrm{h}, \mathrm{k})\) to \((\mathrm{x}\)-axis \(=1), \quad\) (given that) \(|\mathrm{k}|=1\) Distance from \(\phi(\mathrm{h}, \mathrm{k})\) to \(\mathrm{y}\) axis \(=1 \Rightarrow|\mathrm{h}|=1\) According to the question, \(|\mathrm{h}|+|\mathrm{k}|=1\) \(\therefore\) The locus of point \((h, k)\) is \(|x|+|y|=1\) So, It is a square.
AP EAMCET-2020-22.09.2020
Co-Ordinate system
88466
\(A B\) is a line segment moving between the axes such that ' \(A\) ' lies on \(X\)-axis and ' \(B\) ' lies on \(Y\) axis. If \(P\) is a point on \(A B\) such that \(P A=b\) and \(P B=a\), then the equation of locus of \(P\) is
(B) : Let's consider- \(\mathrm{p}(\mathrm{h}, \mathrm{k})\) be any point in the locus \(\mathrm{cm}\). Let \(\quad \mathrm{A}=(\mathrm{x}, 0), \mathrm{B}=(0, \mathrm{y})\) \(\mathrm{PA}=\mathrm{b}, \mathrm{PB}=\mathrm{a}\) (give that) In \(\triangle \mathrm{BNP} \cos \theta=\frac{\mathrm{h}}{\mathrm{a}}\) In \(\triangle \mathrm{PMA}\), \(\sin \theta=\frac{\mathrm{k}}{\mathrm{b}}\) \(\sin ^{2} \theta+\cos ^{2} \theta=1\) (we know) \(\frac{\mathrm{k}^{2}}{\mathrm{~b}^{2}}+\frac{\mathrm{b}^{2}}{\mathrm{a}^{2}}=1\) So, \(\quad \frac{\mathrm{h}^{2}}{\mathrm{a}^{2}}+\frac{\mathrm{k}^{2}}{\mathrm{~b}^{2}}=1\) \(\therefore \quad\) Required locus of point \(\mathrm{p}(\mathrm{h}, \mathrm{k})\) is \(\frac{\mathrm{x}^{2}}{\mathrm{a}^{2}}+\frac{\mathrm{y}^{2}}{\mathrm{~b}^{2}}=1\)
AP EAMCET-2020-22.09.2020
Co-Ordinate system
88467
The locus of midpoints of points of intersection of \(x \cos \theta+y \sin \theta=1\) with the coordinate axes is
(B) : : Given that-locus of mid points of intersection of \(x \cos \theta+y \sin \theta=1\). Let the line cut the axis is \(A\) and \(B\) and if \((h, k)\) be the mid-point of \(\mathrm{AB}\), then- \(2 \mathrm{~h}=\frac{1}{\cos \theta}, 2 \mathrm{k}=\frac{1}{\sin \theta}\) In order to find the locus, eliminate the variable ' \(\theta\) ' by \(\cos ^{2} \theta \sin ^{2} \theta=1\) \(\frac{1}{4 \mathrm{~h}^{2}}+\frac{1}{4 \mathrm{k}^{2}}=1, \frac{1}{\mathrm{x}^{2}}+\frac{1}{\mathrm{y}^{2}}=4\) So, \(\quad\) Coordinate axis \(=\frac{1}{\mathrm{x}^{2}}+\frac{1}{\mathrm{y}^{2}}=4\)
