NEET Test Series from KOTA - 10 Papers In MS WORD
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Co-Ordinate system
88434
A rod of length I slides with its ends on two perpendicular lines. Then, the locus of its mid point is
1 \(x^{2}+y^{2}=\frac{1^{2}}{4}\)
2 \(x^{2}+y^{2}=\frac{1^{2}}{2}\)
3 \(x^{2}-y^{2}=\frac{1^{2}}{4}\)
4 None of these
Explanation:
(A) : Let the intercept \(x\)-axis is \((a, 0)\) and \(y\)-axis is \((0, b)\) and \((\alpha, \beta)\) is the midpoint then, \(\frac{a+0}{2}=\alpha \Rightarrow \mathrm{a}=2 \alpha\) and, \(\frac{b+0}{2}=\beta\) Then, \(\quad l^{2}=(\mathrm{a}-0)^{2}+(0-\mathrm{b})^{2}\) \(\text { ( } l \text { is the length of rod) }\) \(l^{2}=(2 \alpha)^{2}+(2 \beta)^{2}\) \(l^{2}=4 \alpha^{2}+4 \beta^{2}\) \(\alpha^{2}+\beta^{2}=\left(\frac{l}{2}\right)^{2}\) If \((\alpha, \beta)\) be \((x, y)\) then above equation becomes, \(\mathrm{x}^{2}+\mathrm{y}^{2}=\frac{l^{2}}{4}\)
VITEEE-2010
Co-Ordinate system
88435
The coordinates of a point are \(\operatorname{atan}(\theta+\alpha)\) and b \(\tan (\theta+\beta)\), where \(\theta\) is variable, then locus of the point is
1 hyperbola
2 rectangular hyperbola
3 ellipse
4 None of the above
Explanation:
(A) : Given that the co-ordinates point \(\mathrm{x}=\mathrm{a} \tan (\theta+\alpha)\) \(\begin{array}{ll}\Rightarrow \theta+\alpha=\tan ^{-1} \frac{x}{a} \\ y=b \tan (\theta+\beta) \\ \Rightarrow \theta+\beta=\tan ^{-1} \frac{y}{b}\end{array}\) Subtracting equation (i) from (ii) \(\alpha-\beta=\tan ^{-1} \frac{x}{a}-\tan ^{-1} \frac{y}{b}\) \(\Rightarrow \tan ^{-1} \frac{\frac{x}{a}-\frac{y}{b}}{1+\frac{x}{a} \frac{y}{b}}=\alpha-\beta\) \(\Rightarrow x y+a b=(b x-a y) \cot (\alpha-\beta)\) Hence, it is the equation of rectangular hyperbola.
UPSEE-2013
Co-Ordinate system
88436
If the distance of any point \((x, y)\) from the origin is defined as \(d(x, y)=\max \{|x|,|y|\}\) then the locus of the point \((x, y)\) where \(d(x, y)=1\), is
1 A square
2 A circle
3 A triangle
4 None of the above
Explanation:
(A) : Given, \(\mathrm{d}(\mathrm{x}, \mathrm{y})=1\) and, \(\max \{|\mathrm{x}||\mathrm{y}|\}=1\) \(|\mathrm{x}| \leq 1\) and \(|\mathrm{y}| \leq 1\) \(-1 \leq \mathrm{x} \leq 1\) and \(-1 \leq \mathrm{y} \leq 1\) \(\therefore\) locus of \((\mathrm{x}, \mathrm{y})\) is the square having its vertices at \((1\), \(1),(-1,1),(-1,-1)\) and \((1,-1)\)
[JCECE-2016]
Co-Ordinate system
88437
The locus of centre of the circle touching the line \(2 x-y=1\) at \((1,1)\) is
1 \(x+3 y=2\)
2 \(x+2 y=3\)
3 \(x+y=2\)
4 \(2 x-y=1\)
Explanation:
(B) : Given that tine \((2 x-y=1)\) at \((1,1)\) We have to find that the locus centre of the circle touching the line \(=\) ? Let, \((h, k)\) be the centre of circle of the circle touch the line \(2 x-y=1\) at \((1,1)\) \(\therefore \sqrt{(\mathrm{h}-1)^{2}+(\mathrm{k}-1)^{2}}=\frac{2 \mathrm{~h}-\mathrm{k}-1}{\sqrt{4+1}}\) \(\sqrt{5} \sqrt{(\mathrm{h}-1)^{2}+(\mathrm{k}-1)^{2}}=(2 \mathrm{~h}-\mathrm{k}-1)\) On squaring both side, \(5\left(h^{2}+1+k^{2}+1-2 h-2 k\right)=4 h^{2}+k^{2}+1-4 h k-2 h+\) \(2 \mathrm{k}\) \(\mathrm{h}^{2}+4 \mathrm{k}^{2}+9-6 \mathrm{~h}-12 \mathrm{k}+4 \mathrm{hk}=0\) \((\mathrm{h}+2 \mathrm{k}-3)^{2}=0\) \(\mathrm{h}+2 \mathrm{k}-3=0\) \(\mathrm{x}+2 \mathrm{y}=3\) \(=5\left(\mathrm{~h}^{2}+1-2 \mathrm{~h}+\mathrm{k}^{2}-10 \mathrm{~h}-10 \mathrm{k}+10=4 \mathrm{~h}^{2}+\mathrm{k}^{2}-4 \mathrm{hk}\right.\) \(+1-2(2 \mathrm{~h}-\mathrm{k})\) \(\mathrm{h}^{2}+4 \mathrm{k}^{2}-6 \mathrm{~h}-12 \mathrm{k}+9+4 \mathrm{hk}=0\) \((\mathrm{h}+2 \mathrm{k}-3)^{2}=0\) \(\mathrm{h}+2 \mathrm{k}-3=0\) \(\mathrm{h}+2 \mathrm{k}=3\) \(x+2 y=3\)
88434
A rod of length I slides with its ends on two perpendicular lines. Then, the locus of its mid point is
1 \(x^{2}+y^{2}=\frac{1^{2}}{4}\)
2 \(x^{2}+y^{2}=\frac{1^{2}}{2}\)
3 \(x^{2}-y^{2}=\frac{1^{2}}{4}\)
4 None of these
Explanation:
(A) : Let the intercept \(x\)-axis is \((a, 0)\) and \(y\)-axis is \((0, b)\) and \((\alpha, \beta)\) is the midpoint then, \(\frac{a+0}{2}=\alpha \Rightarrow \mathrm{a}=2 \alpha\) and, \(\frac{b+0}{2}=\beta\) Then, \(\quad l^{2}=(\mathrm{a}-0)^{2}+(0-\mathrm{b})^{2}\) \(\text { ( } l \text { is the length of rod) }\) \(l^{2}=(2 \alpha)^{2}+(2 \beta)^{2}\) \(l^{2}=4 \alpha^{2}+4 \beta^{2}\) \(\alpha^{2}+\beta^{2}=\left(\frac{l}{2}\right)^{2}\) If \((\alpha, \beta)\) be \((x, y)\) then above equation becomes, \(\mathrm{x}^{2}+\mathrm{y}^{2}=\frac{l^{2}}{4}\)
VITEEE-2010
Co-Ordinate system
88435
The coordinates of a point are \(\operatorname{atan}(\theta+\alpha)\) and b \(\tan (\theta+\beta)\), where \(\theta\) is variable, then locus of the point is
1 hyperbola
2 rectangular hyperbola
3 ellipse
4 None of the above
Explanation:
(A) : Given that the co-ordinates point \(\mathrm{x}=\mathrm{a} \tan (\theta+\alpha)\) \(\begin{array}{ll}\Rightarrow \theta+\alpha=\tan ^{-1} \frac{x}{a} \\ y=b \tan (\theta+\beta) \\ \Rightarrow \theta+\beta=\tan ^{-1} \frac{y}{b}\end{array}\) Subtracting equation (i) from (ii) \(\alpha-\beta=\tan ^{-1} \frac{x}{a}-\tan ^{-1} \frac{y}{b}\) \(\Rightarrow \tan ^{-1} \frac{\frac{x}{a}-\frac{y}{b}}{1+\frac{x}{a} \frac{y}{b}}=\alpha-\beta\) \(\Rightarrow x y+a b=(b x-a y) \cot (\alpha-\beta)\) Hence, it is the equation of rectangular hyperbola.
UPSEE-2013
Co-Ordinate system
88436
If the distance of any point \((x, y)\) from the origin is defined as \(d(x, y)=\max \{|x|,|y|\}\) then the locus of the point \((x, y)\) where \(d(x, y)=1\), is
1 A square
2 A circle
3 A triangle
4 None of the above
Explanation:
(A) : Given, \(\mathrm{d}(\mathrm{x}, \mathrm{y})=1\) and, \(\max \{|\mathrm{x}||\mathrm{y}|\}=1\) \(|\mathrm{x}| \leq 1\) and \(|\mathrm{y}| \leq 1\) \(-1 \leq \mathrm{x} \leq 1\) and \(-1 \leq \mathrm{y} \leq 1\) \(\therefore\) locus of \((\mathrm{x}, \mathrm{y})\) is the square having its vertices at \((1\), \(1),(-1,1),(-1,-1)\) and \((1,-1)\)
[JCECE-2016]
Co-Ordinate system
88437
The locus of centre of the circle touching the line \(2 x-y=1\) at \((1,1)\) is
1 \(x+3 y=2\)
2 \(x+2 y=3\)
3 \(x+y=2\)
4 \(2 x-y=1\)
Explanation:
(B) : Given that tine \((2 x-y=1)\) at \((1,1)\) We have to find that the locus centre of the circle touching the line \(=\) ? Let, \((h, k)\) be the centre of circle of the circle touch the line \(2 x-y=1\) at \((1,1)\) \(\therefore \sqrt{(\mathrm{h}-1)^{2}+(\mathrm{k}-1)^{2}}=\frac{2 \mathrm{~h}-\mathrm{k}-1}{\sqrt{4+1}}\) \(\sqrt{5} \sqrt{(\mathrm{h}-1)^{2}+(\mathrm{k}-1)^{2}}=(2 \mathrm{~h}-\mathrm{k}-1)\) On squaring both side, \(5\left(h^{2}+1+k^{2}+1-2 h-2 k\right)=4 h^{2}+k^{2}+1-4 h k-2 h+\) \(2 \mathrm{k}\) \(\mathrm{h}^{2}+4 \mathrm{k}^{2}+9-6 \mathrm{~h}-12 \mathrm{k}+4 \mathrm{hk}=0\) \((\mathrm{h}+2 \mathrm{k}-3)^{2}=0\) \(\mathrm{h}+2 \mathrm{k}-3=0\) \(\mathrm{x}+2 \mathrm{y}=3\) \(=5\left(\mathrm{~h}^{2}+1-2 \mathrm{~h}+\mathrm{k}^{2}-10 \mathrm{~h}-10 \mathrm{k}+10=4 \mathrm{~h}^{2}+\mathrm{k}^{2}-4 \mathrm{hk}\right.\) \(+1-2(2 \mathrm{~h}-\mathrm{k})\) \(\mathrm{h}^{2}+4 \mathrm{k}^{2}-6 \mathrm{~h}-12 \mathrm{k}+9+4 \mathrm{hk}=0\) \((\mathrm{h}+2 \mathrm{k}-3)^{2}=0\) \(\mathrm{h}+2 \mathrm{k}-3=0\) \(\mathrm{h}+2 \mathrm{k}=3\) \(x+2 y=3\)
88434
A rod of length I slides with its ends on two perpendicular lines. Then, the locus of its mid point is
1 \(x^{2}+y^{2}=\frac{1^{2}}{4}\)
2 \(x^{2}+y^{2}=\frac{1^{2}}{2}\)
3 \(x^{2}-y^{2}=\frac{1^{2}}{4}\)
4 None of these
Explanation:
(A) : Let the intercept \(x\)-axis is \((a, 0)\) and \(y\)-axis is \((0, b)\) and \((\alpha, \beta)\) is the midpoint then, \(\frac{a+0}{2}=\alpha \Rightarrow \mathrm{a}=2 \alpha\) and, \(\frac{b+0}{2}=\beta\) Then, \(\quad l^{2}=(\mathrm{a}-0)^{2}+(0-\mathrm{b})^{2}\) \(\text { ( } l \text { is the length of rod) }\) \(l^{2}=(2 \alpha)^{2}+(2 \beta)^{2}\) \(l^{2}=4 \alpha^{2}+4 \beta^{2}\) \(\alpha^{2}+\beta^{2}=\left(\frac{l}{2}\right)^{2}\) If \((\alpha, \beta)\) be \((x, y)\) then above equation becomes, \(\mathrm{x}^{2}+\mathrm{y}^{2}=\frac{l^{2}}{4}\)
VITEEE-2010
Co-Ordinate system
88435
The coordinates of a point are \(\operatorname{atan}(\theta+\alpha)\) and b \(\tan (\theta+\beta)\), where \(\theta\) is variable, then locus of the point is
1 hyperbola
2 rectangular hyperbola
3 ellipse
4 None of the above
Explanation:
(A) : Given that the co-ordinates point \(\mathrm{x}=\mathrm{a} \tan (\theta+\alpha)\) \(\begin{array}{ll}\Rightarrow \theta+\alpha=\tan ^{-1} \frac{x}{a} \\ y=b \tan (\theta+\beta) \\ \Rightarrow \theta+\beta=\tan ^{-1} \frac{y}{b}\end{array}\) Subtracting equation (i) from (ii) \(\alpha-\beta=\tan ^{-1} \frac{x}{a}-\tan ^{-1} \frac{y}{b}\) \(\Rightarrow \tan ^{-1} \frac{\frac{x}{a}-\frac{y}{b}}{1+\frac{x}{a} \frac{y}{b}}=\alpha-\beta\) \(\Rightarrow x y+a b=(b x-a y) \cot (\alpha-\beta)\) Hence, it is the equation of rectangular hyperbola.
UPSEE-2013
Co-Ordinate system
88436
If the distance of any point \((x, y)\) from the origin is defined as \(d(x, y)=\max \{|x|,|y|\}\) then the locus of the point \((x, y)\) where \(d(x, y)=1\), is
1 A square
2 A circle
3 A triangle
4 None of the above
Explanation:
(A) : Given, \(\mathrm{d}(\mathrm{x}, \mathrm{y})=1\) and, \(\max \{|\mathrm{x}||\mathrm{y}|\}=1\) \(|\mathrm{x}| \leq 1\) and \(|\mathrm{y}| \leq 1\) \(-1 \leq \mathrm{x} \leq 1\) and \(-1 \leq \mathrm{y} \leq 1\) \(\therefore\) locus of \((\mathrm{x}, \mathrm{y})\) is the square having its vertices at \((1\), \(1),(-1,1),(-1,-1)\) and \((1,-1)\)
[JCECE-2016]
Co-Ordinate system
88437
The locus of centre of the circle touching the line \(2 x-y=1\) at \((1,1)\) is
1 \(x+3 y=2\)
2 \(x+2 y=3\)
3 \(x+y=2\)
4 \(2 x-y=1\)
Explanation:
(B) : Given that tine \((2 x-y=1)\) at \((1,1)\) We have to find that the locus centre of the circle touching the line \(=\) ? Let, \((h, k)\) be the centre of circle of the circle touch the line \(2 x-y=1\) at \((1,1)\) \(\therefore \sqrt{(\mathrm{h}-1)^{2}+(\mathrm{k}-1)^{2}}=\frac{2 \mathrm{~h}-\mathrm{k}-1}{\sqrt{4+1}}\) \(\sqrt{5} \sqrt{(\mathrm{h}-1)^{2}+(\mathrm{k}-1)^{2}}=(2 \mathrm{~h}-\mathrm{k}-1)\) On squaring both side, \(5\left(h^{2}+1+k^{2}+1-2 h-2 k\right)=4 h^{2}+k^{2}+1-4 h k-2 h+\) \(2 \mathrm{k}\) \(\mathrm{h}^{2}+4 \mathrm{k}^{2}+9-6 \mathrm{~h}-12 \mathrm{k}+4 \mathrm{hk}=0\) \((\mathrm{h}+2 \mathrm{k}-3)^{2}=0\) \(\mathrm{h}+2 \mathrm{k}-3=0\) \(\mathrm{x}+2 \mathrm{y}=3\) \(=5\left(\mathrm{~h}^{2}+1-2 \mathrm{~h}+\mathrm{k}^{2}-10 \mathrm{~h}-10 \mathrm{k}+10=4 \mathrm{~h}^{2}+\mathrm{k}^{2}-4 \mathrm{hk}\right.\) \(+1-2(2 \mathrm{~h}-\mathrm{k})\) \(\mathrm{h}^{2}+4 \mathrm{k}^{2}-6 \mathrm{~h}-12 \mathrm{k}+9+4 \mathrm{hk}=0\) \((\mathrm{h}+2 \mathrm{k}-3)^{2}=0\) \(\mathrm{h}+2 \mathrm{k}-3=0\) \(\mathrm{h}+2 \mathrm{k}=3\) \(x+2 y=3\)
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Co-Ordinate system
88434
A rod of length I slides with its ends on two perpendicular lines. Then, the locus of its mid point is
1 \(x^{2}+y^{2}=\frac{1^{2}}{4}\)
2 \(x^{2}+y^{2}=\frac{1^{2}}{2}\)
3 \(x^{2}-y^{2}=\frac{1^{2}}{4}\)
4 None of these
Explanation:
(A) : Let the intercept \(x\)-axis is \((a, 0)\) and \(y\)-axis is \((0, b)\) and \((\alpha, \beta)\) is the midpoint then, \(\frac{a+0}{2}=\alpha \Rightarrow \mathrm{a}=2 \alpha\) and, \(\frac{b+0}{2}=\beta\) Then, \(\quad l^{2}=(\mathrm{a}-0)^{2}+(0-\mathrm{b})^{2}\) \(\text { ( } l \text { is the length of rod) }\) \(l^{2}=(2 \alpha)^{2}+(2 \beta)^{2}\) \(l^{2}=4 \alpha^{2}+4 \beta^{2}\) \(\alpha^{2}+\beta^{2}=\left(\frac{l}{2}\right)^{2}\) If \((\alpha, \beta)\) be \((x, y)\) then above equation becomes, \(\mathrm{x}^{2}+\mathrm{y}^{2}=\frac{l^{2}}{4}\)
VITEEE-2010
Co-Ordinate system
88435
The coordinates of a point are \(\operatorname{atan}(\theta+\alpha)\) and b \(\tan (\theta+\beta)\), where \(\theta\) is variable, then locus of the point is
1 hyperbola
2 rectangular hyperbola
3 ellipse
4 None of the above
Explanation:
(A) : Given that the co-ordinates point \(\mathrm{x}=\mathrm{a} \tan (\theta+\alpha)\) \(\begin{array}{ll}\Rightarrow \theta+\alpha=\tan ^{-1} \frac{x}{a} \\ y=b \tan (\theta+\beta) \\ \Rightarrow \theta+\beta=\tan ^{-1} \frac{y}{b}\end{array}\) Subtracting equation (i) from (ii) \(\alpha-\beta=\tan ^{-1} \frac{x}{a}-\tan ^{-1} \frac{y}{b}\) \(\Rightarrow \tan ^{-1} \frac{\frac{x}{a}-\frac{y}{b}}{1+\frac{x}{a} \frac{y}{b}}=\alpha-\beta\) \(\Rightarrow x y+a b=(b x-a y) \cot (\alpha-\beta)\) Hence, it is the equation of rectangular hyperbola.
UPSEE-2013
Co-Ordinate system
88436
If the distance of any point \((x, y)\) from the origin is defined as \(d(x, y)=\max \{|x|,|y|\}\) then the locus of the point \((x, y)\) where \(d(x, y)=1\), is
1 A square
2 A circle
3 A triangle
4 None of the above
Explanation:
(A) : Given, \(\mathrm{d}(\mathrm{x}, \mathrm{y})=1\) and, \(\max \{|\mathrm{x}||\mathrm{y}|\}=1\) \(|\mathrm{x}| \leq 1\) and \(|\mathrm{y}| \leq 1\) \(-1 \leq \mathrm{x} \leq 1\) and \(-1 \leq \mathrm{y} \leq 1\) \(\therefore\) locus of \((\mathrm{x}, \mathrm{y})\) is the square having its vertices at \((1\), \(1),(-1,1),(-1,-1)\) and \((1,-1)\)
[JCECE-2016]
Co-Ordinate system
88437
The locus of centre of the circle touching the line \(2 x-y=1\) at \((1,1)\) is
1 \(x+3 y=2\)
2 \(x+2 y=3\)
3 \(x+y=2\)
4 \(2 x-y=1\)
Explanation:
(B) : Given that tine \((2 x-y=1)\) at \((1,1)\) We have to find that the locus centre of the circle touching the line \(=\) ? Let, \((h, k)\) be the centre of circle of the circle touch the line \(2 x-y=1\) at \((1,1)\) \(\therefore \sqrt{(\mathrm{h}-1)^{2}+(\mathrm{k}-1)^{2}}=\frac{2 \mathrm{~h}-\mathrm{k}-1}{\sqrt{4+1}}\) \(\sqrt{5} \sqrt{(\mathrm{h}-1)^{2}+(\mathrm{k}-1)^{2}}=(2 \mathrm{~h}-\mathrm{k}-1)\) On squaring both side, \(5\left(h^{2}+1+k^{2}+1-2 h-2 k\right)=4 h^{2}+k^{2}+1-4 h k-2 h+\) \(2 \mathrm{k}\) \(\mathrm{h}^{2}+4 \mathrm{k}^{2}+9-6 \mathrm{~h}-12 \mathrm{k}+4 \mathrm{hk}=0\) \((\mathrm{h}+2 \mathrm{k}-3)^{2}=0\) \(\mathrm{h}+2 \mathrm{k}-3=0\) \(\mathrm{x}+2 \mathrm{y}=3\) \(=5\left(\mathrm{~h}^{2}+1-2 \mathrm{~h}+\mathrm{k}^{2}-10 \mathrm{~h}-10 \mathrm{k}+10=4 \mathrm{~h}^{2}+\mathrm{k}^{2}-4 \mathrm{hk}\right.\) \(+1-2(2 \mathrm{~h}-\mathrm{k})\) \(\mathrm{h}^{2}+4 \mathrm{k}^{2}-6 \mathrm{~h}-12 \mathrm{k}+9+4 \mathrm{hk}=0\) \((\mathrm{h}+2 \mathrm{k}-3)^{2}=0\) \(\mathrm{h}+2 \mathrm{k}-3=0\) \(\mathrm{h}+2 \mathrm{k}=3\) \(x+2 y=3\)
