88477
A point \(P(-3,-2)\) is such that the sum of squares of its distances from the co-ordinate axes is equal to the square of its distance from the line \(x-y=1\). Then the equation of the locus of \(P\) is
1 \(x^{2}+y^{2}-2 x y-2 x-2 y-1=0\)
2 \(x^{2}+y^{2}+2 x y+2 x+2 y+1=0\)
3 \(x^{2}+y^{2}+2 x y+2 x-2 y-1=0\)
4 \(x^{2}+y^{2}-2 x y+2 x-2 y+1=0\)
Explanation:
(C) : It is given that the sum of square of distance of point \(p(x, y)\) is equal to the square of its distance from the line \(\mathrm{x}-\mathrm{y}=1\), So, \(\mathrm{x}^{2}+\mathrm{y}^{2}=\frac{(\mathrm{x}-\mathrm{y}-1)^{2}}{2}\) \(2 x^{2}+2 y^{2}=x^{2}+y^{2}+1-2 x y-2 x+2 y\) \(x^{2}+y^{2}+2 x y+2 x-2 y-1=0\) So, equation of the locus of \(P x^{2}+y^{2}+2 x y+2 x-2 y\) \(-1=0\)
AP EAMCET-2020-22.09.2020
Co-Ordinate system
88478
If the sum of the distances from a variable point \(P\) to the given points \(A(1,0)\) and \(B(0,1)\) is \(\mathbf{2}\), then the locus of \(P\) is
1 \(3 x^{2}+3 y^{2}-4 x-4 y=0\)
2 \(16 x^{2}+7 y^{2}-64 x-48 y=0\)
3 \(3 x^{2}+2 x y+3 y^{2}-4 x-4 y=0\)
4 \(16 x^{2}+38 x y+7 y^{2}-64 x-48 y=0\)
Explanation:
(C) : Given that the sum of the distance a variable point \(P\) the given points \(A(1,0)\) and \(B(0,1)\) is 2 Let, \(\mathrm{P}\) is \((\mathrm{x}, \mathrm{y})\) \(\mathrm{AP}+\mathrm{BP}=2\) Given that \(\sqrt{(x-1)^{2}+y^{2}+} \sqrt{x^{2}+(y-1)^{2}}=2\) \(\sqrt{(x-1)^{2}+y^{2}}=2-\sqrt{x^{2}+(y-1)^{2}}\) \((x-1)^{2}+y^{2}=4+x^{2}+\left(y^{2}+1-2 y\right)\) \(-4 \sqrt{x^{2}+(y-1)^{2}}\) \(x^{2}+y^{2}+1-2 x=4+x^{2}+\left(y^{2}+1-2 y\right)\) \(-4 \sqrt{x^{2}+(y-1)^{2}}\) \(2 y-2 x=4-4 \sqrt{x^{2}+(y-1)^{2}}\) \(y-x-2=-2 \sqrt{x^{2}+(y-1)^{2}}\) \(x^{2}+y^{2}+4-2 x y+4 x-4 y=4\left(x^{2}+y^{2}+1-2 y\right)\) \(x^{2}+y^{2}+4-2 x y+4 x-4 y=4 x^{2}+4 y^{2}+4-8 y\) \(3 x^{2}+3 y^{2}-4 x-4 y+2 x y=0\)
AP EAMCET-2019-21.04.2019
Co-Ordinate system
88479
The locus of the mid-points of the chord of the circle \(x^{2}+y^{2}=4\) which subtends a right angle at the origin, is
1 \(x+y=2\)
2 \(x^{2}+y^{2}=1\)
3 \(x^{2}+y^{2}=2\)
4 \(x+y=1\)
Explanation:
(C) : We have the locus of mid-point of chord and we know perpendicular drawn from centre to the chord, it bisects the chord. Given, Also, \(\mathrm{OC}\) bisects the \(\angle \mathrm{AOB}\) \(\therefore \quad \mathrm{h}^{2}+\mathrm{k}^{2}=\mathrm{OC}^{2}\) \(\Rightarrow \quad \mathrm{h}^{2}+\mathrm{k}^{2}=(\sqrt{2})^{2}\) or \(x^{2}+y^{2}=2\) is required equation of locus of mid-point of chord subtending right angle at centre.
Manipal-2012
Co-Ordinate system
88480
If the equation \(12 x^{2}+7 x y-p y^{2}-18 x+q y+6=0\) represents a pair of perpendicular straight lines, then
1 \(p=12, q=-1\)
2 \(p=-12, q=1\)
3 \(p=12, q=1\)
4 \(p=1, q=1\)
Explanation:
(C) : A general equation of second degree (conic sections) \(a x^{2}+2 h x y+b y^{2}+2 g x+2 f y+c=0\) will represent pair of perpendicular straight lines if- \(\left|\begin{array}{lll}\mathrm{a} & \mathrm{h} & \mathrm{g} \\ \mathrm{h} & \mathrm{b} & \mathrm{f} \\ \mathrm{g} & \mathrm{f} & \mathrm{c}\end{array}\right|=0\) and \(\mathrm{a}+\mathrm{b}=0\) thus \(\left|\begin{array}{ccc}12 & 7 / 2 &-9 \\ 7 / 2& -p &q / 2 \\ -9& q / 2& 6\end{array}\right|=0\) And \(12-\mathrm{p}=0\) \(\Rightarrow \mathrm{p}=12\) put the value of \(p\) in the determinant equation- \(\left|\begin{array}{ccc}12 &7 / 2& -9 \\ 7 / 2& -12 &\mathrm{q} / 2 \\ -9 &\mathrm{q} / 2 &6\end{array}\right|=0\) \(\Rightarrow 12\left(-72+\frac{\mathrm{q}^{2}}{4}\right)-\frac{7}{2}\left(-21 \frac{9 \mathrm{q}}{2}\right)-9\left(\frac{7 \mathrm{q}}{4}-108\right)=0\) \(\Rightarrow-864-3 q^{2}-\frac{147}{2}-\frac{63 q}{4}-\frac{63 q}{4}+972=0\) \(\Rightarrow \quad q=1\) \(\quad p=12\)
Manipal-2011
Co-Ordinate system
88481
The sum of the squares of the distances of a moving point from 2 fixed points \(A(a, 0)\) and \(B\) \((-a, 0)\) is equal to a constant \(2 c^{2}\), then the equation of its locus is
(A): Let \(\mathrm{P}(\mathrm{h}, \mathrm{k})\) be any position of the moving point let \(A(a, o), B(-a, o)\) be the given points then \(\mathrm{PA}^{2}+\mathrm{PB}^{2}=2 \mathrm{C}^{2}\) \(\left(n_{2}-a\right)^{2}+\left(k_{2}-o\right)^{2}+(n+a)^{2}+\left(k_{2}-o\right)^{2}=2 c^{2}\) \(n^{2}-2 a n+a^{2}+k^{2}+n^{2}+2 a n+a^{2}+k^{2}=2 c^{2}\) \(2 \mathrm{n}^{2}+2 \mathrm{a}^{2}+2 \mathrm{k}^{2}=2 \mathrm{c}^{2}\) \(\mathrm{n}^{2}+\mathrm{k}^{2}=\mathrm{c}^{2}-\mathrm{a}^{2}\) Hence, locus of \((n, k)\) is \(x^{2}+y^{2}=c^{2}-a^{2}\).
88477
A point \(P(-3,-2)\) is such that the sum of squares of its distances from the co-ordinate axes is equal to the square of its distance from the line \(x-y=1\). Then the equation of the locus of \(P\) is
1 \(x^{2}+y^{2}-2 x y-2 x-2 y-1=0\)
2 \(x^{2}+y^{2}+2 x y+2 x+2 y+1=0\)
3 \(x^{2}+y^{2}+2 x y+2 x-2 y-1=0\)
4 \(x^{2}+y^{2}-2 x y+2 x-2 y+1=0\)
Explanation:
(C) : It is given that the sum of square of distance of point \(p(x, y)\) is equal to the square of its distance from the line \(\mathrm{x}-\mathrm{y}=1\), So, \(\mathrm{x}^{2}+\mathrm{y}^{2}=\frac{(\mathrm{x}-\mathrm{y}-1)^{2}}{2}\) \(2 x^{2}+2 y^{2}=x^{2}+y^{2}+1-2 x y-2 x+2 y\) \(x^{2}+y^{2}+2 x y+2 x-2 y-1=0\) So, equation of the locus of \(P x^{2}+y^{2}+2 x y+2 x-2 y\) \(-1=0\)
AP EAMCET-2020-22.09.2020
Co-Ordinate system
88478
If the sum of the distances from a variable point \(P\) to the given points \(A(1,0)\) and \(B(0,1)\) is \(\mathbf{2}\), then the locus of \(P\) is
1 \(3 x^{2}+3 y^{2}-4 x-4 y=0\)
2 \(16 x^{2}+7 y^{2}-64 x-48 y=0\)
3 \(3 x^{2}+2 x y+3 y^{2}-4 x-4 y=0\)
4 \(16 x^{2}+38 x y+7 y^{2}-64 x-48 y=0\)
Explanation:
(C) : Given that the sum of the distance a variable point \(P\) the given points \(A(1,0)\) and \(B(0,1)\) is 2 Let, \(\mathrm{P}\) is \((\mathrm{x}, \mathrm{y})\) \(\mathrm{AP}+\mathrm{BP}=2\) Given that \(\sqrt{(x-1)^{2}+y^{2}+} \sqrt{x^{2}+(y-1)^{2}}=2\) \(\sqrt{(x-1)^{2}+y^{2}}=2-\sqrt{x^{2}+(y-1)^{2}}\) \((x-1)^{2}+y^{2}=4+x^{2}+\left(y^{2}+1-2 y\right)\) \(-4 \sqrt{x^{2}+(y-1)^{2}}\) \(x^{2}+y^{2}+1-2 x=4+x^{2}+\left(y^{2}+1-2 y\right)\) \(-4 \sqrt{x^{2}+(y-1)^{2}}\) \(2 y-2 x=4-4 \sqrt{x^{2}+(y-1)^{2}}\) \(y-x-2=-2 \sqrt{x^{2}+(y-1)^{2}}\) \(x^{2}+y^{2}+4-2 x y+4 x-4 y=4\left(x^{2}+y^{2}+1-2 y\right)\) \(x^{2}+y^{2}+4-2 x y+4 x-4 y=4 x^{2}+4 y^{2}+4-8 y\) \(3 x^{2}+3 y^{2}-4 x-4 y+2 x y=0\)
AP EAMCET-2019-21.04.2019
Co-Ordinate system
88479
The locus of the mid-points of the chord of the circle \(x^{2}+y^{2}=4\) which subtends a right angle at the origin, is
1 \(x+y=2\)
2 \(x^{2}+y^{2}=1\)
3 \(x^{2}+y^{2}=2\)
4 \(x+y=1\)
Explanation:
(C) : We have the locus of mid-point of chord and we know perpendicular drawn from centre to the chord, it bisects the chord. Given, Also, \(\mathrm{OC}\) bisects the \(\angle \mathrm{AOB}\) \(\therefore \quad \mathrm{h}^{2}+\mathrm{k}^{2}=\mathrm{OC}^{2}\) \(\Rightarrow \quad \mathrm{h}^{2}+\mathrm{k}^{2}=(\sqrt{2})^{2}\) or \(x^{2}+y^{2}=2\) is required equation of locus of mid-point of chord subtending right angle at centre.
Manipal-2012
Co-Ordinate system
88480
If the equation \(12 x^{2}+7 x y-p y^{2}-18 x+q y+6=0\) represents a pair of perpendicular straight lines, then
1 \(p=12, q=-1\)
2 \(p=-12, q=1\)
3 \(p=12, q=1\)
4 \(p=1, q=1\)
Explanation:
(C) : A general equation of second degree (conic sections) \(a x^{2}+2 h x y+b y^{2}+2 g x+2 f y+c=0\) will represent pair of perpendicular straight lines if- \(\left|\begin{array}{lll}\mathrm{a} & \mathrm{h} & \mathrm{g} \\ \mathrm{h} & \mathrm{b} & \mathrm{f} \\ \mathrm{g} & \mathrm{f} & \mathrm{c}\end{array}\right|=0\) and \(\mathrm{a}+\mathrm{b}=0\) thus \(\left|\begin{array}{ccc}12 & 7 / 2 &-9 \\ 7 / 2& -p &q / 2 \\ -9& q / 2& 6\end{array}\right|=0\) And \(12-\mathrm{p}=0\) \(\Rightarrow \mathrm{p}=12\) put the value of \(p\) in the determinant equation- \(\left|\begin{array}{ccc}12 &7 / 2& -9 \\ 7 / 2& -12 &\mathrm{q} / 2 \\ -9 &\mathrm{q} / 2 &6\end{array}\right|=0\) \(\Rightarrow 12\left(-72+\frac{\mathrm{q}^{2}}{4}\right)-\frac{7}{2}\left(-21 \frac{9 \mathrm{q}}{2}\right)-9\left(\frac{7 \mathrm{q}}{4}-108\right)=0\) \(\Rightarrow-864-3 q^{2}-\frac{147}{2}-\frac{63 q}{4}-\frac{63 q}{4}+972=0\) \(\Rightarrow \quad q=1\) \(\quad p=12\)
Manipal-2011
Co-Ordinate system
88481
The sum of the squares of the distances of a moving point from 2 fixed points \(A(a, 0)\) and \(B\) \((-a, 0)\) is equal to a constant \(2 c^{2}\), then the equation of its locus is
(A): Let \(\mathrm{P}(\mathrm{h}, \mathrm{k})\) be any position of the moving point let \(A(a, o), B(-a, o)\) be the given points then \(\mathrm{PA}^{2}+\mathrm{PB}^{2}=2 \mathrm{C}^{2}\) \(\left(n_{2}-a\right)^{2}+\left(k_{2}-o\right)^{2}+(n+a)^{2}+\left(k_{2}-o\right)^{2}=2 c^{2}\) \(n^{2}-2 a n+a^{2}+k^{2}+n^{2}+2 a n+a^{2}+k^{2}=2 c^{2}\) \(2 \mathrm{n}^{2}+2 \mathrm{a}^{2}+2 \mathrm{k}^{2}=2 \mathrm{c}^{2}\) \(\mathrm{n}^{2}+\mathrm{k}^{2}=\mathrm{c}^{2}-\mathrm{a}^{2}\) Hence, locus of \((n, k)\) is \(x^{2}+y^{2}=c^{2}-a^{2}\).
88477
A point \(P(-3,-2)\) is such that the sum of squares of its distances from the co-ordinate axes is equal to the square of its distance from the line \(x-y=1\). Then the equation of the locus of \(P\) is
1 \(x^{2}+y^{2}-2 x y-2 x-2 y-1=0\)
2 \(x^{2}+y^{2}+2 x y+2 x+2 y+1=0\)
3 \(x^{2}+y^{2}+2 x y+2 x-2 y-1=0\)
4 \(x^{2}+y^{2}-2 x y+2 x-2 y+1=0\)
Explanation:
(C) : It is given that the sum of square of distance of point \(p(x, y)\) is equal to the square of its distance from the line \(\mathrm{x}-\mathrm{y}=1\), So, \(\mathrm{x}^{2}+\mathrm{y}^{2}=\frac{(\mathrm{x}-\mathrm{y}-1)^{2}}{2}\) \(2 x^{2}+2 y^{2}=x^{2}+y^{2}+1-2 x y-2 x+2 y\) \(x^{2}+y^{2}+2 x y+2 x-2 y-1=0\) So, equation of the locus of \(P x^{2}+y^{2}+2 x y+2 x-2 y\) \(-1=0\)
AP EAMCET-2020-22.09.2020
Co-Ordinate system
88478
If the sum of the distances from a variable point \(P\) to the given points \(A(1,0)\) and \(B(0,1)\) is \(\mathbf{2}\), then the locus of \(P\) is
1 \(3 x^{2}+3 y^{2}-4 x-4 y=0\)
2 \(16 x^{2}+7 y^{2}-64 x-48 y=0\)
3 \(3 x^{2}+2 x y+3 y^{2}-4 x-4 y=0\)
4 \(16 x^{2}+38 x y+7 y^{2}-64 x-48 y=0\)
Explanation:
(C) : Given that the sum of the distance a variable point \(P\) the given points \(A(1,0)\) and \(B(0,1)\) is 2 Let, \(\mathrm{P}\) is \((\mathrm{x}, \mathrm{y})\) \(\mathrm{AP}+\mathrm{BP}=2\) Given that \(\sqrt{(x-1)^{2}+y^{2}+} \sqrt{x^{2}+(y-1)^{2}}=2\) \(\sqrt{(x-1)^{2}+y^{2}}=2-\sqrt{x^{2}+(y-1)^{2}}\) \((x-1)^{2}+y^{2}=4+x^{2}+\left(y^{2}+1-2 y\right)\) \(-4 \sqrt{x^{2}+(y-1)^{2}}\) \(x^{2}+y^{2}+1-2 x=4+x^{2}+\left(y^{2}+1-2 y\right)\) \(-4 \sqrt{x^{2}+(y-1)^{2}}\) \(2 y-2 x=4-4 \sqrt{x^{2}+(y-1)^{2}}\) \(y-x-2=-2 \sqrt{x^{2}+(y-1)^{2}}\) \(x^{2}+y^{2}+4-2 x y+4 x-4 y=4\left(x^{2}+y^{2}+1-2 y\right)\) \(x^{2}+y^{2}+4-2 x y+4 x-4 y=4 x^{2}+4 y^{2}+4-8 y\) \(3 x^{2}+3 y^{2}-4 x-4 y+2 x y=0\)
AP EAMCET-2019-21.04.2019
Co-Ordinate system
88479
The locus of the mid-points of the chord of the circle \(x^{2}+y^{2}=4\) which subtends a right angle at the origin, is
1 \(x+y=2\)
2 \(x^{2}+y^{2}=1\)
3 \(x^{2}+y^{2}=2\)
4 \(x+y=1\)
Explanation:
(C) : We have the locus of mid-point of chord and we know perpendicular drawn from centre to the chord, it bisects the chord. Given, Also, \(\mathrm{OC}\) bisects the \(\angle \mathrm{AOB}\) \(\therefore \quad \mathrm{h}^{2}+\mathrm{k}^{2}=\mathrm{OC}^{2}\) \(\Rightarrow \quad \mathrm{h}^{2}+\mathrm{k}^{2}=(\sqrt{2})^{2}\) or \(x^{2}+y^{2}=2\) is required equation of locus of mid-point of chord subtending right angle at centre.
Manipal-2012
Co-Ordinate system
88480
If the equation \(12 x^{2}+7 x y-p y^{2}-18 x+q y+6=0\) represents a pair of perpendicular straight lines, then
1 \(p=12, q=-1\)
2 \(p=-12, q=1\)
3 \(p=12, q=1\)
4 \(p=1, q=1\)
Explanation:
(C) : A general equation of second degree (conic sections) \(a x^{2}+2 h x y+b y^{2}+2 g x+2 f y+c=0\) will represent pair of perpendicular straight lines if- \(\left|\begin{array}{lll}\mathrm{a} & \mathrm{h} & \mathrm{g} \\ \mathrm{h} & \mathrm{b} & \mathrm{f} \\ \mathrm{g} & \mathrm{f} & \mathrm{c}\end{array}\right|=0\) and \(\mathrm{a}+\mathrm{b}=0\) thus \(\left|\begin{array}{ccc}12 & 7 / 2 &-9 \\ 7 / 2& -p &q / 2 \\ -9& q / 2& 6\end{array}\right|=0\) And \(12-\mathrm{p}=0\) \(\Rightarrow \mathrm{p}=12\) put the value of \(p\) in the determinant equation- \(\left|\begin{array}{ccc}12 &7 / 2& -9 \\ 7 / 2& -12 &\mathrm{q} / 2 \\ -9 &\mathrm{q} / 2 &6\end{array}\right|=0\) \(\Rightarrow 12\left(-72+\frac{\mathrm{q}^{2}}{4}\right)-\frac{7}{2}\left(-21 \frac{9 \mathrm{q}}{2}\right)-9\left(\frac{7 \mathrm{q}}{4}-108\right)=0\) \(\Rightarrow-864-3 q^{2}-\frac{147}{2}-\frac{63 q}{4}-\frac{63 q}{4}+972=0\) \(\Rightarrow \quad q=1\) \(\quad p=12\)
Manipal-2011
Co-Ordinate system
88481
The sum of the squares of the distances of a moving point from 2 fixed points \(A(a, 0)\) and \(B\) \((-a, 0)\) is equal to a constant \(2 c^{2}\), then the equation of its locus is
(A): Let \(\mathrm{P}(\mathrm{h}, \mathrm{k})\) be any position of the moving point let \(A(a, o), B(-a, o)\) be the given points then \(\mathrm{PA}^{2}+\mathrm{PB}^{2}=2 \mathrm{C}^{2}\) \(\left(n_{2}-a\right)^{2}+\left(k_{2}-o\right)^{2}+(n+a)^{2}+\left(k_{2}-o\right)^{2}=2 c^{2}\) \(n^{2}-2 a n+a^{2}+k^{2}+n^{2}+2 a n+a^{2}+k^{2}=2 c^{2}\) \(2 \mathrm{n}^{2}+2 \mathrm{a}^{2}+2 \mathrm{k}^{2}=2 \mathrm{c}^{2}\) \(\mathrm{n}^{2}+\mathrm{k}^{2}=\mathrm{c}^{2}-\mathrm{a}^{2}\) Hence, locus of \((n, k)\) is \(x^{2}+y^{2}=c^{2}-a^{2}\).
NEET Test Series from KOTA - 10 Papers In MS WORD
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Co-Ordinate system
88477
A point \(P(-3,-2)\) is such that the sum of squares of its distances from the co-ordinate axes is equal to the square of its distance from the line \(x-y=1\). Then the equation of the locus of \(P\) is
1 \(x^{2}+y^{2}-2 x y-2 x-2 y-1=0\)
2 \(x^{2}+y^{2}+2 x y+2 x+2 y+1=0\)
3 \(x^{2}+y^{2}+2 x y+2 x-2 y-1=0\)
4 \(x^{2}+y^{2}-2 x y+2 x-2 y+1=0\)
Explanation:
(C) : It is given that the sum of square of distance of point \(p(x, y)\) is equal to the square of its distance from the line \(\mathrm{x}-\mathrm{y}=1\), So, \(\mathrm{x}^{2}+\mathrm{y}^{2}=\frac{(\mathrm{x}-\mathrm{y}-1)^{2}}{2}\) \(2 x^{2}+2 y^{2}=x^{2}+y^{2}+1-2 x y-2 x+2 y\) \(x^{2}+y^{2}+2 x y+2 x-2 y-1=0\) So, equation of the locus of \(P x^{2}+y^{2}+2 x y+2 x-2 y\) \(-1=0\)
AP EAMCET-2020-22.09.2020
Co-Ordinate system
88478
If the sum of the distances from a variable point \(P\) to the given points \(A(1,0)\) and \(B(0,1)\) is \(\mathbf{2}\), then the locus of \(P\) is
1 \(3 x^{2}+3 y^{2}-4 x-4 y=0\)
2 \(16 x^{2}+7 y^{2}-64 x-48 y=0\)
3 \(3 x^{2}+2 x y+3 y^{2}-4 x-4 y=0\)
4 \(16 x^{2}+38 x y+7 y^{2}-64 x-48 y=0\)
Explanation:
(C) : Given that the sum of the distance a variable point \(P\) the given points \(A(1,0)\) and \(B(0,1)\) is 2 Let, \(\mathrm{P}\) is \((\mathrm{x}, \mathrm{y})\) \(\mathrm{AP}+\mathrm{BP}=2\) Given that \(\sqrt{(x-1)^{2}+y^{2}+} \sqrt{x^{2}+(y-1)^{2}}=2\) \(\sqrt{(x-1)^{2}+y^{2}}=2-\sqrt{x^{2}+(y-1)^{2}}\) \((x-1)^{2}+y^{2}=4+x^{2}+\left(y^{2}+1-2 y\right)\) \(-4 \sqrt{x^{2}+(y-1)^{2}}\) \(x^{2}+y^{2}+1-2 x=4+x^{2}+\left(y^{2}+1-2 y\right)\) \(-4 \sqrt{x^{2}+(y-1)^{2}}\) \(2 y-2 x=4-4 \sqrt{x^{2}+(y-1)^{2}}\) \(y-x-2=-2 \sqrt{x^{2}+(y-1)^{2}}\) \(x^{2}+y^{2}+4-2 x y+4 x-4 y=4\left(x^{2}+y^{2}+1-2 y\right)\) \(x^{2}+y^{2}+4-2 x y+4 x-4 y=4 x^{2}+4 y^{2}+4-8 y\) \(3 x^{2}+3 y^{2}-4 x-4 y+2 x y=0\)
AP EAMCET-2019-21.04.2019
Co-Ordinate system
88479
The locus of the mid-points of the chord of the circle \(x^{2}+y^{2}=4\) which subtends a right angle at the origin, is
1 \(x+y=2\)
2 \(x^{2}+y^{2}=1\)
3 \(x^{2}+y^{2}=2\)
4 \(x+y=1\)
Explanation:
(C) : We have the locus of mid-point of chord and we know perpendicular drawn from centre to the chord, it bisects the chord. Given, Also, \(\mathrm{OC}\) bisects the \(\angle \mathrm{AOB}\) \(\therefore \quad \mathrm{h}^{2}+\mathrm{k}^{2}=\mathrm{OC}^{2}\) \(\Rightarrow \quad \mathrm{h}^{2}+\mathrm{k}^{2}=(\sqrt{2})^{2}\) or \(x^{2}+y^{2}=2\) is required equation of locus of mid-point of chord subtending right angle at centre.
Manipal-2012
Co-Ordinate system
88480
If the equation \(12 x^{2}+7 x y-p y^{2}-18 x+q y+6=0\) represents a pair of perpendicular straight lines, then
1 \(p=12, q=-1\)
2 \(p=-12, q=1\)
3 \(p=12, q=1\)
4 \(p=1, q=1\)
Explanation:
(C) : A general equation of second degree (conic sections) \(a x^{2}+2 h x y+b y^{2}+2 g x+2 f y+c=0\) will represent pair of perpendicular straight lines if- \(\left|\begin{array}{lll}\mathrm{a} & \mathrm{h} & \mathrm{g} \\ \mathrm{h} & \mathrm{b} & \mathrm{f} \\ \mathrm{g} & \mathrm{f} & \mathrm{c}\end{array}\right|=0\) and \(\mathrm{a}+\mathrm{b}=0\) thus \(\left|\begin{array}{ccc}12 & 7 / 2 &-9 \\ 7 / 2& -p &q / 2 \\ -9& q / 2& 6\end{array}\right|=0\) And \(12-\mathrm{p}=0\) \(\Rightarrow \mathrm{p}=12\) put the value of \(p\) in the determinant equation- \(\left|\begin{array}{ccc}12 &7 / 2& -9 \\ 7 / 2& -12 &\mathrm{q} / 2 \\ -9 &\mathrm{q} / 2 &6\end{array}\right|=0\) \(\Rightarrow 12\left(-72+\frac{\mathrm{q}^{2}}{4}\right)-\frac{7}{2}\left(-21 \frac{9 \mathrm{q}}{2}\right)-9\left(\frac{7 \mathrm{q}}{4}-108\right)=0\) \(\Rightarrow-864-3 q^{2}-\frac{147}{2}-\frac{63 q}{4}-\frac{63 q}{4}+972=0\) \(\Rightarrow \quad q=1\) \(\quad p=12\)
Manipal-2011
Co-Ordinate system
88481
The sum of the squares of the distances of a moving point from 2 fixed points \(A(a, 0)\) and \(B\) \((-a, 0)\) is equal to a constant \(2 c^{2}\), then the equation of its locus is
(A): Let \(\mathrm{P}(\mathrm{h}, \mathrm{k})\) be any position of the moving point let \(A(a, o), B(-a, o)\) be the given points then \(\mathrm{PA}^{2}+\mathrm{PB}^{2}=2 \mathrm{C}^{2}\) \(\left(n_{2}-a\right)^{2}+\left(k_{2}-o\right)^{2}+(n+a)^{2}+\left(k_{2}-o\right)^{2}=2 c^{2}\) \(n^{2}-2 a n+a^{2}+k^{2}+n^{2}+2 a n+a^{2}+k^{2}=2 c^{2}\) \(2 \mathrm{n}^{2}+2 \mathrm{a}^{2}+2 \mathrm{k}^{2}=2 \mathrm{c}^{2}\) \(\mathrm{n}^{2}+\mathrm{k}^{2}=\mathrm{c}^{2}-\mathrm{a}^{2}\) Hence, locus of \((n, k)\) is \(x^{2}+y^{2}=c^{2}-a^{2}\).
88477
A point \(P(-3,-2)\) is such that the sum of squares of its distances from the co-ordinate axes is equal to the square of its distance from the line \(x-y=1\). Then the equation of the locus of \(P\) is
1 \(x^{2}+y^{2}-2 x y-2 x-2 y-1=0\)
2 \(x^{2}+y^{2}+2 x y+2 x+2 y+1=0\)
3 \(x^{2}+y^{2}+2 x y+2 x-2 y-1=0\)
4 \(x^{2}+y^{2}-2 x y+2 x-2 y+1=0\)
Explanation:
(C) : It is given that the sum of square of distance of point \(p(x, y)\) is equal to the square of its distance from the line \(\mathrm{x}-\mathrm{y}=1\), So, \(\mathrm{x}^{2}+\mathrm{y}^{2}=\frac{(\mathrm{x}-\mathrm{y}-1)^{2}}{2}\) \(2 x^{2}+2 y^{2}=x^{2}+y^{2}+1-2 x y-2 x+2 y\) \(x^{2}+y^{2}+2 x y+2 x-2 y-1=0\) So, equation of the locus of \(P x^{2}+y^{2}+2 x y+2 x-2 y\) \(-1=0\)
AP EAMCET-2020-22.09.2020
Co-Ordinate system
88478
If the sum of the distances from a variable point \(P\) to the given points \(A(1,0)\) and \(B(0,1)\) is \(\mathbf{2}\), then the locus of \(P\) is
1 \(3 x^{2}+3 y^{2}-4 x-4 y=0\)
2 \(16 x^{2}+7 y^{2}-64 x-48 y=0\)
3 \(3 x^{2}+2 x y+3 y^{2}-4 x-4 y=0\)
4 \(16 x^{2}+38 x y+7 y^{2}-64 x-48 y=0\)
Explanation:
(C) : Given that the sum of the distance a variable point \(P\) the given points \(A(1,0)\) and \(B(0,1)\) is 2 Let, \(\mathrm{P}\) is \((\mathrm{x}, \mathrm{y})\) \(\mathrm{AP}+\mathrm{BP}=2\) Given that \(\sqrt{(x-1)^{2}+y^{2}+} \sqrt{x^{2}+(y-1)^{2}}=2\) \(\sqrt{(x-1)^{2}+y^{2}}=2-\sqrt{x^{2}+(y-1)^{2}}\) \((x-1)^{2}+y^{2}=4+x^{2}+\left(y^{2}+1-2 y\right)\) \(-4 \sqrt{x^{2}+(y-1)^{2}}\) \(x^{2}+y^{2}+1-2 x=4+x^{2}+\left(y^{2}+1-2 y\right)\) \(-4 \sqrt{x^{2}+(y-1)^{2}}\) \(2 y-2 x=4-4 \sqrt{x^{2}+(y-1)^{2}}\) \(y-x-2=-2 \sqrt{x^{2}+(y-1)^{2}}\) \(x^{2}+y^{2}+4-2 x y+4 x-4 y=4\left(x^{2}+y^{2}+1-2 y\right)\) \(x^{2}+y^{2}+4-2 x y+4 x-4 y=4 x^{2}+4 y^{2}+4-8 y\) \(3 x^{2}+3 y^{2}-4 x-4 y+2 x y=0\)
AP EAMCET-2019-21.04.2019
Co-Ordinate system
88479
The locus of the mid-points of the chord of the circle \(x^{2}+y^{2}=4\) which subtends a right angle at the origin, is
1 \(x+y=2\)
2 \(x^{2}+y^{2}=1\)
3 \(x^{2}+y^{2}=2\)
4 \(x+y=1\)
Explanation:
(C) : We have the locus of mid-point of chord and we know perpendicular drawn from centre to the chord, it bisects the chord. Given, Also, \(\mathrm{OC}\) bisects the \(\angle \mathrm{AOB}\) \(\therefore \quad \mathrm{h}^{2}+\mathrm{k}^{2}=\mathrm{OC}^{2}\) \(\Rightarrow \quad \mathrm{h}^{2}+\mathrm{k}^{2}=(\sqrt{2})^{2}\) or \(x^{2}+y^{2}=2\) is required equation of locus of mid-point of chord subtending right angle at centre.
Manipal-2012
Co-Ordinate system
88480
If the equation \(12 x^{2}+7 x y-p y^{2}-18 x+q y+6=0\) represents a pair of perpendicular straight lines, then
1 \(p=12, q=-1\)
2 \(p=-12, q=1\)
3 \(p=12, q=1\)
4 \(p=1, q=1\)
Explanation:
(C) : A general equation of second degree (conic sections) \(a x^{2}+2 h x y+b y^{2}+2 g x+2 f y+c=0\) will represent pair of perpendicular straight lines if- \(\left|\begin{array}{lll}\mathrm{a} & \mathrm{h} & \mathrm{g} \\ \mathrm{h} & \mathrm{b} & \mathrm{f} \\ \mathrm{g} & \mathrm{f} & \mathrm{c}\end{array}\right|=0\) and \(\mathrm{a}+\mathrm{b}=0\) thus \(\left|\begin{array}{ccc}12 & 7 / 2 &-9 \\ 7 / 2& -p &q / 2 \\ -9& q / 2& 6\end{array}\right|=0\) And \(12-\mathrm{p}=0\) \(\Rightarrow \mathrm{p}=12\) put the value of \(p\) in the determinant equation- \(\left|\begin{array}{ccc}12 &7 / 2& -9 \\ 7 / 2& -12 &\mathrm{q} / 2 \\ -9 &\mathrm{q} / 2 &6\end{array}\right|=0\) \(\Rightarrow 12\left(-72+\frac{\mathrm{q}^{2}}{4}\right)-\frac{7}{2}\left(-21 \frac{9 \mathrm{q}}{2}\right)-9\left(\frac{7 \mathrm{q}}{4}-108\right)=0\) \(\Rightarrow-864-3 q^{2}-\frac{147}{2}-\frac{63 q}{4}-\frac{63 q}{4}+972=0\) \(\Rightarrow \quad q=1\) \(\quad p=12\)
Manipal-2011
Co-Ordinate system
88481
The sum of the squares of the distances of a moving point from 2 fixed points \(A(a, 0)\) and \(B\) \((-a, 0)\) is equal to a constant \(2 c^{2}\), then the equation of its locus is
(A): Let \(\mathrm{P}(\mathrm{h}, \mathrm{k})\) be any position of the moving point let \(A(a, o), B(-a, o)\) be the given points then \(\mathrm{PA}^{2}+\mathrm{PB}^{2}=2 \mathrm{C}^{2}\) \(\left(n_{2}-a\right)^{2}+\left(k_{2}-o\right)^{2}+(n+a)^{2}+\left(k_{2}-o\right)^{2}=2 c^{2}\) \(n^{2}-2 a n+a^{2}+k^{2}+n^{2}+2 a n+a^{2}+k^{2}=2 c^{2}\) \(2 \mathrm{n}^{2}+2 \mathrm{a}^{2}+2 \mathrm{k}^{2}=2 \mathrm{c}^{2}\) \(\mathrm{n}^{2}+\mathrm{k}^{2}=\mathrm{c}^{2}-\mathrm{a}^{2}\) Hence, locus of \((n, k)\) is \(x^{2}+y^{2}=c^{2}-a^{2}\).
