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Co-Ordinate system

88452 A variable line ' \(L\) ' passing through the origin cuts two parallel lines \(x-y+10=0\) and \(x-y+\) \(20=0\) at two points \(A\) and \(B\) Respectively. If \(P\) is a points on line \(L\) such that \(O A, O P, O B\) are in harmonic progression, then the locus of \(P\) is

1 \(3 x+3 y+40=0\)

2 \(3 x+3 y+20=0\)

3 \(3 x-3 y+40=0\)

4 \(3 x-3 y+20=0\)

Explanation:

(C) : Given the equation of line passing through origin is
\(y=m x\) which cut the parallel lines \(x-y+10=0\) and \(x-y+20=0\) at point \(A\) and \(B\) respectively.
Then, Co-ordinate of points \(\mathrm{A}\) and \(\mathrm{B}\) are,
\(y=m x \text { and } x-y+10=0\)
\(x-m x=-10\)
\(x=\frac{-10}{1-m}=\frac{10}{m-1} \text { and } y=\frac{10}{m-1}\)
Therefore A \((\mathrm{x}, \mathrm{y})\)
\(\therefore \quad A\left(\frac{10}{m-1}, \frac{10 m}{m-1}\right)\)
So, \(\quad \mathrm{OA}=\sqrt{\frac{100}{(m-1)^{2}}+\frac{100}{(m-1)^{2}}}\)
\(=\sqrt{\frac{100(1+\mathrm{m})^{2}}{(\mathrm{~m}-1)^{2}}}=\frac{10 \sqrt{\left(1+\mathrm{m}^{2}\right)}}{\mathrm{m}-1}\)
Similary co-ordinate of point \(\mathrm{B}\) is
Therefore \(\mathrm{E}\left(\frac{20}{\mathrm{~m}-1}, \frac{20 \mathrm{~m}}{\mathrm{~m}-1}\right)\)
Then, \(\mathrm{OB}=\sqrt{\frac{400}{(\mathrm{~m}-1)^{2}}+\frac{400 \mathrm{~m}^{2}}{(\mathrm{~m}-1)^{2}}}=\frac{20 \sqrt{1+\mathrm{m}^{2}}}{(\mathrm{~m}-1)}\)
Let point \(\mathrm{p}(\mathrm{h}, \mathrm{k})\) lies on the line \(\mathrm{y}=\mathrm{mx} \mathrm{so}, \mathrm{k}=\mathrm{mh}\),
\(\mathrm{m}=\frac{\mathrm{k}}{\mathrm{h}}\)
and \(\mathrm{OP}=\sqrt{\mathrm{h}^{2}+\mathrm{k}^{2}}\)
Now, OA, OP, OB in H.P
So, \(\quad \frac{1}{\mathrm{OA}}, \frac{1}{\mathrm{OP}}, \frac{1}{\mathrm{OB}}\) in \(\mathrm{AP}\).
Hence, \(\frac{1}{\mathrm{OP}}=\frac{1}{\mathrm{OA}}+\frac{1}{\mathrm{OB}}\)
\(\frac{2}{\sqrt{\mathrm{h}^{2}+\mathrm{k}^{2}}}=\frac{\mathrm{m}-1}{10 \sqrt{1+\mathrm{m}^{2}}}+\frac{\mathrm{m}-1}{20 \sqrt{1+\mathrm{m}^{2}}}\)
\(\frac{2}{\sqrt{\mathrm{h}^{2}+\mathrm{k}^{2}}}=\frac{\mathrm{m}-1}{10 \sqrt{1+\mathrm{m}^{2}}}\left[1+\frac{1}{2}\right]\)
Putting the value of \(\mathrm{m}\), we get-
\(\frac{2}{\sqrt{\mathrm{h}^{2}+\mathrm{k}^{2}}}=\frac{\left(\frac{\mathrm{k}}{\mathrm{h}}-1\right)}{10 \sqrt{1+\frac{\mathrm{k}^{2}}{\mathrm{~h}^{2}}} \times \frac{3}{2}}\)
\(\frac{2}{\sqrt{\mathrm{h}^{2}+\mathrm{k}^{2}}}=\frac{3(\mathrm{k}-\mathrm{h}) / \mathrm{h}}{20 \frac{\sqrt{\mathrm{h}^{2}+\mathrm{k}^{2}}}{\mathrm{~h}}}\)
\(\frac{2}{\sqrt{\mathrm{h}^{2}+\mathrm{k}^{2}}}=\frac{3(\mathrm{k}-\mathrm{h})}{20 \sqrt{\mathrm{h}^{2}+\mathrm{k}^{2}}}\)
\(3(\mathrm{k}-\mathrm{h})=40\)
\(3 \mathrm{~h}-3 \mathrm{k}+40=0\)
So, locus the point is \(3 x-3 y+40=0\)

AP EAMCET-2018-24.04.2018

Co-Ordinate system

88453 The locus of the centroid of the triangle with vertices at \((\operatorname{acos} \theta, \operatorname{asin} \theta)(b \sin \theta,-b \cos \theta)\) and
\((1,0)\) is (here, \(\theta\) is a parameter)

1 \((3 x+1)^{2}+9 y=a^{2}+b^{2}\)

2 \((3 x-1)^{2}+9 y^{2}=a^{2}-b^{2}\)

3 \((3 x-1)^{2}+9 y^{2}=a^{2}+b^{2}\)

4 \((3 x+1)^{2}+9 y^{2}=a^{2}-b^{2}\)

AP EAMCET-2014

Co-Ordinate system

88454 If equation to the locus of points equidistant from the points \((-2,3),(6,-5)\) is \(a x+b y+c=0\), where a \(>0\), then the ascending order of \(a, b, c\) is

1 a, b, c

2 c, b, a

3 b, c, a

4 a, c, b

AP EAMCET-2015

Co-Ordinate system

88455 For different values of \(\alpha\), the locus of the point of intersection of the two straight lines \(\sqrt{3} x-y-4 \sqrt{3} \alpha=0\) and \(\sqrt{3} \alpha x+\alpha y-4 \sqrt{3}=0\) is

1 A hyperbola with eccentricity 2

2 An ellipse with eccentricity \(\sqrt{\frac{2}{3}}\)

3 A hyperbola with eccentricity \(\sqrt{\frac{19}{16}}\)

4 An ellipse with eccentricity \(\frac{3}{4}\)

WB JEE-2010

Co-Ordinate system

88452 A variable line ' \(L\) ' passing through the origin cuts two parallel lines \(x-y+10=0\) and \(x-y+\) \(20=0\) at two points \(A\) and \(B\) Respectively. If \(P\) is a points on line \(L\) such that \(O A, O P, O B\) are in harmonic progression, then the locus of \(P\) is

1 \(3 x+3 y+40=0\)

2 \(3 x+3 y+20=0\)

3 \(3 x-3 y+40=0\)

4 \(3 x-3 y+20=0\)

Explanation:

(C) : Given the equation of line passing through origin is
\(y=m x\) which cut the parallel lines \(x-y+10=0\) and \(x-y+20=0\) at point \(A\) and \(B\) respectively.
Then, Co-ordinate of points \(\mathrm{A}\) and \(\mathrm{B}\) are,
\(y=m x \text { and } x-y+10=0\)
\(x-m x=-10\)
\(x=\frac{-10}{1-m}=\frac{10}{m-1} \text { and } y=\frac{10}{m-1}\)
Therefore A \((\mathrm{x}, \mathrm{y})\)
\(\therefore \quad A\left(\frac{10}{m-1}, \frac{10 m}{m-1}\right)\)
So, \(\quad \mathrm{OA}=\sqrt{\frac{100}{(m-1)^{2}}+\frac{100}{(m-1)^{2}}}\)
\(=\sqrt{\frac{100(1+\mathrm{m})^{2}}{(\mathrm{~m}-1)^{2}}}=\frac{10 \sqrt{\left(1+\mathrm{m}^{2}\right)}}{\mathrm{m}-1}\)
Similary co-ordinate of point \(\mathrm{B}\) is
Therefore \(\mathrm{E}\left(\frac{20}{\mathrm{~m}-1}, \frac{20 \mathrm{~m}}{\mathrm{~m}-1}\right)\)
Then, \(\mathrm{OB}=\sqrt{\frac{400}{(\mathrm{~m}-1)^{2}}+\frac{400 \mathrm{~m}^{2}}{(\mathrm{~m}-1)^{2}}}=\frac{20 \sqrt{1+\mathrm{m}^{2}}}{(\mathrm{~m}-1)}\)
Let point \(\mathrm{p}(\mathrm{h}, \mathrm{k})\) lies on the line \(\mathrm{y}=\mathrm{mx} \mathrm{so}, \mathrm{k}=\mathrm{mh}\),
\(\mathrm{m}=\frac{\mathrm{k}}{\mathrm{h}}\)
and \(\mathrm{OP}=\sqrt{\mathrm{h}^{2}+\mathrm{k}^{2}}\)
Now, OA, OP, OB in H.P
So, \(\quad \frac{1}{\mathrm{OA}}, \frac{1}{\mathrm{OP}}, \frac{1}{\mathrm{OB}}\) in \(\mathrm{AP}\).
Hence, \(\frac{1}{\mathrm{OP}}=\frac{1}{\mathrm{OA}}+\frac{1}{\mathrm{OB}}\)
\(\frac{2}{\sqrt{\mathrm{h}^{2}+\mathrm{k}^{2}}}=\frac{\mathrm{m}-1}{10 \sqrt{1+\mathrm{m}^{2}}}+\frac{\mathrm{m}-1}{20 \sqrt{1+\mathrm{m}^{2}}}\)
\(\frac{2}{\sqrt{\mathrm{h}^{2}+\mathrm{k}^{2}}}=\frac{\mathrm{m}-1}{10 \sqrt{1+\mathrm{m}^{2}}}\left[1+\frac{1}{2}\right]\)
Putting the value of \(\mathrm{m}\), we get-
\(\frac{2}{\sqrt{\mathrm{h}^{2}+\mathrm{k}^{2}}}=\frac{\left(\frac{\mathrm{k}}{\mathrm{h}}-1\right)}{10 \sqrt{1+\frac{\mathrm{k}^{2}}{\mathrm{~h}^{2}}} \times \frac{3}{2}}\)
\(\frac{2}{\sqrt{\mathrm{h}^{2}+\mathrm{k}^{2}}}=\frac{3(\mathrm{k}-\mathrm{h}) / \mathrm{h}}{20 \frac{\sqrt{\mathrm{h}^{2}+\mathrm{k}^{2}}}{\mathrm{~h}}}\)
\(\frac{2}{\sqrt{\mathrm{h}^{2}+\mathrm{k}^{2}}}=\frac{3(\mathrm{k}-\mathrm{h})}{20 \sqrt{\mathrm{h}^{2}+\mathrm{k}^{2}}}\)
\(3(\mathrm{k}-\mathrm{h})=40\)
\(3 \mathrm{~h}-3 \mathrm{k}+40=0\)
So, locus the point is \(3 x-3 y+40=0\)

AP EAMCET-2018-24.04.2018

Co-Ordinate system

88453 The locus of the centroid of the triangle with vertices at \((\operatorname{acos} \theta, \operatorname{asin} \theta)(b \sin \theta,-b \cos \theta)\) and
\((1,0)\) is (here, \(\theta\) is a parameter)

1 \((3 x+1)^{2}+9 y=a^{2}+b^{2}\)

2 \((3 x-1)^{2}+9 y^{2}=a^{2}-b^{2}\)

3 \((3 x-1)^{2}+9 y^{2}=a^{2}+b^{2}\)

4 \((3 x+1)^{2}+9 y^{2}=a^{2}-b^{2}\)

AP EAMCET-2014

Co-Ordinate system

88454 If equation to the locus of points equidistant from the points \((-2,3),(6,-5)\) is \(a x+b y+c=0\), where a \(>0\), then the ascending order of \(a, b, c\) is

1 a, b, c

2 c, b, a

3 b, c, a

4 a, c, b

AP EAMCET-2015

Co-Ordinate system

88455 For different values of \(\alpha\), the locus of the point of intersection of the two straight lines \(\sqrt{3} x-y-4 \sqrt{3} \alpha=0\) and \(\sqrt{3} \alpha x+\alpha y-4 \sqrt{3}=0\) is

1 A hyperbola with eccentricity 2

2 An ellipse with eccentricity \(\sqrt{\frac{2}{3}}\)

3 A hyperbola with eccentricity \(\sqrt{\frac{19}{16}}\)

4 An ellipse with eccentricity \(\frac{3}{4}\)

WB JEE-2010

Co-Ordinate system

88452 A variable line ' \(L\) ' passing through the origin cuts two parallel lines \(x-y+10=0\) and \(x-y+\) \(20=0\) at two points \(A\) and \(B\) Respectively. If \(P\) is a points on line \(L\) such that \(O A, O P, O B\) are in harmonic progression, then the locus of \(P\) is

1 \(3 x+3 y+40=0\)

2 \(3 x+3 y+20=0\)

3 \(3 x-3 y+40=0\)

4 \(3 x-3 y+20=0\)

Explanation:

(C) : Given the equation of line passing through origin is
\(y=m x\) which cut the parallel lines \(x-y+10=0\) and \(x-y+20=0\) at point \(A\) and \(B\) respectively.
Then, Co-ordinate of points \(\mathrm{A}\) and \(\mathrm{B}\) are,
\(y=m x \text { and } x-y+10=0\)
\(x-m x=-10\)
\(x=\frac{-10}{1-m}=\frac{10}{m-1} \text { and } y=\frac{10}{m-1}\)
Therefore A \((\mathrm{x}, \mathrm{y})\)
\(\therefore \quad A\left(\frac{10}{m-1}, \frac{10 m}{m-1}\right)\)
So, \(\quad \mathrm{OA}=\sqrt{\frac{100}{(m-1)^{2}}+\frac{100}{(m-1)^{2}}}\)
\(=\sqrt{\frac{100(1+\mathrm{m})^{2}}{(\mathrm{~m}-1)^{2}}}=\frac{10 \sqrt{\left(1+\mathrm{m}^{2}\right)}}{\mathrm{m}-1}\)
Similary co-ordinate of point \(\mathrm{B}\) is
Therefore \(\mathrm{E}\left(\frac{20}{\mathrm{~m}-1}, \frac{20 \mathrm{~m}}{\mathrm{~m}-1}\right)\)
Then, \(\mathrm{OB}=\sqrt{\frac{400}{(\mathrm{~m}-1)^{2}}+\frac{400 \mathrm{~m}^{2}}{(\mathrm{~m}-1)^{2}}}=\frac{20 \sqrt{1+\mathrm{m}^{2}}}{(\mathrm{~m}-1)}\)
Let point \(\mathrm{p}(\mathrm{h}, \mathrm{k})\) lies on the line \(\mathrm{y}=\mathrm{mx} \mathrm{so}, \mathrm{k}=\mathrm{mh}\),
\(\mathrm{m}=\frac{\mathrm{k}}{\mathrm{h}}\)
and \(\mathrm{OP}=\sqrt{\mathrm{h}^{2}+\mathrm{k}^{2}}\)
Now, OA, OP, OB in H.P
So, \(\quad \frac{1}{\mathrm{OA}}, \frac{1}{\mathrm{OP}}, \frac{1}{\mathrm{OB}}\) in \(\mathrm{AP}\).
Hence, \(\frac{1}{\mathrm{OP}}=\frac{1}{\mathrm{OA}}+\frac{1}{\mathrm{OB}}\)
\(\frac{2}{\sqrt{\mathrm{h}^{2}+\mathrm{k}^{2}}}=\frac{\mathrm{m}-1}{10 \sqrt{1+\mathrm{m}^{2}}}+\frac{\mathrm{m}-1}{20 \sqrt{1+\mathrm{m}^{2}}}\)
\(\frac{2}{\sqrt{\mathrm{h}^{2}+\mathrm{k}^{2}}}=\frac{\mathrm{m}-1}{10 \sqrt{1+\mathrm{m}^{2}}}\left[1+\frac{1}{2}\right]\)
Putting the value of \(\mathrm{m}\), we get-
\(\frac{2}{\sqrt{\mathrm{h}^{2}+\mathrm{k}^{2}}}=\frac{\left(\frac{\mathrm{k}}{\mathrm{h}}-1\right)}{10 \sqrt{1+\frac{\mathrm{k}^{2}}{\mathrm{~h}^{2}}} \times \frac{3}{2}}\)
\(\frac{2}{\sqrt{\mathrm{h}^{2}+\mathrm{k}^{2}}}=\frac{3(\mathrm{k}-\mathrm{h}) / \mathrm{h}}{20 \frac{\sqrt{\mathrm{h}^{2}+\mathrm{k}^{2}}}{\mathrm{~h}}}\)
\(\frac{2}{\sqrt{\mathrm{h}^{2}+\mathrm{k}^{2}}}=\frac{3(\mathrm{k}-\mathrm{h})}{20 \sqrt{\mathrm{h}^{2}+\mathrm{k}^{2}}}\)
\(3(\mathrm{k}-\mathrm{h})=40\)
\(3 \mathrm{~h}-3 \mathrm{k}+40=0\)
So, locus the point is \(3 x-3 y+40=0\)

AP EAMCET-2018-24.04.2018

Co-Ordinate system

88453 The locus of the centroid of the triangle with vertices at \((\operatorname{acos} \theta, \operatorname{asin} \theta)(b \sin \theta,-b \cos \theta)\) and
\((1,0)\) is (here, \(\theta\) is a parameter)

1 \((3 x+1)^{2}+9 y=a^{2}+b^{2}\)

2 \((3 x-1)^{2}+9 y^{2}=a^{2}-b^{2}\)

3 \((3 x-1)^{2}+9 y^{2}=a^{2}+b^{2}\)

4 \((3 x+1)^{2}+9 y^{2}=a^{2}-b^{2}\)

AP EAMCET-2014

Co-Ordinate system

88454 If equation to the locus of points equidistant from the points \((-2,3),(6,-5)\) is \(a x+b y+c=0\), where a \(>0\), then the ascending order of \(a, b, c\) is

1 a, b, c

2 c, b, a

3 b, c, a

4 a, c, b

AP EAMCET-2015

Co-Ordinate system

88455 For different values of \(\alpha\), the locus of the point of intersection of the two straight lines \(\sqrt{3} x-y-4 \sqrt{3} \alpha=0\) and \(\sqrt{3} \alpha x+\alpha y-4 \sqrt{3}=0\) is

1 A hyperbola with eccentricity 2

2 An ellipse with eccentricity \(\sqrt{\frac{2}{3}}\)

3 A hyperbola with eccentricity \(\sqrt{\frac{19}{16}}\)

4 An ellipse with eccentricity \(\frac{3}{4}\)

WB JEE-2010

Co-Ordinate system

88452 A variable line ' \(L\) ' passing through the origin cuts two parallel lines \(x-y+10=0\) and \(x-y+\) \(20=0\) at two points \(A\) and \(B\) Respectively. If \(P\) is a points on line \(L\) such that \(O A, O P, O B\) are in harmonic progression, then the locus of \(P\) is

1 \(3 x+3 y+40=0\)

2 \(3 x+3 y+20=0\)

3 \(3 x-3 y+40=0\)

4 \(3 x-3 y+20=0\)

Explanation:

(C) : Given the equation of line passing through origin is
\(y=m x\) which cut the parallel lines \(x-y+10=0\) and \(x-y+20=0\) at point \(A\) and \(B\) respectively.
Then, Co-ordinate of points \(\mathrm{A}\) and \(\mathrm{B}\) are,
\(y=m x \text { and } x-y+10=0\)
\(x-m x=-10\)
\(x=\frac{-10}{1-m}=\frac{10}{m-1} \text { and } y=\frac{10}{m-1}\)
Therefore A \((\mathrm{x}, \mathrm{y})\)
\(\therefore \quad A\left(\frac{10}{m-1}, \frac{10 m}{m-1}\right)\)
So, \(\quad \mathrm{OA}=\sqrt{\frac{100}{(m-1)^{2}}+\frac{100}{(m-1)^{2}}}\)
\(=\sqrt{\frac{100(1+\mathrm{m})^{2}}{(\mathrm{~m}-1)^{2}}}=\frac{10 \sqrt{\left(1+\mathrm{m}^{2}\right)}}{\mathrm{m}-1}\)
Similary co-ordinate of point \(\mathrm{B}\) is
Therefore \(\mathrm{E}\left(\frac{20}{\mathrm{~m}-1}, \frac{20 \mathrm{~m}}{\mathrm{~m}-1}\right)\)
Then, \(\mathrm{OB}=\sqrt{\frac{400}{(\mathrm{~m}-1)^{2}}+\frac{400 \mathrm{~m}^{2}}{(\mathrm{~m}-1)^{2}}}=\frac{20 \sqrt{1+\mathrm{m}^{2}}}{(\mathrm{~m}-1)}\)
Let point \(\mathrm{p}(\mathrm{h}, \mathrm{k})\) lies on the line \(\mathrm{y}=\mathrm{mx} \mathrm{so}, \mathrm{k}=\mathrm{mh}\),
\(\mathrm{m}=\frac{\mathrm{k}}{\mathrm{h}}\)
and \(\mathrm{OP}=\sqrt{\mathrm{h}^{2}+\mathrm{k}^{2}}\)
Now, OA, OP, OB in H.P
So, \(\quad \frac{1}{\mathrm{OA}}, \frac{1}{\mathrm{OP}}, \frac{1}{\mathrm{OB}}\) in \(\mathrm{AP}\).
Hence, \(\frac{1}{\mathrm{OP}}=\frac{1}{\mathrm{OA}}+\frac{1}{\mathrm{OB}}\)
\(\frac{2}{\sqrt{\mathrm{h}^{2}+\mathrm{k}^{2}}}=\frac{\mathrm{m}-1}{10 \sqrt{1+\mathrm{m}^{2}}}+\frac{\mathrm{m}-1}{20 \sqrt{1+\mathrm{m}^{2}}}\)
\(\frac{2}{\sqrt{\mathrm{h}^{2}+\mathrm{k}^{2}}}=\frac{\mathrm{m}-1}{10 \sqrt{1+\mathrm{m}^{2}}}\left[1+\frac{1}{2}\right]\)
Putting the value of \(\mathrm{m}\), we get-
\(\frac{2}{\sqrt{\mathrm{h}^{2}+\mathrm{k}^{2}}}=\frac{\left(\frac{\mathrm{k}}{\mathrm{h}}-1\right)}{10 \sqrt{1+\frac{\mathrm{k}^{2}}{\mathrm{~h}^{2}}} \times \frac{3}{2}}\)
\(\frac{2}{\sqrt{\mathrm{h}^{2}+\mathrm{k}^{2}}}=\frac{3(\mathrm{k}-\mathrm{h}) / \mathrm{h}}{20 \frac{\sqrt{\mathrm{h}^{2}+\mathrm{k}^{2}}}{\mathrm{~h}}}\)
\(\frac{2}{\sqrt{\mathrm{h}^{2}+\mathrm{k}^{2}}}=\frac{3(\mathrm{k}-\mathrm{h})}{20 \sqrt{\mathrm{h}^{2}+\mathrm{k}^{2}}}\)
\(3(\mathrm{k}-\mathrm{h})=40\)
\(3 \mathrm{~h}-3 \mathrm{k}+40=0\)
So, locus the point is \(3 x-3 y+40=0\)

AP EAMCET-2018-24.04.2018

Co-Ordinate system

88453 The locus of the centroid of the triangle with vertices at \((\operatorname{acos} \theta, \operatorname{asin} \theta)(b \sin \theta,-b \cos \theta)\) and
\((1,0)\) is (here, \(\theta\) is a parameter)

1 \((3 x+1)^{2}+9 y=a^{2}+b^{2}\)

2 \((3 x-1)^{2}+9 y^{2}=a^{2}-b^{2}\)

3 \((3 x-1)^{2}+9 y^{2}=a^{2}+b^{2}\)

4 \((3 x+1)^{2}+9 y^{2}=a^{2}-b^{2}\)

AP EAMCET-2014

Co-Ordinate system

88454 If equation to the locus of points equidistant from the points \((-2,3),(6,-5)\) is \(a x+b y+c=0\), where a \(>0\), then the ascending order of \(a, b, c\) is

1 a, b, c

2 c, b, a

3 b, c, a

4 a, c, b

AP EAMCET-2015

Co-Ordinate system

88455 For different values of \(\alpha\), the locus of the point of intersection of the two straight lines \(\sqrt{3} x-y-4 \sqrt{3} \alpha=0\) and \(\sqrt{3} \alpha x+\alpha y-4 \sqrt{3}=0\) is

1 A hyperbola with eccentricity 2

2 An ellipse with eccentricity \(\sqrt{\frac{2}{3}}\)

3 A hyperbola with eccentricity \(\sqrt{\frac{19}{16}}\)

4 An ellipse with eccentricity \(\frac{3}{4}\)

WB JEE-2010