NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Straight Line
88810
If the lines \(a y^{2}+b x y+e x+d y=0\) represents a pair of lines then
1 \(\mathrm{bd}-\mathrm{ae}=0 \quad\) or \(\quad \mathrm{e}=0\)
2 \(\mathrm{be}-\mathrm{ad}=0 \quad\) or \(\quad \mathrm{e}=0\)
3 \(\mathrm{ad}-\mathrm{eb}=0 \quad\) or \(\quad \mathrm{e}=0\)
4 \(\mathrm{ad}+\mathrm{be}=0 \quad\) or \(\quad \mathrm{e}=0\)
Explanation:
(A) : Comparing given equation with
\(\mathrm{Ax}^{2}+2 \mathrm{Hxy}+\mathrm{By}^{2}+2 \mathrm{Gx}+2 \mathrm{Fy}+\mathrm{C}=0\), we get
Given equation represents a pair of lines if
\(\mathrm{ABC}+2 \mathrm{FGH}-\mathrm{AF}^{2}-\mathrm{BG}^{2}-\mathrm{CH}^{2}=0\)
\((0)(\mathrm{a})(0)+2\left(\frac{\mathrm{d}}{2}\right)\left(\frac{\mathrm{e}}{2}\right)\left(\frac{\mathrm{b}}{2}\right)-0\left(\frac{\mathrm{d}}{2}\right)^{2}-\mathrm{a}\left(\frac{\mathrm{e}}{2}\right)^{2}-0\left(\frac{\mathrm{b}}{2}\right)^{2}=p\)
\(0+\frac{\mathrm{deb}}{4}-0-\frac{\mathrm{ae}^{2}}{4}-0=0 \Rightarrow \mathrm{deb}-\mathrm{ae}^{2}=0\)
\(\Rightarrow \mathrm{e}(\mathrm{bd}-\mathrm{ae})=0\)
\(\therefore \mathrm{e}=0\) or \(\mathrm{bd}-\mathrm{ae}=0\)
MHT CET-2013
Straight Line
88811
If a pair of lines \(x^{2}-2 p x y-y^{2}=0\) and \(x^{2}-2 q x y-y^{2}=0\) is such that each pair bisects the angle between the other pair, then
1 \(\mathrm{pq}=-1\)
2 \(\mathrm{pq}=1\)
3 \(\frac{1}{\mathrm{p}}+\frac{1}{\mathrm{q}}=0\)
4 \(\frac{1}{\mathrm{p}}-\frac{1}{\mathrm{q}}=0\)
Explanation:
(A) :
We have two pairs of lines bisecting angle between each other.
Let \(x^{2}-2 p x y-y^{2}=0\)
be such that \(A=1, H=-p, B=-1\)
Equation of angle bisector for (i) is
\(\frac{x^{2}-y^{2}}{A-B}=\frac{x y}{H} \Rightarrow \frac{x^{2}-y^{2}}{1-(-1)}=\frac{x y}{-p}\) i.e. \(-p x^{2}+p y^{2}-2 x y=0\)
We have \(x^{2}-2 q x y-y^{2}=0\)
(iii) as equation of
bisector of (i).
Since equations (ii) and (iii) represent same line, we write
\(\frac{-\mathrm{p}}{1}=-\frac{-2}{-2 \mathrm{q}} \Rightarrow-\mathrm{p}-\frac{1}{\mathrm{q}}=0 \Rightarrow \mathrm{pq}=-1\)
MHT CET-2011
Straight Line
88812
If one of the lines of the pair \(a x^{2}+2 h x y+b y^{2}=0\) bisects the angle between positive direction of the axes, then \(a, b\) and \(h\) satisfy the relation
1 \(a+b=2|\mathrm{~h}|\)
2 \(a+b=-2 h\)
3 \(\mathrm{a}-\mathrm{b}=2 \mid \mathrm{h}\)
4 \((a-b)^{2}=4 h^{2}\)
Explanation:
(B) :
As one of the lines of the pair of lines \(a x^{2}+2 h x y+b y^{2}=0\) bisect the angle between positive direction of the axes, the line \(y=x\) satisfies the pair of lines.
\(\therefore \mathrm{ax}^{2}+2 \mathrm{hx}(\mathrm{x})+(\mathrm{x})^{2}=0 \Rightarrow \mathrm{a}+2 \mathrm{~h}+\mathrm{b}=0\)
\(\Rightarrow \mathrm{a}+\mathrm{b}=-2 \mathrm{~h}\)
MHT CET-2011
Straight Line
88813
Find the equation of pair of lines at a distance of 5 units from the line \(y=1\).
1 \(\mathrm{y}^{2}-\mathrm{y}-24=0\)
2 \(\mathrm{y}^{2}-2 \mathrm{y}-24=0\)
3 \(\mathrm{y}^{2}+\mathrm{y}+24=0\)
4 \(\mathrm{y}^{2}+2 \mathrm{y}-24=0\)
Explanation:
(B) :
The equation of the lines at a distance of 5 units from the line \(y=1\) are \(y=6\) and \(y=-4\) i.e. \(y-6=0\) and \(\mathrm{y}+4=0\)
Their joint equation is
\((y-6)(y+4)=0 \Rightarrow y^{2}-2 y-24=0\)
88810
If the lines \(a y^{2}+b x y+e x+d y=0\) represents a pair of lines then
1 \(\mathrm{bd}-\mathrm{ae}=0 \quad\) or \(\quad \mathrm{e}=0\)
2 \(\mathrm{be}-\mathrm{ad}=0 \quad\) or \(\quad \mathrm{e}=0\)
3 \(\mathrm{ad}-\mathrm{eb}=0 \quad\) or \(\quad \mathrm{e}=0\)
4 \(\mathrm{ad}+\mathrm{be}=0 \quad\) or \(\quad \mathrm{e}=0\)
Explanation:
(A) : Comparing given equation with
\(\mathrm{Ax}^{2}+2 \mathrm{Hxy}+\mathrm{By}^{2}+2 \mathrm{Gx}+2 \mathrm{Fy}+\mathrm{C}=0\), we get
Given equation represents a pair of lines if
\(\mathrm{ABC}+2 \mathrm{FGH}-\mathrm{AF}^{2}-\mathrm{BG}^{2}-\mathrm{CH}^{2}=0\)
\((0)(\mathrm{a})(0)+2\left(\frac{\mathrm{d}}{2}\right)\left(\frac{\mathrm{e}}{2}\right)\left(\frac{\mathrm{b}}{2}\right)-0\left(\frac{\mathrm{d}}{2}\right)^{2}-\mathrm{a}\left(\frac{\mathrm{e}}{2}\right)^{2}-0\left(\frac{\mathrm{b}}{2}\right)^{2}=p\)
\(0+\frac{\mathrm{deb}}{4}-0-\frac{\mathrm{ae}^{2}}{4}-0=0 \Rightarrow \mathrm{deb}-\mathrm{ae}^{2}=0\)
\(\Rightarrow \mathrm{e}(\mathrm{bd}-\mathrm{ae})=0\)
\(\therefore \mathrm{e}=0\) or \(\mathrm{bd}-\mathrm{ae}=0\)
MHT CET-2013
Straight Line
88811
If a pair of lines \(x^{2}-2 p x y-y^{2}=0\) and \(x^{2}-2 q x y-y^{2}=0\) is such that each pair bisects the angle between the other pair, then
1 \(\mathrm{pq}=-1\)
2 \(\mathrm{pq}=1\)
3 \(\frac{1}{\mathrm{p}}+\frac{1}{\mathrm{q}}=0\)
4 \(\frac{1}{\mathrm{p}}-\frac{1}{\mathrm{q}}=0\)
Explanation:
(A) :
We have two pairs of lines bisecting angle between each other.
Let \(x^{2}-2 p x y-y^{2}=0\)
be such that \(A=1, H=-p, B=-1\)
Equation of angle bisector for (i) is
\(\frac{x^{2}-y^{2}}{A-B}=\frac{x y}{H} \Rightarrow \frac{x^{2}-y^{2}}{1-(-1)}=\frac{x y}{-p}\) i.e. \(-p x^{2}+p y^{2}-2 x y=0\)
We have \(x^{2}-2 q x y-y^{2}=0\)
(iii) as equation of
bisector of (i).
Since equations (ii) and (iii) represent same line, we write
\(\frac{-\mathrm{p}}{1}=-\frac{-2}{-2 \mathrm{q}} \Rightarrow-\mathrm{p}-\frac{1}{\mathrm{q}}=0 \Rightarrow \mathrm{pq}=-1\)
MHT CET-2011
Straight Line
88812
If one of the lines of the pair \(a x^{2}+2 h x y+b y^{2}=0\) bisects the angle between positive direction of the axes, then \(a, b\) and \(h\) satisfy the relation
1 \(a+b=2|\mathrm{~h}|\)
2 \(a+b=-2 h\)
3 \(\mathrm{a}-\mathrm{b}=2 \mid \mathrm{h}\)
4 \((a-b)^{2}=4 h^{2}\)
Explanation:
(B) :
As one of the lines of the pair of lines \(a x^{2}+2 h x y+b y^{2}=0\) bisect the angle between positive direction of the axes, the line \(y=x\) satisfies the pair of lines.
\(\therefore \mathrm{ax}^{2}+2 \mathrm{hx}(\mathrm{x})+(\mathrm{x})^{2}=0 \Rightarrow \mathrm{a}+2 \mathrm{~h}+\mathrm{b}=0\)
\(\Rightarrow \mathrm{a}+\mathrm{b}=-2 \mathrm{~h}\)
MHT CET-2011
Straight Line
88813
Find the equation of pair of lines at a distance of 5 units from the line \(y=1\).
1 \(\mathrm{y}^{2}-\mathrm{y}-24=0\)
2 \(\mathrm{y}^{2}-2 \mathrm{y}-24=0\)
3 \(\mathrm{y}^{2}+\mathrm{y}+24=0\)
4 \(\mathrm{y}^{2}+2 \mathrm{y}-24=0\)
Explanation:
(B) :
The equation of the lines at a distance of 5 units from the line \(y=1\) are \(y=6\) and \(y=-4\) i.e. \(y-6=0\) and \(\mathrm{y}+4=0\)
Their joint equation is
\((y-6)(y+4)=0 \Rightarrow y^{2}-2 y-24=0\)
88810
If the lines \(a y^{2}+b x y+e x+d y=0\) represents a pair of lines then
1 \(\mathrm{bd}-\mathrm{ae}=0 \quad\) or \(\quad \mathrm{e}=0\)
2 \(\mathrm{be}-\mathrm{ad}=0 \quad\) or \(\quad \mathrm{e}=0\)
3 \(\mathrm{ad}-\mathrm{eb}=0 \quad\) or \(\quad \mathrm{e}=0\)
4 \(\mathrm{ad}+\mathrm{be}=0 \quad\) or \(\quad \mathrm{e}=0\)
Explanation:
(A) : Comparing given equation with
\(\mathrm{Ax}^{2}+2 \mathrm{Hxy}+\mathrm{By}^{2}+2 \mathrm{Gx}+2 \mathrm{Fy}+\mathrm{C}=0\), we get
Given equation represents a pair of lines if
\(\mathrm{ABC}+2 \mathrm{FGH}-\mathrm{AF}^{2}-\mathrm{BG}^{2}-\mathrm{CH}^{2}=0\)
\((0)(\mathrm{a})(0)+2\left(\frac{\mathrm{d}}{2}\right)\left(\frac{\mathrm{e}}{2}\right)\left(\frac{\mathrm{b}}{2}\right)-0\left(\frac{\mathrm{d}}{2}\right)^{2}-\mathrm{a}\left(\frac{\mathrm{e}}{2}\right)^{2}-0\left(\frac{\mathrm{b}}{2}\right)^{2}=p\)
\(0+\frac{\mathrm{deb}}{4}-0-\frac{\mathrm{ae}^{2}}{4}-0=0 \Rightarrow \mathrm{deb}-\mathrm{ae}^{2}=0\)
\(\Rightarrow \mathrm{e}(\mathrm{bd}-\mathrm{ae})=0\)
\(\therefore \mathrm{e}=0\) or \(\mathrm{bd}-\mathrm{ae}=0\)
MHT CET-2013
Straight Line
88811
If a pair of lines \(x^{2}-2 p x y-y^{2}=0\) and \(x^{2}-2 q x y-y^{2}=0\) is such that each pair bisects the angle between the other pair, then
1 \(\mathrm{pq}=-1\)
2 \(\mathrm{pq}=1\)
3 \(\frac{1}{\mathrm{p}}+\frac{1}{\mathrm{q}}=0\)
4 \(\frac{1}{\mathrm{p}}-\frac{1}{\mathrm{q}}=0\)
Explanation:
(A) :
We have two pairs of lines bisecting angle between each other.
Let \(x^{2}-2 p x y-y^{2}=0\)
be such that \(A=1, H=-p, B=-1\)
Equation of angle bisector for (i) is
\(\frac{x^{2}-y^{2}}{A-B}=\frac{x y}{H} \Rightarrow \frac{x^{2}-y^{2}}{1-(-1)}=\frac{x y}{-p}\) i.e. \(-p x^{2}+p y^{2}-2 x y=0\)
We have \(x^{2}-2 q x y-y^{2}=0\)
(iii) as equation of
bisector of (i).
Since equations (ii) and (iii) represent same line, we write
\(\frac{-\mathrm{p}}{1}=-\frac{-2}{-2 \mathrm{q}} \Rightarrow-\mathrm{p}-\frac{1}{\mathrm{q}}=0 \Rightarrow \mathrm{pq}=-1\)
MHT CET-2011
Straight Line
88812
If one of the lines of the pair \(a x^{2}+2 h x y+b y^{2}=0\) bisects the angle between positive direction of the axes, then \(a, b\) and \(h\) satisfy the relation
1 \(a+b=2|\mathrm{~h}|\)
2 \(a+b=-2 h\)
3 \(\mathrm{a}-\mathrm{b}=2 \mid \mathrm{h}\)
4 \((a-b)^{2}=4 h^{2}\)
Explanation:
(B) :
As one of the lines of the pair of lines \(a x^{2}+2 h x y+b y^{2}=0\) bisect the angle between positive direction of the axes, the line \(y=x\) satisfies the pair of lines.
\(\therefore \mathrm{ax}^{2}+2 \mathrm{hx}(\mathrm{x})+(\mathrm{x})^{2}=0 \Rightarrow \mathrm{a}+2 \mathrm{~h}+\mathrm{b}=0\)
\(\Rightarrow \mathrm{a}+\mathrm{b}=-2 \mathrm{~h}\)
MHT CET-2011
Straight Line
88813
Find the equation of pair of lines at a distance of 5 units from the line \(y=1\).
1 \(\mathrm{y}^{2}-\mathrm{y}-24=0\)
2 \(\mathrm{y}^{2}-2 \mathrm{y}-24=0\)
3 \(\mathrm{y}^{2}+\mathrm{y}+24=0\)
4 \(\mathrm{y}^{2}+2 \mathrm{y}-24=0\)
Explanation:
(B) :
The equation of the lines at a distance of 5 units from the line \(y=1\) are \(y=6\) and \(y=-4\) i.e. \(y-6=0\) and \(\mathrm{y}+4=0\)
Their joint equation is
\((y-6)(y+4)=0 \Rightarrow y^{2}-2 y-24=0\)
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Straight Line
88810
If the lines \(a y^{2}+b x y+e x+d y=0\) represents a pair of lines then
1 \(\mathrm{bd}-\mathrm{ae}=0 \quad\) or \(\quad \mathrm{e}=0\)
2 \(\mathrm{be}-\mathrm{ad}=0 \quad\) or \(\quad \mathrm{e}=0\)
3 \(\mathrm{ad}-\mathrm{eb}=0 \quad\) or \(\quad \mathrm{e}=0\)
4 \(\mathrm{ad}+\mathrm{be}=0 \quad\) or \(\quad \mathrm{e}=0\)
Explanation:
(A) : Comparing given equation with
\(\mathrm{Ax}^{2}+2 \mathrm{Hxy}+\mathrm{By}^{2}+2 \mathrm{Gx}+2 \mathrm{Fy}+\mathrm{C}=0\), we get
Given equation represents a pair of lines if
\(\mathrm{ABC}+2 \mathrm{FGH}-\mathrm{AF}^{2}-\mathrm{BG}^{2}-\mathrm{CH}^{2}=0\)
\((0)(\mathrm{a})(0)+2\left(\frac{\mathrm{d}}{2}\right)\left(\frac{\mathrm{e}}{2}\right)\left(\frac{\mathrm{b}}{2}\right)-0\left(\frac{\mathrm{d}}{2}\right)^{2}-\mathrm{a}\left(\frac{\mathrm{e}}{2}\right)^{2}-0\left(\frac{\mathrm{b}}{2}\right)^{2}=p\)
\(0+\frac{\mathrm{deb}}{4}-0-\frac{\mathrm{ae}^{2}}{4}-0=0 \Rightarrow \mathrm{deb}-\mathrm{ae}^{2}=0\)
\(\Rightarrow \mathrm{e}(\mathrm{bd}-\mathrm{ae})=0\)
\(\therefore \mathrm{e}=0\) or \(\mathrm{bd}-\mathrm{ae}=0\)
MHT CET-2013
Straight Line
88811
If a pair of lines \(x^{2}-2 p x y-y^{2}=0\) and \(x^{2}-2 q x y-y^{2}=0\) is such that each pair bisects the angle between the other pair, then
1 \(\mathrm{pq}=-1\)
2 \(\mathrm{pq}=1\)
3 \(\frac{1}{\mathrm{p}}+\frac{1}{\mathrm{q}}=0\)
4 \(\frac{1}{\mathrm{p}}-\frac{1}{\mathrm{q}}=0\)
Explanation:
(A) :
We have two pairs of lines bisecting angle between each other.
Let \(x^{2}-2 p x y-y^{2}=0\)
be such that \(A=1, H=-p, B=-1\)
Equation of angle bisector for (i) is
\(\frac{x^{2}-y^{2}}{A-B}=\frac{x y}{H} \Rightarrow \frac{x^{2}-y^{2}}{1-(-1)}=\frac{x y}{-p}\) i.e. \(-p x^{2}+p y^{2}-2 x y=0\)
We have \(x^{2}-2 q x y-y^{2}=0\)
(iii) as equation of
bisector of (i).
Since equations (ii) and (iii) represent same line, we write
\(\frac{-\mathrm{p}}{1}=-\frac{-2}{-2 \mathrm{q}} \Rightarrow-\mathrm{p}-\frac{1}{\mathrm{q}}=0 \Rightarrow \mathrm{pq}=-1\)
MHT CET-2011
Straight Line
88812
If one of the lines of the pair \(a x^{2}+2 h x y+b y^{2}=0\) bisects the angle between positive direction of the axes, then \(a, b\) and \(h\) satisfy the relation
1 \(a+b=2|\mathrm{~h}|\)
2 \(a+b=-2 h\)
3 \(\mathrm{a}-\mathrm{b}=2 \mid \mathrm{h}\)
4 \((a-b)^{2}=4 h^{2}\)
Explanation:
(B) :
As one of the lines of the pair of lines \(a x^{2}+2 h x y+b y^{2}=0\) bisect the angle between positive direction of the axes, the line \(y=x\) satisfies the pair of lines.
\(\therefore \mathrm{ax}^{2}+2 \mathrm{hx}(\mathrm{x})+(\mathrm{x})^{2}=0 \Rightarrow \mathrm{a}+2 \mathrm{~h}+\mathrm{b}=0\)
\(\Rightarrow \mathrm{a}+\mathrm{b}=-2 \mathrm{~h}\)
MHT CET-2011
Straight Line
88813
Find the equation of pair of lines at a distance of 5 units from the line \(y=1\).
1 \(\mathrm{y}^{2}-\mathrm{y}-24=0\)
2 \(\mathrm{y}^{2}-2 \mathrm{y}-24=0\)
3 \(\mathrm{y}^{2}+\mathrm{y}+24=0\)
4 \(\mathrm{y}^{2}+2 \mathrm{y}-24=0\)
Explanation:
(B) :
The equation of the lines at a distance of 5 units from the line \(y=1\) are \(y=6\) and \(y=-4\) i.e. \(y-6=0\) and \(\mathrm{y}+4=0\)
Their joint equation is
\((y-6)(y+4)=0 \Rightarrow y^{2}-2 y-24=0\)