88864
The slopes of the lines represented by \(x^{2}+2 h x y\) \(+2 y^{2}=0\) are in the ratio \((1: 2)\) then \(h=\)
1 \(\pm \frac{1}{2}\)
2 \(\pm \frac{3}{2}\)
3 \(\pm 1\)
4 \(\pm 3\)
Explanation:
(B): Given,
\(\mathrm{x}^{2}+2 \mathrm{hxy}+2 \mathrm{y}^{2}=0\)
On comparing ax \(+2 \mathrm{hxy}+\mathrm{by}^{2}=0\), we get -
Let \(\mathrm{y}=\mathrm{m}_{1} \mathrm{x}\) and \(\mathrm{y}=\mathrm{m}_{2} \mathrm{x}\) be the lines,
Whose combined from is eq. (i) -
\(\therefore \mathrm{m}_{1}+\mathrm{m}_{2}=\frac{-2 \mathrm{~h}}{\mathrm{~b}}=-\frac{2 \mathrm{~h}}{2}=-\mathrm{h} \tag{ii}\)
\(\text { and } \mathrm{m}_{1} \cdot \mathrm{m}_{2}=\frac{\mathrm{a}}{\mathrm{b}}=\frac{1}{2} \tag{iii}\)
\(\because \mathrm{m}_{1}: \mathrm{m}_{2}=1: 2\)
\(\frac{\mathrm{m}_{1}}{\mathrm{~m}_{2}}=\frac{1}{2} \Rightarrow \mathrm{m}_{2}=2 \mathrm{~m}_{1}\)
from eq \(\mathrm{q}^{\mathrm{n}}\). (ii), we get \(\mathrm{m}_{1}+\mathrm{m}_{2}=-\mathrm{h}\) and
from eq \({ }^{n}\). (iii), we get \(m_{1} m_{2}=\frac{1}{2}\)
\(\mathrm{m}_{1}+2 \mathrm{~m}_{1}=-\mathrm{h}\) and \(\mathrm{m}_{1} \mathrm{~m}_{2}=\frac{1}{2}\)
\(\mathrm{m}_{1}=\frac{-\mathrm{h}}{3} \quad\) and \(\quad 2 \mathrm{~m}_{1}{ }^{2}=\frac{1}{2}\) \(-\frac{\mathrm{h}}{3}= \pm \frac{1}{3} \quad\) and \(\quad \mathrm{m}_{1}= \pm \frac{1}{2} \Rightarrow \mathrm{h}= \pm \frac{3}{2}\)
APEAPCET-2021-20.08.2021
Straight Line
88865
The acute angle between lines \(6 x^{2}+11 x y-10 y^{2}=\) 0 is
(D) : Given,
\(6 x^{2}+11 x y-10 y^{2}=0\)
On comparing ax \({ }^{2}+2 h x y+y^{2}=0\) we get -
\(\mathrm{a}=\mathrm{b}, \mathrm{b}=-10, \mathrm{~h}=\frac{11}{2}\)
\(\tan \theta=\left|\frac{2 \sqrt{\mathrm{h}^{2}-\mathrm{ab}}}{\mathrm{a}+\mathrm{b}}\right| \Rightarrow \theta=\tan ^{-1}\left|\frac{2 \sqrt{\frac{121}{4}+60}}{-4}\right|\)
\(\theta=\tan ^{-1}\left|\frac{\sqrt{361}}{-4}\right| \Rightarrow \theta=\tan ^{-1} \frac{\sqrt{361}}{4}\)
Shift-II]
Straight Line
88866
If \(p_{1} \cdot p_{2}\) denote the lengths of perpendiculars from \((2,3)\) onto the lines given by \(15 x^{2}+31 x y+\)
\(14 y^{2}=0 \text { and if } p_{1}>p_{2} \text {. then } p_{1}^{2} \frac{1}{-}-p_{2}{ }^{2}+\frac{1}{}\)
1 -1
2 0
3 12
4 1
Explanation:
(C): Lines given by \(15 x^{2}+31 x y+14 y^{2}=0\) are \(5 x\) \(+7 y=0 \& 3 y+2 y=0\)
Length of perpendicular from \(\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)\) from line \(\mathrm{ax}+\) by \(+\mathrm{c}=\) is \(\left|\frac{\mathrm{ax}_{1}+\mathrm{by}_{1}+\mathrm{c}}{\sqrt{\mathrm{a}^{2}+\mathrm{b}^{2}}}\right|\)
So, distance from point \((2,3) \_\)from the lines
\(\mathrm{p}_{1}=\left|\frac{5(2)+7(3)}{\sqrt{5^{2}+7^{2}}}\right| \& \mathrm{p}_{2}=\left|\frac{3(2)+2(3)}{\sqrt{3^{2}+2^{2}}}\right|\)
\(\mathrm{p}_{1}=\frac{31}{\sqrt{74}} \quad \& \mathrm{p}_{2}=\frac{12}{\sqrt{13}} \quad\left[\mathrm{p}_{1}>\mathrm{p}_{2}\right]\)
\(=\mathrm{p}_{1}^{2}+\frac{1}{74}-\mathrm{p}_{2}^{2}+\frac{1}{13}\)
\(=\frac{961}{74}+\frac{1}{74}-\frac{144}{13}+\frac{1}{13} \Rightarrow \frac{962}{74}-\frac{143}{13}=13-11=2\)
AP EAMCET-2021-23.08.2021
Straight Line
88867
If the anlge between two lines represented by \(2 x^{2}+5 x y+3 y^{2}+7 y+4=0\) is \(\tan ^{-1} \mathrm{~m}\), then \(m\) is equal to
1 \(1 / 5\)
2 1
3 \(7 / 5\)
4 7
Explanation:
(A) : Given,
\(2 x^{2}+5 x y+3 y^{2}+7 y+4=0\)
On comparing general equation of conic section,
\(a=2, b=3, h=\frac{5}{2}\)
\(\therefore \tan \theta=\left|\frac{2 \sqrt{(5 / 2)^{2}-6}}{2+3}\right| \Rightarrow m=\frac{2 \cdot 1 / 2}{5}=\frac{1}{5}\)
88864
The slopes of the lines represented by \(x^{2}+2 h x y\) \(+2 y^{2}=0\) are in the ratio \((1: 2)\) then \(h=\)
1 \(\pm \frac{1}{2}\)
2 \(\pm \frac{3}{2}\)
3 \(\pm 1\)
4 \(\pm 3\)
Explanation:
(B): Given,
\(\mathrm{x}^{2}+2 \mathrm{hxy}+2 \mathrm{y}^{2}=0\)
On comparing ax \(+2 \mathrm{hxy}+\mathrm{by}^{2}=0\), we get -
Let \(\mathrm{y}=\mathrm{m}_{1} \mathrm{x}\) and \(\mathrm{y}=\mathrm{m}_{2} \mathrm{x}\) be the lines,
Whose combined from is eq. (i) -
\(\therefore \mathrm{m}_{1}+\mathrm{m}_{2}=\frac{-2 \mathrm{~h}}{\mathrm{~b}}=-\frac{2 \mathrm{~h}}{2}=-\mathrm{h} \tag{ii}\)
\(\text { and } \mathrm{m}_{1} \cdot \mathrm{m}_{2}=\frac{\mathrm{a}}{\mathrm{b}}=\frac{1}{2} \tag{iii}\)
\(\because \mathrm{m}_{1}: \mathrm{m}_{2}=1: 2\)
\(\frac{\mathrm{m}_{1}}{\mathrm{~m}_{2}}=\frac{1}{2} \Rightarrow \mathrm{m}_{2}=2 \mathrm{~m}_{1}\)
from eq \(\mathrm{q}^{\mathrm{n}}\). (ii), we get \(\mathrm{m}_{1}+\mathrm{m}_{2}=-\mathrm{h}\) and
from eq \({ }^{n}\). (iii), we get \(m_{1} m_{2}=\frac{1}{2}\)
\(\mathrm{m}_{1}+2 \mathrm{~m}_{1}=-\mathrm{h}\) and \(\mathrm{m}_{1} \mathrm{~m}_{2}=\frac{1}{2}\)
\(\mathrm{m}_{1}=\frac{-\mathrm{h}}{3} \quad\) and \(\quad 2 \mathrm{~m}_{1}{ }^{2}=\frac{1}{2}\) \(-\frac{\mathrm{h}}{3}= \pm \frac{1}{3} \quad\) and \(\quad \mathrm{m}_{1}= \pm \frac{1}{2} \Rightarrow \mathrm{h}= \pm \frac{3}{2}\)
APEAPCET-2021-20.08.2021
Straight Line
88865
The acute angle between lines \(6 x^{2}+11 x y-10 y^{2}=\) 0 is
(D) : Given,
\(6 x^{2}+11 x y-10 y^{2}=0\)
On comparing ax \({ }^{2}+2 h x y+y^{2}=0\) we get -
\(\mathrm{a}=\mathrm{b}, \mathrm{b}=-10, \mathrm{~h}=\frac{11}{2}\)
\(\tan \theta=\left|\frac{2 \sqrt{\mathrm{h}^{2}-\mathrm{ab}}}{\mathrm{a}+\mathrm{b}}\right| \Rightarrow \theta=\tan ^{-1}\left|\frac{2 \sqrt{\frac{121}{4}+60}}{-4}\right|\)
\(\theta=\tan ^{-1}\left|\frac{\sqrt{361}}{-4}\right| \Rightarrow \theta=\tan ^{-1} \frac{\sqrt{361}}{4}\)
Shift-II]
Straight Line
88866
If \(p_{1} \cdot p_{2}\) denote the lengths of perpendiculars from \((2,3)\) onto the lines given by \(15 x^{2}+31 x y+\)
\(14 y^{2}=0 \text { and if } p_{1}>p_{2} \text {. then } p_{1}^{2} \frac{1}{-}-p_{2}{ }^{2}+\frac{1}{}\)
1 -1
2 0
3 12
4 1
Explanation:
(C): Lines given by \(15 x^{2}+31 x y+14 y^{2}=0\) are \(5 x\) \(+7 y=0 \& 3 y+2 y=0\)
Length of perpendicular from \(\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)\) from line \(\mathrm{ax}+\) by \(+\mathrm{c}=\) is \(\left|\frac{\mathrm{ax}_{1}+\mathrm{by}_{1}+\mathrm{c}}{\sqrt{\mathrm{a}^{2}+\mathrm{b}^{2}}}\right|\)
So, distance from point \((2,3) \_\)from the lines
\(\mathrm{p}_{1}=\left|\frac{5(2)+7(3)}{\sqrt{5^{2}+7^{2}}}\right| \& \mathrm{p}_{2}=\left|\frac{3(2)+2(3)}{\sqrt{3^{2}+2^{2}}}\right|\)
\(\mathrm{p}_{1}=\frac{31}{\sqrt{74}} \quad \& \mathrm{p}_{2}=\frac{12}{\sqrt{13}} \quad\left[\mathrm{p}_{1}>\mathrm{p}_{2}\right]\)
\(=\mathrm{p}_{1}^{2}+\frac{1}{74}-\mathrm{p}_{2}^{2}+\frac{1}{13}\)
\(=\frac{961}{74}+\frac{1}{74}-\frac{144}{13}+\frac{1}{13} \Rightarrow \frac{962}{74}-\frac{143}{13}=13-11=2\)
AP EAMCET-2021-23.08.2021
Straight Line
88867
If the anlge between two lines represented by \(2 x^{2}+5 x y+3 y^{2}+7 y+4=0\) is \(\tan ^{-1} \mathrm{~m}\), then \(m\) is equal to
1 \(1 / 5\)
2 1
3 \(7 / 5\)
4 7
Explanation:
(A) : Given,
\(2 x^{2}+5 x y+3 y^{2}+7 y+4=0\)
On comparing general equation of conic section,
\(a=2, b=3, h=\frac{5}{2}\)
\(\therefore \tan \theta=\left|\frac{2 \sqrt{(5 / 2)^{2}-6}}{2+3}\right| \Rightarrow m=\frac{2 \cdot 1 / 2}{5}=\frac{1}{5}\)
88864
The slopes of the lines represented by \(x^{2}+2 h x y\) \(+2 y^{2}=0\) are in the ratio \((1: 2)\) then \(h=\)
1 \(\pm \frac{1}{2}\)
2 \(\pm \frac{3}{2}\)
3 \(\pm 1\)
4 \(\pm 3\)
Explanation:
(B): Given,
\(\mathrm{x}^{2}+2 \mathrm{hxy}+2 \mathrm{y}^{2}=0\)
On comparing ax \(+2 \mathrm{hxy}+\mathrm{by}^{2}=0\), we get -
Let \(\mathrm{y}=\mathrm{m}_{1} \mathrm{x}\) and \(\mathrm{y}=\mathrm{m}_{2} \mathrm{x}\) be the lines,
Whose combined from is eq. (i) -
\(\therefore \mathrm{m}_{1}+\mathrm{m}_{2}=\frac{-2 \mathrm{~h}}{\mathrm{~b}}=-\frac{2 \mathrm{~h}}{2}=-\mathrm{h} \tag{ii}\)
\(\text { and } \mathrm{m}_{1} \cdot \mathrm{m}_{2}=\frac{\mathrm{a}}{\mathrm{b}}=\frac{1}{2} \tag{iii}\)
\(\because \mathrm{m}_{1}: \mathrm{m}_{2}=1: 2\)
\(\frac{\mathrm{m}_{1}}{\mathrm{~m}_{2}}=\frac{1}{2} \Rightarrow \mathrm{m}_{2}=2 \mathrm{~m}_{1}\)
from eq \(\mathrm{q}^{\mathrm{n}}\). (ii), we get \(\mathrm{m}_{1}+\mathrm{m}_{2}=-\mathrm{h}\) and
from eq \({ }^{n}\). (iii), we get \(m_{1} m_{2}=\frac{1}{2}\)
\(\mathrm{m}_{1}+2 \mathrm{~m}_{1}=-\mathrm{h}\) and \(\mathrm{m}_{1} \mathrm{~m}_{2}=\frac{1}{2}\)
\(\mathrm{m}_{1}=\frac{-\mathrm{h}}{3} \quad\) and \(\quad 2 \mathrm{~m}_{1}{ }^{2}=\frac{1}{2}\) \(-\frac{\mathrm{h}}{3}= \pm \frac{1}{3} \quad\) and \(\quad \mathrm{m}_{1}= \pm \frac{1}{2} \Rightarrow \mathrm{h}= \pm \frac{3}{2}\)
APEAPCET-2021-20.08.2021
Straight Line
88865
The acute angle between lines \(6 x^{2}+11 x y-10 y^{2}=\) 0 is
(D) : Given,
\(6 x^{2}+11 x y-10 y^{2}=0\)
On comparing ax \({ }^{2}+2 h x y+y^{2}=0\) we get -
\(\mathrm{a}=\mathrm{b}, \mathrm{b}=-10, \mathrm{~h}=\frac{11}{2}\)
\(\tan \theta=\left|\frac{2 \sqrt{\mathrm{h}^{2}-\mathrm{ab}}}{\mathrm{a}+\mathrm{b}}\right| \Rightarrow \theta=\tan ^{-1}\left|\frac{2 \sqrt{\frac{121}{4}+60}}{-4}\right|\)
\(\theta=\tan ^{-1}\left|\frac{\sqrt{361}}{-4}\right| \Rightarrow \theta=\tan ^{-1} \frac{\sqrt{361}}{4}\)
Shift-II]
Straight Line
88866
If \(p_{1} \cdot p_{2}\) denote the lengths of perpendiculars from \((2,3)\) onto the lines given by \(15 x^{2}+31 x y+\)
\(14 y^{2}=0 \text { and if } p_{1}>p_{2} \text {. then } p_{1}^{2} \frac{1}{-}-p_{2}{ }^{2}+\frac{1}{}\)
1 -1
2 0
3 12
4 1
Explanation:
(C): Lines given by \(15 x^{2}+31 x y+14 y^{2}=0\) are \(5 x\) \(+7 y=0 \& 3 y+2 y=0\)
Length of perpendicular from \(\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)\) from line \(\mathrm{ax}+\) by \(+\mathrm{c}=\) is \(\left|\frac{\mathrm{ax}_{1}+\mathrm{by}_{1}+\mathrm{c}}{\sqrt{\mathrm{a}^{2}+\mathrm{b}^{2}}}\right|\)
So, distance from point \((2,3) \_\)from the lines
\(\mathrm{p}_{1}=\left|\frac{5(2)+7(3)}{\sqrt{5^{2}+7^{2}}}\right| \& \mathrm{p}_{2}=\left|\frac{3(2)+2(3)}{\sqrt{3^{2}+2^{2}}}\right|\)
\(\mathrm{p}_{1}=\frac{31}{\sqrt{74}} \quad \& \mathrm{p}_{2}=\frac{12}{\sqrt{13}} \quad\left[\mathrm{p}_{1}>\mathrm{p}_{2}\right]\)
\(=\mathrm{p}_{1}^{2}+\frac{1}{74}-\mathrm{p}_{2}^{2}+\frac{1}{13}\)
\(=\frac{961}{74}+\frac{1}{74}-\frac{144}{13}+\frac{1}{13} \Rightarrow \frac{962}{74}-\frac{143}{13}=13-11=2\)
AP EAMCET-2021-23.08.2021
Straight Line
88867
If the anlge between two lines represented by \(2 x^{2}+5 x y+3 y^{2}+7 y+4=0\) is \(\tan ^{-1} \mathrm{~m}\), then \(m\) is equal to
1 \(1 / 5\)
2 1
3 \(7 / 5\)
4 7
Explanation:
(A) : Given,
\(2 x^{2}+5 x y+3 y^{2}+7 y+4=0\)
On comparing general equation of conic section,
\(a=2, b=3, h=\frac{5}{2}\)
\(\therefore \tan \theta=\left|\frac{2 \sqrt{(5 / 2)^{2}-6}}{2+3}\right| \Rightarrow m=\frac{2 \cdot 1 / 2}{5}=\frac{1}{5}\)
NEET Test Series from KOTA - 10 Papers In MS WORD
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Straight Line
88864
The slopes of the lines represented by \(x^{2}+2 h x y\) \(+2 y^{2}=0\) are in the ratio \((1: 2)\) then \(h=\)
1 \(\pm \frac{1}{2}\)
2 \(\pm \frac{3}{2}\)
3 \(\pm 1\)
4 \(\pm 3\)
Explanation:
(B): Given,
\(\mathrm{x}^{2}+2 \mathrm{hxy}+2 \mathrm{y}^{2}=0\)
On comparing ax \(+2 \mathrm{hxy}+\mathrm{by}^{2}=0\), we get -
Let \(\mathrm{y}=\mathrm{m}_{1} \mathrm{x}\) and \(\mathrm{y}=\mathrm{m}_{2} \mathrm{x}\) be the lines,
Whose combined from is eq. (i) -
\(\therefore \mathrm{m}_{1}+\mathrm{m}_{2}=\frac{-2 \mathrm{~h}}{\mathrm{~b}}=-\frac{2 \mathrm{~h}}{2}=-\mathrm{h} \tag{ii}\)
\(\text { and } \mathrm{m}_{1} \cdot \mathrm{m}_{2}=\frac{\mathrm{a}}{\mathrm{b}}=\frac{1}{2} \tag{iii}\)
\(\because \mathrm{m}_{1}: \mathrm{m}_{2}=1: 2\)
\(\frac{\mathrm{m}_{1}}{\mathrm{~m}_{2}}=\frac{1}{2} \Rightarrow \mathrm{m}_{2}=2 \mathrm{~m}_{1}\)
from eq \(\mathrm{q}^{\mathrm{n}}\). (ii), we get \(\mathrm{m}_{1}+\mathrm{m}_{2}=-\mathrm{h}\) and
from eq \({ }^{n}\). (iii), we get \(m_{1} m_{2}=\frac{1}{2}\)
\(\mathrm{m}_{1}+2 \mathrm{~m}_{1}=-\mathrm{h}\) and \(\mathrm{m}_{1} \mathrm{~m}_{2}=\frac{1}{2}\)
\(\mathrm{m}_{1}=\frac{-\mathrm{h}}{3} \quad\) and \(\quad 2 \mathrm{~m}_{1}{ }^{2}=\frac{1}{2}\) \(-\frac{\mathrm{h}}{3}= \pm \frac{1}{3} \quad\) and \(\quad \mathrm{m}_{1}= \pm \frac{1}{2} \Rightarrow \mathrm{h}= \pm \frac{3}{2}\)
APEAPCET-2021-20.08.2021
Straight Line
88865
The acute angle between lines \(6 x^{2}+11 x y-10 y^{2}=\) 0 is
(D) : Given,
\(6 x^{2}+11 x y-10 y^{2}=0\)
On comparing ax \({ }^{2}+2 h x y+y^{2}=0\) we get -
\(\mathrm{a}=\mathrm{b}, \mathrm{b}=-10, \mathrm{~h}=\frac{11}{2}\)
\(\tan \theta=\left|\frac{2 \sqrt{\mathrm{h}^{2}-\mathrm{ab}}}{\mathrm{a}+\mathrm{b}}\right| \Rightarrow \theta=\tan ^{-1}\left|\frac{2 \sqrt{\frac{121}{4}+60}}{-4}\right|\)
\(\theta=\tan ^{-1}\left|\frac{\sqrt{361}}{-4}\right| \Rightarrow \theta=\tan ^{-1} \frac{\sqrt{361}}{4}\)
Shift-II]
Straight Line
88866
If \(p_{1} \cdot p_{2}\) denote the lengths of perpendiculars from \((2,3)\) onto the lines given by \(15 x^{2}+31 x y+\)
\(14 y^{2}=0 \text { and if } p_{1}>p_{2} \text {. then } p_{1}^{2} \frac{1}{-}-p_{2}{ }^{2}+\frac{1}{}\)
1 -1
2 0
3 12
4 1
Explanation:
(C): Lines given by \(15 x^{2}+31 x y+14 y^{2}=0\) are \(5 x\) \(+7 y=0 \& 3 y+2 y=0\)
Length of perpendicular from \(\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)\) from line \(\mathrm{ax}+\) by \(+\mathrm{c}=\) is \(\left|\frac{\mathrm{ax}_{1}+\mathrm{by}_{1}+\mathrm{c}}{\sqrt{\mathrm{a}^{2}+\mathrm{b}^{2}}}\right|\)
So, distance from point \((2,3) \_\)from the lines
\(\mathrm{p}_{1}=\left|\frac{5(2)+7(3)}{\sqrt{5^{2}+7^{2}}}\right| \& \mathrm{p}_{2}=\left|\frac{3(2)+2(3)}{\sqrt{3^{2}+2^{2}}}\right|\)
\(\mathrm{p}_{1}=\frac{31}{\sqrt{74}} \quad \& \mathrm{p}_{2}=\frac{12}{\sqrt{13}} \quad\left[\mathrm{p}_{1}>\mathrm{p}_{2}\right]\)
\(=\mathrm{p}_{1}^{2}+\frac{1}{74}-\mathrm{p}_{2}^{2}+\frac{1}{13}\)
\(=\frac{961}{74}+\frac{1}{74}-\frac{144}{13}+\frac{1}{13} \Rightarrow \frac{962}{74}-\frac{143}{13}=13-11=2\)
AP EAMCET-2021-23.08.2021
Straight Line
88867
If the anlge between two lines represented by \(2 x^{2}+5 x y+3 y^{2}+7 y+4=0\) is \(\tan ^{-1} \mathrm{~m}\), then \(m\) is equal to
1 \(1 / 5\)
2 1
3 \(7 / 5\)
4 7
Explanation:
(A) : Given,
\(2 x^{2}+5 x y+3 y^{2}+7 y+4=0\)
On comparing general equation of conic section,
\(a=2, b=3, h=\frac{5}{2}\)
\(\therefore \tan \theta=\left|\frac{2 \sqrt{(5 / 2)^{2}-6}}{2+3}\right| \Rightarrow m=\frac{2 \cdot 1 / 2}{5}=\frac{1}{5}\)