88394
If the acute angle between the lines \(a x^{2}+2 h x y+b y^{2}=0\) is \(60^{\circ}\) then \((\mathbf{a}+3 \mathrm{~b})(3 \mathbf{a}+\mathbf{b})=\)
88395
If lines represented by \(\left(1+\sin ^{2} \theta\right) x^{2}+2 h x y+2 \sin \theta y^{2}\) \(=0=, \theta \in[0,2 \pi]\) are perpendicular to each other then, \(\theta=\)
1 \(\pi\)
2 \(\frac{\pi}{6}\)
3 \(\frac{\pi}{2}\)
4 \(\frac{3 \pi}{2}\)
Explanation:
(D) : The equation of line is \(\left(1+\sin ^{2} \theta\right) \mathrm{x}^{2}+2 \mathrm{hxy}+2 \sin \theta \mathrm{y}^{2}=0\) \(\Rightarrow \mathrm{a}=1+\sin ^{2} \theta, \mathrm{b}=2 \sin \theta\) The lines represented by equation (i) are perpendicular. \(\therefore \mathrm{a}+\mathrm{b}=0\) i.e. \(1+\sin ^{2} \theta+2 \sin \theta=0 \Rightarrow(\sin \theta+1)^{2}=0\) \(\sin \theta=-1 \Rightarrow \sin \theta=-\sin \frac{\pi}{2}=\sin \left(\pi+\frac{\pi}{2}\right)\) \(\Rightarrow \sin \theta=\sin \frac{3 \pi}{2} \Rightarrow \theta=\frac{3 \pi}{2}\)
MHT CET-2019
Co-Ordinate system
88396
The angle between the lines \(\sin ^{2} \alpha \cdot y^{2}-2 x y\). \(\cos ^{2} \alpha+\left(\cos ^{2} \alpha-1\right) x^{2}=0\) is
1 \(90^{\circ}\)
2 \(\alpha\)
3 \(\alpha / 2\)
4 \(2 \alpha\)
Explanation:
(A) : Given, The equation of straight lines \(\sin ^{2} \alpha \cdot y^{2}-2 x y \cos ^{2} \alpha+\left(\cos ^{2} \alpha-1\right) x^{2}=0\). We know that, The general equation of straight lines \(a x^{2}+2 h x y+b y^{2}=0\) Now, Comparing equation (i) and (ii) \(b=\sin ^{2} \alpha, a=\left(\cos ^{2} \alpha-1\right) \tag{ii}\\ h=-\cos ^{2} \alpha\) \(a+b=\cos ^{2} \alpha-1+\sin ^{2} \alpha=1-1=0\) Line are orthogonal to each other Therefore, \(\quad \theta=90^{\circ}\)
Karnataka CET-2013
Co-Ordinate system
88397
The angle between lines joining the origin to the point of intersection of the line \(\sqrt{3} x+y=2\) and the curve \(y^{2}-x^{2}=4\) is
1 \(\tan ^{-1} \frac{2}{\sqrt{3}}\)
2 \(\frac{\pi}{6}\)
3 \(\tan ^{-1}\left(\frac{\sqrt{3}}{2}\right)\)
4 \(\frac{\pi}{2}\)
Explanation:
(C): On homogenizing, \(\mathrm{y}^{2}-\mathrm{x}^{2}=4\) with the help of the line \(\sqrt{3} x+y=2\), \(\frac{\sqrt{3} \mathrm{x}+\mathrm{y}}{2}=1\) Now, we get \(y^{2}-x^{2}=4 \frac{(\sqrt{3} x+y)^{2}}{4}\) \(y^{2}-x^{2}=(\sqrt{3} x+y)^{2}\) \(y^{2}-x^{2}=3 x^{2}+y^{2}+2 \sqrt{3} x y\) \(3 x^{2}+x^{2}+y^{2}-y^{2}+2 \sqrt{3} x y=0\) \(4 x^{2}+2 \sqrt{3} x y=0\) On comparing with \(\mathrm{ax}^{2}+2 \mathrm{hxy}+\mathrm{by}^{2}=0\) we get, \(\mathrm{a}=4, \mathrm{~h}=\sqrt{3} \text { and } \mathrm{b}=0\) We know that, \(\tan \theta=2 \frac{\sqrt{\mathrm{h}^{2}-\mathrm{ab}}}{\mathrm{a}+\mathrm{b}}=\frac{2 \cdot \sqrt{3-0}}{4+0}\) \(\therefore\) The angle between the lines is \(\theta=\tan ^{-1}\left(\frac{\sqrt{3}}{2}\right)\)
88394
If the acute angle between the lines \(a x^{2}+2 h x y+b y^{2}=0\) is \(60^{\circ}\) then \((\mathbf{a}+3 \mathrm{~b})(3 \mathbf{a}+\mathbf{b})=\)
88395
If lines represented by \(\left(1+\sin ^{2} \theta\right) x^{2}+2 h x y+2 \sin \theta y^{2}\) \(=0=, \theta \in[0,2 \pi]\) are perpendicular to each other then, \(\theta=\)
1 \(\pi\)
2 \(\frac{\pi}{6}\)
3 \(\frac{\pi}{2}\)
4 \(\frac{3 \pi}{2}\)
Explanation:
(D) : The equation of line is \(\left(1+\sin ^{2} \theta\right) \mathrm{x}^{2}+2 \mathrm{hxy}+2 \sin \theta \mathrm{y}^{2}=0\) \(\Rightarrow \mathrm{a}=1+\sin ^{2} \theta, \mathrm{b}=2 \sin \theta\) The lines represented by equation (i) are perpendicular. \(\therefore \mathrm{a}+\mathrm{b}=0\) i.e. \(1+\sin ^{2} \theta+2 \sin \theta=0 \Rightarrow(\sin \theta+1)^{2}=0\) \(\sin \theta=-1 \Rightarrow \sin \theta=-\sin \frac{\pi}{2}=\sin \left(\pi+\frac{\pi}{2}\right)\) \(\Rightarrow \sin \theta=\sin \frac{3 \pi}{2} \Rightarrow \theta=\frac{3 \pi}{2}\)
MHT CET-2019
Co-Ordinate system
88396
The angle between the lines \(\sin ^{2} \alpha \cdot y^{2}-2 x y\). \(\cos ^{2} \alpha+\left(\cos ^{2} \alpha-1\right) x^{2}=0\) is
1 \(90^{\circ}\)
2 \(\alpha\)
3 \(\alpha / 2\)
4 \(2 \alpha\)
Explanation:
(A) : Given, The equation of straight lines \(\sin ^{2} \alpha \cdot y^{2}-2 x y \cos ^{2} \alpha+\left(\cos ^{2} \alpha-1\right) x^{2}=0\). We know that, The general equation of straight lines \(a x^{2}+2 h x y+b y^{2}=0\) Now, Comparing equation (i) and (ii) \(b=\sin ^{2} \alpha, a=\left(\cos ^{2} \alpha-1\right) \tag{ii}\\ h=-\cos ^{2} \alpha\) \(a+b=\cos ^{2} \alpha-1+\sin ^{2} \alpha=1-1=0\) Line are orthogonal to each other Therefore, \(\quad \theta=90^{\circ}\)
Karnataka CET-2013
Co-Ordinate system
88397
The angle between lines joining the origin to the point of intersection of the line \(\sqrt{3} x+y=2\) and the curve \(y^{2}-x^{2}=4\) is
1 \(\tan ^{-1} \frac{2}{\sqrt{3}}\)
2 \(\frac{\pi}{6}\)
3 \(\tan ^{-1}\left(\frac{\sqrt{3}}{2}\right)\)
4 \(\frac{\pi}{2}\)
Explanation:
(C): On homogenizing, \(\mathrm{y}^{2}-\mathrm{x}^{2}=4\) with the help of the line \(\sqrt{3} x+y=2\), \(\frac{\sqrt{3} \mathrm{x}+\mathrm{y}}{2}=1\) Now, we get \(y^{2}-x^{2}=4 \frac{(\sqrt{3} x+y)^{2}}{4}\) \(y^{2}-x^{2}=(\sqrt{3} x+y)^{2}\) \(y^{2}-x^{2}=3 x^{2}+y^{2}+2 \sqrt{3} x y\) \(3 x^{2}+x^{2}+y^{2}-y^{2}+2 \sqrt{3} x y=0\) \(4 x^{2}+2 \sqrt{3} x y=0\) On comparing with \(\mathrm{ax}^{2}+2 \mathrm{hxy}+\mathrm{by}^{2}=0\) we get, \(\mathrm{a}=4, \mathrm{~h}=\sqrt{3} \text { and } \mathrm{b}=0\) We know that, \(\tan \theta=2 \frac{\sqrt{\mathrm{h}^{2}-\mathrm{ab}}}{\mathrm{a}+\mathrm{b}}=\frac{2 \cdot \sqrt{3-0}}{4+0}\) \(\therefore\) The angle between the lines is \(\theta=\tan ^{-1}\left(\frac{\sqrt{3}}{2}\right)\)
88394
If the acute angle between the lines \(a x^{2}+2 h x y+b y^{2}=0\) is \(60^{\circ}\) then \((\mathbf{a}+3 \mathrm{~b})(3 \mathbf{a}+\mathbf{b})=\)
88395
If lines represented by \(\left(1+\sin ^{2} \theta\right) x^{2}+2 h x y+2 \sin \theta y^{2}\) \(=0=, \theta \in[0,2 \pi]\) are perpendicular to each other then, \(\theta=\)
1 \(\pi\)
2 \(\frac{\pi}{6}\)
3 \(\frac{\pi}{2}\)
4 \(\frac{3 \pi}{2}\)
Explanation:
(D) : The equation of line is \(\left(1+\sin ^{2} \theta\right) \mathrm{x}^{2}+2 \mathrm{hxy}+2 \sin \theta \mathrm{y}^{2}=0\) \(\Rightarrow \mathrm{a}=1+\sin ^{2} \theta, \mathrm{b}=2 \sin \theta\) The lines represented by equation (i) are perpendicular. \(\therefore \mathrm{a}+\mathrm{b}=0\) i.e. \(1+\sin ^{2} \theta+2 \sin \theta=0 \Rightarrow(\sin \theta+1)^{2}=0\) \(\sin \theta=-1 \Rightarrow \sin \theta=-\sin \frac{\pi}{2}=\sin \left(\pi+\frac{\pi}{2}\right)\) \(\Rightarrow \sin \theta=\sin \frac{3 \pi}{2} \Rightarrow \theta=\frac{3 \pi}{2}\)
MHT CET-2019
Co-Ordinate system
88396
The angle between the lines \(\sin ^{2} \alpha \cdot y^{2}-2 x y\). \(\cos ^{2} \alpha+\left(\cos ^{2} \alpha-1\right) x^{2}=0\) is
1 \(90^{\circ}\)
2 \(\alpha\)
3 \(\alpha / 2\)
4 \(2 \alpha\)
Explanation:
(A) : Given, The equation of straight lines \(\sin ^{2} \alpha \cdot y^{2}-2 x y \cos ^{2} \alpha+\left(\cos ^{2} \alpha-1\right) x^{2}=0\). We know that, The general equation of straight lines \(a x^{2}+2 h x y+b y^{2}=0\) Now, Comparing equation (i) and (ii) \(b=\sin ^{2} \alpha, a=\left(\cos ^{2} \alpha-1\right) \tag{ii}\\ h=-\cos ^{2} \alpha\) \(a+b=\cos ^{2} \alpha-1+\sin ^{2} \alpha=1-1=0\) Line are orthogonal to each other Therefore, \(\quad \theta=90^{\circ}\)
Karnataka CET-2013
Co-Ordinate system
88397
The angle between lines joining the origin to the point of intersection of the line \(\sqrt{3} x+y=2\) and the curve \(y^{2}-x^{2}=4\) is
1 \(\tan ^{-1} \frac{2}{\sqrt{3}}\)
2 \(\frac{\pi}{6}\)
3 \(\tan ^{-1}\left(\frac{\sqrt{3}}{2}\right)\)
4 \(\frac{\pi}{2}\)
Explanation:
(C): On homogenizing, \(\mathrm{y}^{2}-\mathrm{x}^{2}=4\) with the help of the line \(\sqrt{3} x+y=2\), \(\frac{\sqrt{3} \mathrm{x}+\mathrm{y}}{2}=1\) Now, we get \(y^{2}-x^{2}=4 \frac{(\sqrt{3} x+y)^{2}}{4}\) \(y^{2}-x^{2}=(\sqrt{3} x+y)^{2}\) \(y^{2}-x^{2}=3 x^{2}+y^{2}+2 \sqrt{3} x y\) \(3 x^{2}+x^{2}+y^{2}-y^{2}+2 \sqrt{3} x y=0\) \(4 x^{2}+2 \sqrt{3} x y=0\) On comparing with \(\mathrm{ax}^{2}+2 \mathrm{hxy}+\mathrm{by}^{2}=0\) we get, \(\mathrm{a}=4, \mathrm{~h}=\sqrt{3} \text { and } \mathrm{b}=0\) We know that, \(\tan \theta=2 \frac{\sqrt{\mathrm{h}^{2}-\mathrm{ab}}}{\mathrm{a}+\mathrm{b}}=\frac{2 \cdot \sqrt{3-0}}{4+0}\) \(\therefore\) The angle between the lines is \(\theta=\tan ^{-1}\left(\frac{\sqrt{3}}{2}\right)\)
88394
If the acute angle between the lines \(a x^{2}+2 h x y+b y^{2}=0\) is \(60^{\circ}\) then \((\mathbf{a}+3 \mathrm{~b})(3 \mathbf{a}+\mathbf{b})=\)
88395
If lines represented by \(\left(1+\sin ^{2} \theta\right) x^{2}+2 h x y+2 \sin \theta y^{2}\) \(=0=, \theta \in[0,2 \pi]\) are perpendicular to each other then, \(\theta=\)
1 \(\pi\)
2 \(\frac{\pi}{6}\)
3 \(\frac{\pi}{2}\)
4 \(\frac{3 \pi}{2}\)
Explanation:
(D) : The equation of line is \(\left(1+\sin ^{2} \theta\right) \mathrm{x}^{2}+2 \mathrm{hxy}+2 \sin \theta \mathrm{y}^{2}=0\) \(\Rightarrow \mathrm{a}=1+\sin ^{2} \theta, \mathrm{b}=2 \sin \theta\) The lines represented by equation (i) are perpendicular. \(\therefore \mathrm{a}+\mathrm{b}=0\) i.e. \(1+\sin ^{2} \theta+2 \sin \theta=0 \Rightarrow(\sin \theta+1)^{2}=0\) \(\sin \theta=-1 \Rightarrow \sin \theta=-\sin \frac{\pi}{2}=\sin \left(\pi+\frac{\pi}{2}\right)\) \(\Rightarrow \sin \theta=\sin \frac{3 \pi}{2} \Rightarrow \theta=\frac{3 \pi}{2}\)
MHT CET-2019
Co-Ordinate system
88396
The angle between the lines \(\sin ^{2} \alpha \cdot y^{2}-2 x y\). \(\cos ^{2} \alpha+\left(\cos ^{2} \alpha-1\right) x^{2}=0\) is
1 \(90^{\circ}\)
2 \(\alpha\)
3 \(\alpha / 2\)
4 \(2 \alpha\)
Explanation:
(A) : Given, The equation of straight lines \(\sin ^{2} \alpha \cdot y^{2}-2 x y \cos ^{2} \alpha+\left(\cos ^{2} \alpha-1\right) x^{2}=0\). We know that, The general equation of straight lines \(a x^{2}+2 h x y+b y^{2}=0\) Now, Comparing equation (i) and (ii) \(b=\sin ^{2} \alpha, a=\left(\cos ^{2} \alpha-1\right) \tag{ii}\\ h=-\cos ^{2} \alpha\) \(a+b=\cos ^{2} \alpha-1+\sin ^{2} \alpha=1-1=0\) Line are orthogonal to each other Therefore, \(\quad \theta=90^{\circ}\)
Karnataka CET-2013
Co-Ordinate system
88397
The angle between lines joining the origin to the point of intersection of the line \(\sqrt{3} x+y=2\) and the curve \(y^{2}-x^{2}=4\) is
1 \(\tan ^{-1} \frac{2}{\sqrt{3}}\)
2 \(\frac{\pi}{6}\)
3 \(\tan ^{-1}\left(\frac{\sqrt{3}}{2}\right)\)
4 \(\frac{\pi}{2}\)
Explanation:
(C): On homogenizing, \(\mathrm{y}^{2}-\mathrm{x}^{2}=4\) with the help of the line \(\sqrt{3} x+y=2\), \(\frac{\sqrt{3} \mathrm{x}+\mathrm{y}}{2}=1\) Now, we get \(y^{2}-x^{2}=4 \frac{(\sqrt{3} x+y)^{2}}{4}\) \(y^{2}-x^{2}=(\sqrt{3} x+y)^{2}\) \(y^{2}-x^{2}=3 x^{2}+y^{2}+2 \sqrt{3} x y\) \(3 x^{2}+x^{2}+y^{2}-y^{2}+2 \sqrt{3} x y=0\) \(4 x^{2}+2 \sqrt{3} x y=0\) On comparing with \(\mathrm{ax}^{2}+2 \mathrm{hxy}+\mathrm{by}^{2}=0\) we get, \(\mathrm{a}=4, \mathrm{~h}=\sqrt{3} \text { and } \mathrm{b}=0\) We know that, \(\tan \theta=2 \frac{\sqrt{\mathrm{h}^{2}-\mathrm{ab}}}{\mathrm{a}+\mathrm{b}}=\frac{2 \cdot \sqrt{3-0}}{4+0}\) \(\therefore\) The angle between the lines is \(\theta=\tan ^{-1}\left(\frac{\sqrt{3}}{2}\right)\)
