88410
\((-\sqrt{2}, \sqrt{2})\) are Cartesian co-ordinates of the point, then its polar co-ordinates are
1 \(\left(4, \frac{5 \pi}{4}\right)\)
2 \(\left(3, \frac{7 \pi}{4}\right)\)
3 \(\left(1, \frac{4 \pi}{3}\right)\)
4 \(\left(2, \frac{3 \pi}{4}\right)\)
Explanation:
(D) : Given, The Cartesian co-ordinates of the point is \((-\sqrt{2}, \sqrt{2})\) We know that, \(r \cos \theta=-\sqrt{2} \tag{i}\) \(\mathrm{r} \sin \theta=\sqrt{2}\) Squaring the both equation and adding them, \(r^{2} \cos ^{2} \theta+r^{2} \sin ^{2} \theta=2+2 \tag{ii}\) \(r^{2}=4\) \(r= \pm 2\) \(\tan \theta=\left(\frac{\sqrt{2}}{-\sqrt{2}}\right)\) \(\tan \theta=-1\) \(\theta=3 \pi / 4\) \(\therefore\) Required polar co-ordinate \((\mathrm{r}, \theta)=\left(2, \frac{3 \pi}{4}\right)\)
MHT CET-2019
Co-Ordinate system
88411
The polar coordinates of \(P\) are \(\left(2, \frac{\pi}{6}\right)\). If \(Q\) is the image of \(P\) about the \(X\)-axis, then the polar coordinates of \(Q\) are
1 \(\left(2, \frac{\pi}{6}\right)\)
2 \(\left(2, \frac{11 \pi}{6}\right)\)
3 \(\left(2, \frac{5 \pi}{6}\right)\)
4 \(\left(2, \frac{\pi}{3}\right)\)
Explanation:
(B) : Given, \(\mathrm{P} \equiv\left(2, \frac{\pi}{6}\right)\) The Cartesian coordinates of \(\mathrm{P}\) are \(\mathrm{x}=2 \cos \frac{\pi}{6}=2\left(\frac{\sqrt{3}}{2}\right)=\sqrt{3} \quad\) and \(\mathrm{y}=2 \sin \frac{\pi}{6}=2\left(\frac{1}{2}\right)=1 \Rightarrow \mathrm{p}=(\sqrt{3}, 1)\) As \(\mathrm{Q}\) is the image of \(\mathrm{P}\) about the \(\mathrm{X}\)-axis, \(\therefore \mathrm{Q} \equiv(\sqrt{3},-1)\) i.e. \(\mathrm{Q}\) lies in \(4^{\text {th }}\) quadrant Now \(\mathrm{r}=\sqrt{3+1}=2\) and \(\theta=\tan ^{-1}\left(\frac{-1}{\sqrt{3}}\right)=-\tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)=2 \pi-\frac{\pi}{6}=\frac{11 \pi}{6}\) Hence, the polar coordinates of \(\mathrm{Q}\) are \(\left(2, \frac{11 \pi}{6}\right)\)
MHT CET-2019
Co-Ordinate system
88412
If the point \(A(5, k), B(-3,1)\) and \(C(-7,-2)\) are collinear, then \(k=\)
1 \(\frac{1}{7}\)
2 7
3 -7
4 \(\frac{-1}{7}\)
Explanation:
(B) : Slope of \(A B=\) slope of \(B C\) \(\frac{1-k}{-3-5}=\frac{-2-1}{-7+3} \Rightarrow \frac{1-k}{-8}=\frac{-3}{-4}\) \(\frac{1-k}{2}=-3 \Rightarrow k=7\)
MHT CET-2020
Co-Ordinate system
88413
The Cartesian co-ordinates of the point whose polar co-ordinates are \(\left(\frac{1}{\mathbf{2}}, 120^{\circ}\right)\) are
(C) : Given, \(P(r, \theta)=\left(\frac{1}{2}, 120^{\circ}\right) \Rightarrow r=\frac{1}{2}, \theta=120^{\circ}\) We have, \(x=r \cos \theta=\frac{1}{2} \cos 120^{\circ}=\frac{1}{2}\left(-\frac{1}{2}\right) \Rightarrow x=-\frac{1}{4}\) And \(y=r \sin \theta=\frac{1}{2} \sin 120^{\circ}\) And \(=\frac{1}{2}\left(\frac{\sqrt{3}}{2}\right)=\frac{\sqrt{3}}{4}\) \(\therefore\) Required point is \(P\left(-\frac{1}{4}, \frac{\sqrt{3}}{4}\right)\)
MHT CET-2020
Co-Ordinate system
88414
A normal to the curve \(2 x^{2}-y^{2}=14\) at the point \(\left(x_{1}, y_{1}\right)\) is parallel to the straight line \(x+3 y=4\). Then the point \(\left(\mathrm{x}_{1}, \mathrm{y}_{\mathbf{1}}\right)\) is
1 \((3,3)\)
2 \((-4,-2)\)
3 \((2,3)\)
4 \((3,2)\)
Explanation:
(D) : Given, \(2 \mathrm{x}^{2}-\mathrm{y}^{2}=14\) Differentiating (i) w.r.t \(x\), we get- \(4 x-2 y \frac{d y}{d x}=0 \Rightarrow \frac{d y}{d x}=\frac{2 x}{y}\) Let \(\mathrm{m},=\) slope of normal at \((\mathrm{x}, \mathrm{y})=,-\frac{\mathrm{y} \text {, }}{2 \mathrm{x} \text {, }}\) Also, slope of line \(x+3 y=4\) is \(m_{2}=-\frac{1}{3}\) Since the line and normal to the curve are parallel. \(\mathrm{m}_{1} \mathrm{~m}_{2}\) \(\Rightarrow-\frac{\mathrm{y}_{1}}{2 \mathrm{x}_{1}}=-\frac{1}{3} \Rightarrow 2 \mathrm{x}_{1}=3 \mathrm{y}_{1}\) \(\Rightarrow \mathrm{y}_{1}=\frac{2}{3} \mathrm{x}_{1}\) Since \(\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)\) lies on (i) \(\therefore 2 \mathrm{x}_{1}^{2}-\mathrm{y}_{1}^{2}=14 \Rightarrow 2 \mathrm{x}_{1}^{2}-\frac{4}{9} \mathrm{x}_{1}^{2}=14\) \(14 x_{1}^{2}=126\) \(\mathrm{x}_{1}= \pm 3\) \(\therefore \mathrm{y}_{1}=20 \mathrm{r}-2\) So, point \(\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)\) is \((3,2)\)
88410
\((-\sqrt{2}, \sqrt{2})\) are Cartesian co-ordinates of the point, then its polar co-ordinates are
1 \(\left(4, \frac{5 \pi}{4}\right)\)
2 \(\left(3, \frac{7 \pi}{4}\right)\)
3 \(\left(1, \frac{4 \pi}{3}\right)\)
4 \(\left(2, \frac{3 \pi}{4}\right)\)
Explanation:
(D) : Given, The Cartesian co-ordinates of the point is \((-\sqrt{2}, \sqrt{2})\) We know that, \(r \cos \theta=-\sqrt{2} \tag{i}\) \(\mathrm{r} \sin \theta=\sqrt{2}\) Squaring the both equation and adding them, \(r^{2} \cos ^{2} \theta+r^{2} \sin ^{2} \theta=2+2 \tag{ii}\) \(r^{2}=4\) \(r= \pm 2\) \(\tan \theta=\left(\frac{\sqrt{2}}{-\sqrt{2}}\right)\) \(\tan \theta=-1\) \(\theta=3 \pi / 4\) \(\therefore\) Required polar co-ordinate \((\mathrm{r}, \theta)=\left(2, \frac{3 \pi}{4}\right)\)
MHT CET-2019
Co-Ordinate system
88411
The polar coordinates of \(P\) are \(\left(2, \frac{\pi}{6}\right)\). If \(Q\) is the image of \(P\) about the \(X\)-axis, then the polar coordinates of \(Q\) are
1 \(\left(2, \frac{\pi}{6}\right)\)
2 \(\left(2, \frac{11 \pi}{6}\right)\)
3 \(\left(2, \frac{5 \pi}{6}\right)\)
4 \(\left(2, \frac{\pi}{3}\right)\)
Explanation:
(B) : Given, \(\mathrm{P} \equiv\left(2, \frac{\pi}{6}\right)\) The Cartesian coordinates of \(\mathrm{P}\) are \(\mathrm{x}=2 \cos \frac{\pi}{6}=2\left(\frac{\sqrt{3}}{2}\right)=\sqrt{3} \quad\) and \(\mathrm{y}=2 \sin \frac{\pi}{6}=2\left(\frac{1}{2}\right)=1 \Rightarrow \mathrm{p}=(\sqrt{3}, 1)\) As \(\mathrm{Q}\) is the image of \(\mathrm{P}\) about the \(\mathrm{X}\)-axis, \(\therefore \mathrm{Q} \equiv(\sqrt{3},-1)\) i.e. \(\mathrm{Q}\) lies in \(4^{\text {th }}\) quadrant Now \(\mathrm{r}=\sqrt{3+1}=2\) and \(\theta=\tan ^{-1}\left(\frac{-1}{\sqrt{3}}\right)=-\tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)=2 \pi-\frac{\pi}{6}=\frac{11 \pi}{6}\) Hence, the polar coordinates of \(\mathrm{Q}\) are \(\left(2, \frac{11 \pi}{6}\right)\)
MHT CET-2019
Co-Ordinate system
88412
If the point \(A(5, k), B(-3,1)\) and \(C(-7,-2)\) are collinear, then \(k=\)
1 \(\frac{1}{7}\)
2 7
3 -7
4 \(\frac{-1}{7}\)
Explanation:
(B) : Slope of \(A B=\) slope of \(B C\) \(\frac{1-k}{-3-5}=\frac{-2-1}{-7+3} \Rightarrow \frac{1-k}{-8}=\frac{-3}{-4}\) \(\frac{1-k}{2}=-3 \Rightarrow k=7\)
MHT CET-2020
Co-Ordinate system
88413
The Cartesian co-ordinates of the point whose polar co-ordinates are \(\left(\frac{1}{\mathbf{2}}, 120^{\circ}\right)\) are
(C) : Given, \(P(r, \theta)=\left(\frac{1}{2}, 120^{\circ}\right) \Rightarrow r=\frac{1}{2}, \theta=120^{\circ}\) We have, \(x=r \cos \theta=\frac{1}{2} \cos 120^{\circ}=\frac{1}{2}\left(-\frac{1}{2}\right) \Rightarrow x=-\frac{1}{4}\) And \(y=r \sin \theta=\frac{1}{2} \sin 120^{\circ}\) And \(=\frac{1}{2}\left(\frac{\sqrt{3}}{2}\right)=\frac{\sqrt{3}}{4}\) \(\therefore\) Required point is \(P\left(-\frac{1}{4}, \frac{\sqrt{3}}{4}\right)\)
MHT CET-2020
Co-Ordinate system
88414
A normal to the curve \(2 x^{2}-y^{2}=14\) at the point \(\left(x_{1}, y_{1}\right)\) is parallel to the straight line \(x+3 y=4\). Then the point \(\left(\mathrm{x}_{1}, \mathrm{y}_{\mathbf{1}}\right)\) is
1 \((3,3)\)
2 \((-4,-2)\)
3 \((2,3)\)
4 \((3,2)\)
Explanation:
(D) : Given, \(2 \mathrm{x}^{2}-\mathrm{y}^{2}=14\) Differentiating (i) w.r.t \(x\), we get- \(4 x-2 y \frac{d y}{d x}=0 \Rightarrow \frac{d y}{d x}=\frac{2 x}{y}\) Let \(\mathrm{m},=\) slope of normal at \((\mathrm{x}, \mathrm{y})=,-\frac{\mathrm{y} \text {, }}{2 \mathrm{x} \text {, }}\) Also, slope of line \(x+3 y=4\) is \(m_{2}=-\frac{1}{3}\) Since the line and normal to the curve are parallel. \(\mathrm{m}_{1} \mathrm{~m}_{2}\) \(\Rightarrow-\frac{\mathrm{y}_{1}}{2 \mathrm{x}_{1}}=-\frac{1}{3} \Rightarrow 2 \mathrm{x}_{1}=3 \mathrm{y}_{1}\) \(\Rightarrow \mathrm{y}_{1}=\frac{2}{3} \mathrm{x}_{1}\) Since \(\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)\) lies on (i) \(\therefore 2 \mathrm{x}_{1}^{2}-\mathrm{y}_{1}^{2}=14 \Rightarrow 2 \mathrm{x}_{1}^{2}-\frac{4}{9} \mathrm{x}_{1}^{2}=14\) \(14 x_{1}^{2}=126\) \(\mathrm{x}_{1}= \pm 3\) \(\therefore \mathrm{y}_{1}=20 \mathrm{r}-2\) So, point \(\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)\) is \((3,2)\)
88410
\((-\sqrt{2}, \sqrt{2})\) are Cartesian co-ordinates of the point, then its polar co-ordinates are
1 \(\left(4, \frac{5 \pi}{4}\right)\)
2 \(\left(3, \frac{7 \pi}{4}\right)\)
3 \(\left(1, \frac{4 \pi}{3}\right)\)
4 \(\left(2, \frac{3 \pi}{4}\right)\)
Explanation:
(D) : Given, The Cartesian co-ordinates of the point is \((-\sqrt{2}, \sqrt{2})\) We know that, \(r \cos \theta=-\sqrt{2} \tag{i}\) \(\mathrm{r} \sin \theta=\sqrt{2}\) Squaring the both equation and adding them, \(r^{2} \cos ^{2} \theta+r^{2} \sin ^{2} \theta=2+2 \tag{ii}\) \(r^{2}=4\) \(r= \pm 2\) \(\tan \theta=\left(\frac{\sqrt{2}}{-\sqrt{2}}\right)\) \(\tan \theta=-1\) \(\theta=3 \pi / 4\) \(\therefore\) Required polar co-ordinate \((\mathrm{r}, \theta)=\left(2, \frac{3 \pi}{4}\right)\)
MHT CET-2019
Co-Ordinate system
88411
The polar coordinates of \(P\) are \(\left(2, \frac{\pi}{6}\right)\). If \(Q\) is the image of \(P\) about the \(X\)-axis, then the polar coordinates of \(Q\) are
1 \(\left(2, \frac{\pi}{6}\right)\)
2 \(\left(2, \frac{11 \pi}{6}\right)\)
3 \(\left(2, \frac{5 \pi}{6}\right)\)
4 \(\left(2, \frac{\pi}{3}\right)\)
Explanation:
(B) : Given, \(\mathrm{P} \equiv\left(2, \frac{\pi}{6}\right)\) The Cartesian coordinates of \(\mathrm{P}\) are \(\mathrm{x}=2 \cos \frac{\pi}{6}=2\left(\frac{\sqrt{3}}{2}\right)=\sqrt{3} \quad\) and \(\mathrm{y}=2 \sin \frac{\pi}{6}=2\left(\frac{1}{2}\right)=1 \Rightarrow \mathrm{p}=(\sqrt{3}, 1)\) As \(\mathrm{Q}\) is the image of \(\mathrm{P}\) about the \(\mathrm{X}\)-axis, \(\therefore \mathrm{Q} \equiv(\sqrt{3},-1)\) i.e. \(\mathrm{Q}\) lies in \(4^{\text {th }}\) quadrant Now \(\mathrm{r}=\sqrt{3+1}=2\) and \(\theta=\tan ^{-1}\left(\frac{-1}{\sqrt{3}}\right)=-\tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)=2 \pi-\frac{\pi}{6}=\frac{11 \pi}{6}\) Hence, the polar coordinates of \(\mathrm{Q}\) are \(\left(2, \frac{11 \pi}{6}\right)\)
MHT CET-2019
Co-Ordinate system
88412
If the point \(A(5, k), B(-3,1)\) and \(C(-7,-2)\) are collinear, then \(k=\)
1 \(\frac{1}{7}\)
2 7
3 -7
4 \(\frac{-1}{7}\)
Explanation:
(B) : Slope of \(A B=\) slope of \(B C\) \(\frac{1-k}{-3-5}=\frac{-2-1}{-7+3} \Rightarrow \frac{1-k}{-8}=\frac{-3}{-4}\) \(\frac{1-k}{2}=-3 \Rightarrow k=7\)
MHT CET-2020
Co-Ordinate system
88413
The Cartesian co-ordinates of the point whose polar co-ordinates are \(\left(\frac{1}{\mathbf{2}}, 120^{\circ}\right)\) are
(C) : Given, \(P(r, \theta)=\left(\frac{1}{2}, 120^{\circ}\right) \Rightarrow r=\frac{1}{2}, \theta=120^{\circ}\) We have, \(x=r \cos \theta=\frac{1}{2} \cos 120^{\circ}=\frac{1}{2}\left(-\frac{1}{2}\right) \Rightarrow x=-\frac{1}{4}\) And \(y=r \sin \theta=\frac{1}{2} \sin 120^{\circ}\) And \(=\frac{1}{2}\left(\frac{\sqrt{3}}{2}\right)=\frac{\sqrt{3}}{4}\) \(\therefore\) Required point is \(P\left(-\frac{1}{4}, \frac{\sqrt{3}}{4}\right)\)
MHT CET-2020
Co-Ordinate system
88414
A normal to the curve \(2 x^{2}-y^{2}=14\) at the point \(\left(x_{1}, y_{1}\right)\) is parallel to the straight line \(x+3 y=4\). Then the point \(\left(\mathrm{x}_{1}, \mathrm{y}_{\mathbf{1}}\right)\) is
1 \((3,3)\)
2 \((-4,-2)\)
3 \((2,3)\)
4 \((3,2)\)
Explanation:
(D) : Given, \(2 \mathrm{x}^{2}-\mathrm{y}^{2}=14\) Differentiating (i) w.r.t \(x\), we get- \(4 x-2 y \frac{d y}{d x}=0 \Rightarrow \frac{d y}{d x}=\frac{2 x}{y}\) Let \(\mathrm{m},=\) slope of normal at \((\mathrm{x}, \mathrm{y})=,-\frac{\mathrm{y} \text {, }}{2 \mathrm{x} \text {, }}\) Also, slope of line \(x+3 y=4\) is \(m_{2}=-\frac{1}{3}\) Since the line and normal to the curve are parallel. \(\mathrm{m}_{1} \mathrm{~m}_{2}\) \(\Rightarrow-\frac{\mathrm{y}_{1}}{2 \mathrm{x}_{1}}=-\frac{1}{3} \Rightarrow 2 \mathrm{x}_{1}=3 \mathrm{y}_{1}\) \(\Rightarrow \mathrm{y}_{1}=\frac{2}{3} \mathrm{x}_{1}\) Since \(\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)\) lies on (i) \(\therefore 2 \mathrm{x}_{1}^{2}-\mathrm{y}_{1}^{2}=14 \Rightarrow 2 \mathrm{x}_{1}^{2}-\frac{4}{9} \mathrm{x}_{1}^{2}=14\) \(14 x_{1}^{2}=126\) \(\mathrm{x}_{1}= \pm 3\) \(\therefore \mathrm{y}_{1}=20 \mathrm{r}-2\) So, point \(\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)\) is \((3,2)\)
NEET Test Series from KOTA - 10 Papers In MS WORD
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Co-Ordinate system
88410
\((-\sqrt{2}, \sqrt{2})\) are Cartesian co-ordinates of the point, then its polar co-ordinates are
1 \(\left(4, \frac{5 \pi}{4}\right)\)
2 \(\left(3, \frac{7 \pi}{4}\right)\)
3 \(\left(1, \frac{4 \pi}{3}\right)\)
4 \(\left(2, \frac{3 \pi}{4}\right)\)
Explanation:
(D) : Given, The Cartesian co-ordinates of the point is \((-\sqrt{2}, \sqrt{2})\) We know that, \(r \cos \theta=-\sqrt{2} \tag{i}\) \(\mathrm{r} \sin \theta=\sqrt{2}\) Squaring the both equation and adding them, \(r^{2} \cos ^{2} \theta+r^{2} \sin ^{2} \theta=2+2 \tag{ii}\) \(r^{2}=4\) \(r= \pm 2\) \(\tan \theta=\left(\frac{\sqrt{2}}{-\sqrt{2}}\right)\) \(\tan \theta=-1\) \(\theta=3 \pi / 4\) \(\therefore\) Required polar co-ordinate \((\mathrm{r}, \theta)=\left(2, \frac{3 \pi}{4}\right)\)
MHT CET-2019
Co-Ordinate system
88411
The polar coordinates of \(P\) are \(\left(2, \frac{\pi}{6}\right)\). If \(Q\) is the image of \(P\) about the \(X\)-axis, then the polar coordinates of \(Q\) are
1 \(\left(2, \frac{\pi}{6}\right)\)
2 \(\left(2, \frac{11 \pi}{6}\right)\)
3 \(\left(2, \frac{5 \pi}{6}\right)\)
4 \(\left(2, \frac{\pi}{3}\right)\)
Explanation:
(B) : Given, \(\mathrm{P} \equiv\left(2, \frac{\pi}{6}\right)\) The Cartesian coordinates of \(\mathrm{P}\) are \(\mathrm{x}=2 \cos \frac{\pi}{6}=2\left(\frac{\sqrt{3}}{2}\right)=\sqrt{3} \quad\) and \(\mathrm{y}=2 \sin \frac{\pi}{6}=2\left(\frac{1}{2}\right)=1 \Rightarrow \mathrm{p}=(\sqrt{3}, 1)\) As \(\mathrm{Q}\) is the image of \(\mathrm{P}\) about the \(\mathrm{X}\)-axis, \(\therefore \mathrm{Q} \equiv(\sqrt{3},-1)\) i.e. \(\mathrm{Q}\) lies in \(4^{\text {th }}\) quadrant Now \(\mathrm{r}=\sqrt{3+1}=2\) and \(\theta=\tan ^{-1}\left(\frac{-1}{\sqrt{3}}\right)=-\tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)=2 \pi-\frac{\pi}{6}=\frac{11 \pi}{6}\) Hence, the polar coordinates of \(\mathrm{Q}\) are \(\left(2, \frac{11 \pi}{6}\right)\)
MHT CET-2019
Co-Ordinate system
88412
If the point \(A(5, k), B(-3,1)\) and \(C(-7,-2)\) are collinear, then \(k=\)
1 \(\frac{1}{7}\)
2 7
3 -7
4 \(\frac{-1}{7}\)
Explanation:
(B) : Slope of \(A B=\) slope of \(B C\) \(\frac{1-k}{-3-5}=\frac{-2-1}{-7+3} \Rightarrow \frac{1-k}{-8}=\frac{-3}{-4}\) \(\frac{1-k}{2}=-3 \Rightarrow k=7\)
MHT CET-2020
Co-Ordinate system
88413
The Cartesian co-ordinates of the point whose polar co-ordinates are \(\left(\frac{1}{\mathbf{2}}, 120^{\circ}\right)\) are
(C) : Given, \(P(r, \theta)=\left(\frac{1}{2}, 120^{\circ}\right) \Rightarrow r=\frac{1}{2}, \theta=120^{\circ}\) We have, \(x=r \cos \theta=\frac{1}{2} \cos 120^{\circ}=\frac{1}{2}\left(-\frac{1}{2}\right) \Rightarrow x=-\frac{1}{4}\) And \(y=r \sin \theta=\frac{1}{2} \sin 120^{\circ}\) And \(=\frac{1}{2}\left(\frac{\sqrt{3}}{2}\right)=\frac{\sqrt{3}}{4}\) \(\therefore\) Required point is \(P\left(-\frac{1}{4}, \frac{\sqrt{3}}{4}\right)\)
MHT CET-2020
Co-Ordinate system
88414
A normal to the curve \(2 x^{2}-y^{2}=14\) at the point \(\left(x_{1}, y_{1}\right)\) is parallel to the straight line \(x+3 y=4\). Then the point \(\left(\mathrm{x}_{1}, \mathrm{y}_{\mathbf{1}}\right)\) is
1 \((3,3)\)
2 \((-4,-2)\)
3 \((2,3)\)
4 \((3,2)\)
Explanation:
(D) : Given, \(2 \mathrm{x}^{2}-\mathrm{y}^{2}=14\) Differentiating (i) w.r.t \(x\), we get- \(4 x-2 y \frac{d y}{d x}=0 \Rightarrow \frac{d y}{d x}=\frac{2 x}{y}\) Let \(\mathrm{m},=\) slope of normal at \((\mathrm{x}, \mathrm{y})=,-\frac{\mathrm{y} \text {, }}{2 \mathrm{x} \text {, }}\) Also, slope of line \(x+3 y=4\) is \(m_{2}=-\frac{1}{3}\) Since the line and normal to the curve are parallel. \(\mathrm{m}_{1} \mathrm{~m}_{2}\) \(\Rightarrow-\frac{\mathrm{y}_{1}}{2 \mathrm{x}_{1}}=-\frac{1}{3} \Rightarrow 2 \mathrm{x}_{1}=3 \mathrm{y}_{1}\) \(\Rightarrow \mathrm{y}_{1}=\frac{2}{3} \mathrm{x}_{1}\) Since \(\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)\) lies on (i) \(\therefore 2 \mathrm{x}_{1}^{2}-\mathrm{y}_{1}^{2}=14 \Rightarrow 2 \mathrm{x}_{1}^{2}-\frac{4}{9} \mathrm{x}_{1}^{2}=14\) \(14 x_{1}^{2}=126\) \(\mathrm{x}_{1}= \pm 3\) \(\therefore \mathrm{y}_{1}=20 \mathrm{r}-2\) So, point \(\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)\) is \((3,2)\)
88410
\((-\sqrt{2}, \sqrt{2})\) are Cartesian co-ordinates of the point, then its polar co-ordinates are
1 \(\left(4, \frac{5 \pi}{4}\right)\)
2 \(\left(3, \frac{7 \pi}{4}\right)\)
3 \(\left(1, \frac{4 \pi}{3}\right)\)
4 \(\left(2, \frac{3 \pi}{4}\right)\)
Explanation:
(D) : Given, The Cartesian co-ordinates of the point is \((-\sqrt{2}, \sqrt{2})\) We know that, \(r \cos \theta=-\sqrt{2} \tag{i}\) \(\mathrm{r} \sin \theta=\sqrt{2}\) Squaring the both equation and adding them, \(r^{2} \cos ^{2} \theta+r^{2} \sin ^{2} \theta=2+2 \tag{ii}\) \(r^{2}=4\) \(r= \pm 2\) \(\tan \theta=\left(\frac{\sqrt{2}}{-\sqrt{2}}\right)\) \(\tan \theta=-1\) \(\theta=3 \pi / 4\) \(\therefore\) Required polar co-ordinate \((\mathrm{r}, \theta)=\left(2, \frac{3 \pi}{4}\right)\)
MHT CET-2019
Co-Ordinate system
88411
The polar coordinates of \(P\) are \(\left(2, \frac{\pi}{6}\right)\). If \(Q\) is the image of \(P\) about the \(X\)-axis, then the polar coordinates of \(Q\) are
1 \(\left(2, \frac{\pi}{6}\right)\)
2 \(\left(2, \frac{11 \pi}{6}\right)\)
3 \(\left(2, \frac{5 \pi}{6}\right)\)
4 \(\left(2, \frac{\pi}{3}\right)\)
Explanation:
(B) : Given, \(\mathrm{P} \equiv\left(2, \frac{\pi}{6}\right)\) The Cartesian coordinates of \(\mathrm{P}\) are \(\mathrm{x}=2 \cos \frac{\pi}{6}=2\left(\frac{\sqrt{3}}{2}\right)=\sqrt{3} \quad\) and \(\mathrm{y}=2 \sin \frac{\pi}{6}=2\left(\frac{1}{2}\right)=1 \Rightarrow \mathrm{p}=(\sqrt{3}, 1)\) As \(\mathrm{Q}\) is the image of \(\mathrm{P}\) about the \(\mathrm{X}\)-axis, \(\therefore \mathrm{Q} \equiv(\sqrt{3},-1)\) i.e. \(\mathrm{Q}\) lies in \(4^{\text {th }}\) quadrant Now \(\mathrm{r}=\sqrt{3+1}=2\) and \(\theta=\tan ^{-1}\left(\frac{-1}{\sqrt{3}}\right)=-\tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)=2 \pi-\frac{\pi}{6}=\frac{11 \pi}{6}\) Hence, the polar coordinates of \(\mathrm{Q}\) are \(\left(2, \frac{11 \pi}{6}\right)\)
MHT CET-2019
Co-Ordinate system
88412
If the point \(A(5, k), B(-3,1)\) and \(C(-7,-2)\) are collinear, then \(k=\)
1 \(\frac{1}{7}\)
2 7
3 -7
4 \(\frac{-1}{7}\)
Explanation:
(B) : Slope of \(A B=\) slope of \(B C\) \(\frac{1-k}{-3-5}=\frac{-2-1}{-7+3} \Rightarrow \frac{1-k}{-8}=\frac{-3}{-4}\) \(\frac{1-k}{2}=-3 \Rightarrow k=7\)
MHT CET-2020
Co-Ordinate system
88413
The Cartesian co-ordinates of the point whose polar co-ordinates are \(\left(\frac{1}{\mathbf{2}}, 120^{\circ}\right)\) are
(C) : Given, \(P(r, \theta)=\left(\frac{1}{2}, 120^{\circ}\right) \Rightarrow r=\frac{1}{2}, \theta=120^{\circ}\) We have, \(x=r \cos \theta=\frac{1}{2} \cos 120^{\circ}=\frac{1}{2}\left(-\frac{1}{2}\right) \Rightarrow x=-\frac{1}{4}\) And \(y=r \sin \theta=\frac{1}{2} \sin 120^{\circ}\) And \(=\frac{1}{2}\left(\frac{\sqrt{3}}{2}\right)=\frac{\sqrt{3}}{4}\) \(\therefore\) Required point is \(P\left(-\frac{1}{4}, \frac{\sqrt{3}}{4}\right)\)
MHT CET-2020
Co-Ordinate system
88414
A normal to the curve \(2 x^{2}-y^{2}=14\) at the point \(\left(x_{1}, y_{1}\right)\) is parallel to the straight line \(x+3 y=4\). Then the point \(\left(\mathrm{x}_{1}, \mathrm{y}_{\mathbf{1}}\right)\) is
1 \((3,3)\)
2 \((-4,-2)\)
3 \((2,3)\)
4 \((3,2)\)
Explanation:
(D) : Given, \(2 \mathrm{x}^{2}-\mathrm{y}^{2}=14\) Differentiating (i) w.r.t \(x\), we get- \(4 x-2 y \frac{d y}{d x}=0 \Rightarrow \frac{d y}{d x}=\frac{2 x}{y}\) Let \(\mathrm{m},=\) slope of normal at \((\mathrm{x}, \mathrm{y})=,-\frac{\mathrm{y} \text {, }}{2 \mathrm{x} \text {, }}\) Also, slope of line \(x+3 y=4\) is \(m_{2}=-\frac{1}{3}\) Since the line and normal to the curve are parallel. \(\mathrm{m}_{1} \mathrm{~m}_{2}\) \(\Rightarrow-\frac{\mathrm{y}_{1}}{2 \mathrm{x}_{1}}=-\frac{1}{3} \Rightarrow 2 \mathrm{x}_{1}=3 \mathrm{y}_{1}\) \(\Rightarrow \mathrm{y}_{1}=\frac{2}{3} \mathrm{x}_{1}\) Since \(\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)\) lies on (i) \(\therefore 2 \mathrm{x}_{1}^{2}-\mathrm{y}_{1}^{2}=14 \Rightarrow 2 \mathrm{x}_{1}^{2}-\frac{4}{9} \mathrm{x}_{1}^{2}=14\) \(14 x_{1}^{2}=126\) \(\mathrm{x}_{1}= \pm 3\) \(\therefore \mathrm{y}_{1}=20 \mathrm{r}-2\) So, point \(\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)\) is \((3,2)\)