119768
The co-ordinate of a point on the auxiliary circle of the ellipse \(x^2+2 y^2=4\) corresponding to the point on the ellipse whose eccentric angle is \(60^{\circ}\) will be
1 \((\sqrt{3}, 1)\)
2 \((1, \sqrt{3})\)
3 \((1,1)\)
4 \((1,2)\)
Explanation:
B \(\mathrm{x}=\mathrm{a} \cos \theta, \mathrm{y}=\mathrm{b} \sin \theta\). By the formula of \(\frac{\mathrm{x}^2}{\mathrm{a}^2}+\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1\) \(x^2+2 y^2=4\) \(\frac{\mathrm{x}^2}{4}+\frac{\mathrm{y}^2}{2}=1\) Compare eq \({ }^{\mathrm{n}}\). (i) \& (ii) \(a^2=4 \Rightarrow a=2,\) \(b^2=2 \Rightarrow b=\sqrt{2}\) \(x=a \cos \theta, y=a \sin \theta\) \(x=2 \cos 60^{\circ}, y=2 \sin \theta\) \(x=2 \times \frac{1}{2}=1, y=2 \times \frac{\sqrt{3}}{2}\) \(x=1, y=\sqrt{3}\) Parametric form \(\mathrm{x}=1, \quad \mathrm{y}=\sqrt{3}\) Coordinate of the point \((1, \sqrt{3})\) Hence option (b) is correct.
WB JEE-2021
Conic Section
119769
If the lines \(2 x+3 y+1=0\) and \(3 x-y-4=0\) lie along diameters of a circle of circumference \(10 \pi\), then the equation of the circle is
1 \(x^2+y^2-2 x+2 y-23=0\)
2 \(x^2+y^2-2 x-2 y-23=0\)
3 \(x^2+y^2+2 x+2 y-23=0\)
4 \(x^2+y^2+2 x-2 y-23=0\)
Explanation:
A Let centre of circle be \((\mathrm{h}, \mathrm{k})\) and radius \(=\mathrm{r}\) Circumference of circle \(=2 \pi \mathrm{r}\) \(2 \pi \mathrm{r}=10 \pi\) \(\mathrm{r}=5 \text { units }\) Also point \((\mathrm{h}, \mathrm{k})\) lies on both lines \(2 \mathrm{~h}+3 \mathrm{k}+1=0\) And \(3 \mathrm{~h}-\mathrm{k}-4=0\) Solving these two equations \(\mathrm{h}=1 \text { and } \mathrm{k}=-1\) Equation of circle becomes \((x-1)^2+(y+1)^2=(5)^2\) \(x^2+1-2 x+y^2+1+2 y=25\) \(x^2+y^2-2 x+2 y+2-25=0\) \(x^2+y^2-2 x+2 y-23=0\)
Jamia Millia Islamia-2007
Conic Section
119770
The equation of the circle passing through (1, 1) and the points of intersection of \(x^2+y^2+13 x-3 y=0\) and \(11 x+\frac{1}{2} y+\frac{25}{2}=0\) is
1 \(4 x^2+4 y^2-30 x-10 y=25\)
2 \(4 x^2+4 y^2-30 x-13 y-25=0\)
3 \(4 x^2+4 y^2-17 x-10 y+25=0\)
4 None of the above
Explanation:
D The required equation of circle is \(\left(x^2+y^2+13 x-3 y\right)+\lambda\left(11 x+\frac{1}{2} y+\frac{25}{2}\right)=0\) This circle passes through \((1,1)\). \(\left(1^2+1^2+13-3\right)+\lambda\left(11+\frac{1}{2}+\frac{25}{2}\right)=0\) \(\therefore \quad 12+\lambda(24)=0\) \(\Rightarrow \quad \lambda=-\frac{1}{2}\) On putting this value of \(\lambda\) in Eq. (i), we get \(x^2+y^2+13 x-3 y-\frac{11}{2} x-\frac{1}{4} y-\frac{25}{4}=0\) \(\Rightarrow 4 x^2+4 y^2+52 x-12 y-22 x-y-25=0\) \(\Rightarrow 4 x^2+4 y^2+30 x-13 y-25=0\)
Manipal UGET-2012
Conic Section
119771
The \(\triangle \mathrm{PQR}\) is inscribed in the circle \(\mathrm{x}^2+\mathrm{y}^2=25\) If \(Q\) and \(R\) have coordinates \((3,4)\) and \((-4,3)\), respectively. Then, \(\angle \mathrm{QPR}\) is equal to
1 \(\frac{\pi}{2}\)
2 \(\frac{\pi}{3}\)
3 \(\frac{\pi}{4}\)
4 \(\frac{\pi}{6}\)
Explanation:
C Let \(\mathrm{O}\) is the point on centre and \(\mathrm{P}\) is the point on circumference. Therefore, angle QOR is double the angle QPR. So, it is sufficient to find the angle QOR. Now, slope of OQ, \(\mathrm{m}_1=\frac{4-0}{3-0}=\frac{4}{3}\) Slope of OR, \(\mathrm{m}_2=\frac{3-0}{-4-0}=-\frac{3}{4}\) Again, \(\mathrm{m}_1 \mathrm{~m}_2=-1\) Therefore, \(\quad \angle \mathrm{QPR}=\frac{90^{\circ}}{2}=45^{\circ}\)
Manipal UGET-2012
Conic Section
119772
One diagonal of a square is along the line \(8 x-\) \(15 y=0\) and one of its vertex is \((1,2)\). Then, the equation of the sides of the square passing though this vertex are
B Slope of \(\mathrm{BD}=\frac{- \text { Coeffcient of } \mathrm{x}}{\text { Coffcient } \mathrm{y}}=\frac{8}{15} \text { angle }\) \(\text { made by } \mathrm{BD} \text { with } \mathrm{AD} \text { and } \mathrm{DC} \text { is } 45^{\circ} \text {. }\) \(\tan 45^{\circ}= \pm \frac{\mathrm{m}-8 / 15}{1+\mathrm{m} \cdot 8 / 15}\) \(\text { or }(15+8 m)= \pm(15 m-8)\) \(\mathrm{m}=\frac{23}{7} \text { and } \frac{-7}{23}\) \(\mathrm{y}-2=\frac{23}{7}(\mathrm{x}-1)\) \(\text { or } 23 x-7 y-9=0\) \(\mathrm{y}-2=-\frac{7}{23}(\mathrm{x}-1)\) \(7 \mathrm{x}+23 \mathrm{y}-53=0\) Let slope of DC be \(\mathrm{m}\), then- \(\therefore\) Equation of line DC and equation of line \(\mathrm{BC}\) is
119768
The co-ordinate of a point on the auxiliary circle of the ellipse \(x^2+2 y^2=4\) corresponding to the point on the ellipse whose eccentric angle is \(60^{\circ}\) will be
1 \((\sqrt{3}, 1)\)
2 \((1, \sqrt{3})\)
3 \((1,1)\)
4 \((1,2)\)
Explanation:
B \(\mathrm{x}=\mathrm{a} \cos \theta, \mathrm{y}=\mathrm{b} \sin \theta\). By the formula of \(\frac{\mathrm{x}^2}{\mathrm{a}^2}+\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1\) \(x^2+2 y^2=4\) \(\frac{\mathrm{x}^2}{4}+\frac{\mathrm{y}^2}{2}=1\) Compare eq \({ }^{\mathrm{n}}\). (i) \& (ii) \(a^2=4 \Rightarrow a=2,\) \(b^2=2 \Rightarrow b=\sqrt{2}\) \(x=a \cos \theta, y=a \sin \theta\) \(x=2 \cos 60^{\circ}, y=2 \sin \theta\) \(x=2 \times \frac{1}{2}=1, y=2 \times \frac{\sqrt{3}}{2}\) \(x=1, y=\sqrt{3}\) Parametric form \(\mathrm{x}=1, \quad \mathrm{y}=\sqrt{3}\) Coordinate of the point \((1, \sqrt{3})\) Hence option (b) is correct.
WB JEE-2021
Conic Section
119769
If the lines \(2 x+3 y+1=0\) and \(3 x-y-4=0\) lie along diameters of a circle of circumference \(10 \pi\), then the equation of the circle is
1 \(x^2+y^2-2 x+2 y-23=0\)
2 \(x^2+y^2-2 x-2 y-23=0\)
3 \(x^2+y^2+2 x+2 y-23=0\)
4 \(x^2+y^2+2 x-2 y-23=0\)
Explanation:
A Let centre of circle be \((\mathrm{h}, \mathrm{k})\) and radius \(=\mathrm{r}\) Circumference of circle \(=2 \pi \mathrm{r}\) \(2 \pi \mathrm{r}=10 \pi\) \(\mathrm{r}=5 \text { units }\) Also point \((\mathrm{h}, \mathrm{k})\) lies on both lines \(2 \mathrm{~h}+3 \mathrm{k}+1=0\) And \(3 \mathrm{~h}-\mathrm{k}-4=0\) Solving these two equations \(\mathrm{h}=1 \text { and } \mathrm{k}=-1\) Equation of circle becomes \((x-1)^2+(y+1)^2=(5)^2\) \(x^2+1-2 x+y^2+1+2 y=25\) \(x^2+y^2-2 x+2 y+2-25=0\) \(x^2+y^2-2 x+2 y-23=0\)
Jamia Millia Islamia-2007
Conic Section
119770
The equation of the circle passing through (1, 1) and the points of intersection of \(x^2+y^2+13 x-3 y=0\) and \(11 x+\frac{1}{2} y+\frac{25}{2}=0\) is
1 \(4 x^2+4 y^2-30 x-10 y=25\)
2 \(4 x^2+4 y^2-30 x-13 y-25=0\)
3 \(4 x^2+4 y^2-17 x-10 y+25=0\)
4 None of the above
Explanation:
D The required equation of circle is \(\left(x^2+y^2+13 x-3 y\right)+\lambda\left(11 x+\frac{1}{2} y+\frac{25}{2}\right)=0\) This circle passes through \((1,1)\). \(\left(1^2+1^2+13-3\right)+\lambda\left(11+\frac{1}{2}+\frac{25}{2}\right)=0\) \(\therefore \quad 12+\lambda(24)=0\) \(\Rightarrow \quad \lambda=-\frac{1}{2}\) On putting this value of \(\lambda\) in Eq. (i), we get \(x^2+y^2+13 x-3 y-\frac{11}{2} x-\frac{1}{4} y-\frac{25}{4}=0\) \(\Rightarrow 4 x^2+4 y^2+52 x-12 y-22 x-y-25=0\) \(\Rightarrow 4 x^2+4 y^2+30 x-13 y-25=0\)
Manipal UGET-2012
Conic Section
119771
The \(\triangle \mathrm{PQR}\) is inscribed in the circle \(\mathrm{x}^2+\mathrm{y}^2=25\) If \(Q\) and \(R\) have coordinates \((3,4)\) and \((-4,3)\), respectively. Then, \(\angle \mathrm{QPR}\) is equal to
1 \(\frac{\pi}{2}\)
2 \(\frac{\pi}{3}\)
3 \(\frac{\pi}{4}\)
4 \(\frac{\pi}{6}\)
Explanation:
C Let \(\mathrm{O}\) is the point on centre and \(\mathrm{P}\) is the point on circumference. Therefore, angle QOR is double the angle QPR. So, it is sufficient to find the angle QOR. Now, slope of OQ, \(\mathrm{m}_1=\frac{4-0}{3-0}=\frac{4}{3}\) Slope of OR, \(\mathrm{m}_2=\frac{3-0}{-4-0}=-\frac{3}{4}\) Again, \(\mathrm{m}_1 \mathrm{~m}_2=-1\) Therefore, \(\quad \angle \mathrm{QPR}=\frac{90^{\circ}}{2}=45^{\circ}\)
Manipal UGET-2012
Conic Section
119772
One diagonal of a square is along the line \(8 x-\) \(15 y=0\) and one of its vertex is \((1,2)\). Then, the equation of the sides of the square passing though this vertex are
B Slope of \(\mathrm{BD}=\frac{- \text { Coeffcient of } \mathrm{x}}{\text { Coffcient } \mathrm{y}}=\frac{8}{15} \text { angle }\) \(\text { made by } \mathrm{BD} \text { with } \mathrm{AD} \text { and } \mathrm{DC} \text { is } 45^{\circ} \text {. }\) \(\tan 45^{\circ}= \pm \frac{\mathrm{m}-8 / 15}{1+\mathrm{m} \cdot 8 / 15}\) \(\text { or }(15+8 m)= \pm(15 m-8)\) \(\mathrm{m}=\frac{23}{7} \text { and } \frac{-7}{23}\) \(\mathrm{y}-2=\frac{23}{7}(\mathrm{x}-1)\) \(\text { or } 23 x-7 y-9=0\) \(\mathrm{y}-2=-\frac{7}{23}(\mathrm{x}-1)\) \(7 \mathrm{x}+23 \mathrm{y}-53=0\) Let slope of DC be \(\mathrm{m}\), then- \(\therefore\) Equation of line DC and equation of line \(\mathrm{BC}\) is
NEET Test Series from KOTA - 10 Papers In MS WORD
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Conic Section
119768
The co-ordinate of a point on the auxiliary circle of the ellipse \(x^2+2 y^2=4\) corresponding to the point on the ellipse whose eccentric angle is \(60^{\circ}\) will be
1 \((\sqrt{3}, 1)\)
2 \((1, \sqrt{3})\)
3 \((1,1)\)
4 \((1,2)\)
Explanation:
B \(\mathrm{x}=\mathrm{a} \cos \theta, \mathrm{y}=\mathrm{b} \sin \theta\). By the formula of \(\frac{\mathrm{x}^2}{\mathrm{a}^2}+\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1\) \(x^2+2 y^2=4\) \(\frac{\mathrm{x}^2}{4}+\frac{\mathrm{y}^2}{2}=1\) Compare eq \({ }^{\mathrm{n}}\). (i) \& (ii) \(a^2=4 \Rightarrow a=2,\) \(b^2=2 \Rightarrow b=\sqrt{2}\) \(x=a \cos \theta, y=a \sin \theta\) \(x=2 \cos 60^{\circ}, y=2 \sin \theta\) \(x=2 \times \frac{1}{2}=1, y=2 \times \frac{\sqrt{3}}{2}\) \(x=1, y=\sqrt{3}\) Parametric form \(\mathrm{x}=1, \quad \mathrm{y}=\sqrt{3}\) Coordinate of the point \((1, \sqrt{3})\) Hence option (b) is correct.
WB JEE-2021
Conic Section
119769
If the lines \(2 x+3 y+1=0\) and \(3 x-y-4=0\) lie along diameters of a circle of circumference \(10 \pi\), then the equation of the circle is
1 \(x^2+y^2-2 x+2 y-23=0\)
2 \(x^2+y^2-2 x-2 y-23=0\)
3 \(x^2+y^2+2 x+2 y-23=0\)
4 \(x^2+y^2+2 x-2 y-23=0\)
Explanation:
A Let centre of circle be \((\mathrm{h}, \mathrm{k})\) and radius \(=\mathrm{r}\) Circumference of circle \(=2 \pi \mathrm{r}\) \(2 \pi \mathrm{r}=10 \pi\) \(\mathrm{r}=5 \text { units }\) Also point \((\mathrm{h}, \mathrm{k})\) lies on both lines \(2 \mathrm{~h}+3 \mathrm{k}+1=0\) And \(3 \mathrm{~h}-\mathrm{k}-4=0\) Solving these two equations \(\mathrm{h}=1 \text { and } \mathrm{k}=-1\) Equation of circle becomes \((x-1)^2+(y+1)^2=(5)^2\) \(x^2+1-2 x+y^2+1+2 y=25\) \(x^2+y^2-2 x+2 y+2-25=0\) \(x^2+y^2-2 x+2 y-23=0\)
Jamia Millia Islamia-2007
Conic Section
119770
The equation of the circle passing through (1, 1) and the points of intersection of \(x^2+y^2+13 x-3 y=0\) and \(11 x+\frac{1}{2} y+\frac{25}{2}=0\) is
1 \(4 x^2+4 y^2-30 x-10 y=25\)
2 \(4 x^2+4 y^2-30 x-13 y-25=0\)
3 \(4 x^2+4 y^2-17 x-10 y+25=0\)
4 None of the above
Explanation:
D The required equation of circle is \(\left(x^2+y^2+13 x-3 y\right)+\lambda\left(11 x+\frac{1}{2} y+\frac{25}{2}\right)=0\) This circle passes through \((1,1)\). \(\left(1^2+1^2+13-3\right)+\lambda\left(11+\frac{1}{2}+\frac{25}{2}\right)=0\) \(\therefore \quad 12+\lambda(24)=0\) \(\Rightarrow \quad \lambda=-\frac{1}{2}\) On putting this value of \(\lambda\) in Eq. (i), we get \(x^2+y^2+13 x-3 y-\frac{11}{2} x-\frac{1}{4} y-\frac{25}{4}=0\) \(\Rightarrow 4 x^2+4 y^2+52 x-12 y-22 x-y-25=0\) \(\Rightarrow 4 x^2+4 y^2+30 x-13 y-25=0\)
Manipal UGET-2012
Conic Section
119771
The \(\triangle \mathrm{PQR}\) is inscribed in the circle \(\mathrm{x}^2+\mathrm{y}^2=25\) If \(Q\) and \(R\) have coordinates \((3,4)\) and \((-4,3)\), respectively. Then, \(\angle \mathrm{QPR}\) is equal to
1 \(\frac{\pi}{2}\)
2 \(\frac{\pi}{3}\)
3 \(\frac{\pi}{4}\)
4 \(\frac{\pi}{6}\)
Explanation:
C Let \(\mathrm{O}\) is the point on centre and \(\mathrm{P}\) is the point on circumference. Therefore, angle QOR is double the angle QPR. So, it is sufficient to find the angle QOR. Now, slope of OQ, \(\mathrm{m}_1=\frac{4-0}{3-0}=\frac{4}{3}\) Slope of OR, \(\mathrm{m}_2=\frac{3-0}{-4-0}=-\frac{3}{4}\) Again, \(\mathrm{m}_1 \mathrm{~m}_2=-1\) Therefore, \(\quad \angle \mathrm{QPR}=\frac{90^{\circ}}{2}=45^{\circ}\)
Manipal UGET-2012
Conic Section
119772
One diagonal of a square is along the line \(8 x-\) \(15 y=0\) and one of its vertex is \((1,2)\). Then, the equation of the sides of the square passing though this vertex are
B Slope of \(\mathrm{BD}=\frac{- \text { Coeffcient of } \mathrm{x}}{\text { Coffcient } \mathrm{y}}=\frac{8}{15} \text { angle }\) \(\text { made by } \mathrm{BD} \text { with } \mathrm{AD} \text { and } \mathrm{DC} \text { is } 45^{\circ} \text {. }\) \(\tan 45^{\circ}= \pm \frac{\mathrm{m}-8 / 15}{1+\mathrm{m} \cdot 8 / 15}\) \(\text { or }(15+8 m)= \pm(15 m-8)\) \(\mathrm{m}=\frac{23}{7} \text { and } \frac{-7}{23}\) \(\mathrm{y}-2=\frac{23}{7}(\mathrm{x}-1)\) \(\text { or } 23 x-7 y-9=0\) \(\mathrm{y}-2=-\frac{7}{23}(\mathrm{x}-1)\) \(7 \mathrm{x}+23 \mathrm{y}-53=0\) Let slope of DC be \(\mathrm{m}\), then- \(\therefore\) Equation of line DC and equation of line \(\mathrm{BC}\) is
119768
The co-ordinate of a point on the auxiliary circle of the ellipse \(x^2+2 y^2=4\) corresponding to the point on the ellipse whose eccentric angle is \(60^{\circ}\) will be
1 \((\sqrt{3}, 1)\)
2 \((1, \sqrt{3})\)
3 \((1,1)\)
4 \((1,2)\)
Explanation:
B \(\mathrm{x}=\mathrm{a} \cos \theta, \mathrm{y}=\mathrm{b} \sin \theta\). By the formula of \(\frac{\mathrm{x}^2}{\mathrm{a}^2}+\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1\) \(x^2+2 y^2=4\) \(\frac{\mathrm{x}^2}{4}+\frac{\mathrm{y}^2}{2}=1\) Compare eq \({ }^{\mathrm{n}}\). (i) \& (ii) \(a^2=4 \Rightarrow a=2,\) \(b^2=2 \Rightarrow b=\sqrt{2}\) \(x=a \cos \theta, y=a \sin \theta\) \(x=2 \cos 60^{\circ}, y=2 \sin \theta\) \(x=2 \times \frac{1}{2}=1, y=2 \times \frac{\sqrt{3}}{2}\) \(x=1, y=\sqrt{3}\) Parametric form \(\mathrm{x}=1, \quad \mathrm{y}=\sqrt{3}\) Coordinate of the point \((1, \sqrt{3})\) Hence option (b) is correct.
WB JEE-2021
Conic Section
119769
If the lines \(2 x+3 y+1=0\) and \(3 x-y-4=0\) lie along diameters of a circle of circumference \(10 \pi\), then the equation of the circle is
1 \(x^2+y^2-2 x+2 y-23=0\)
2 \(x^2+y^2-2 x-2 y-23=0\)
3 \(x^2+y^2+2 x+2 y-23=0\)
4 \(x^2+y^2+2 x-2 y-23=0\)
Explanation:
A Let centre of circle be \((\mathrm{h}, \mathrm{k})\) and radius \(=\mathrm{r}\) Circumference of circle \(=2 \pi \mathrm{r}\) \(2 \pi \mathrm{r}=10 \pi\) \(\mathrm{r}=5 \text { units }\) Also point \((\mathrm{h}, \mathrm{k})\) lies on both lines \(2 \mathrm{~h}+3 \mathrm{k}+1=0\) And \(3 \mathrm{~h}-\mathrm{k}-4=0\) Solving these two equations \(\mathrm{h}=1 \text { and } \mathrm{k}=-1\) Equation of circle becomes \((x-1)^2+(y+1)^2=(5)^2\) \(x^2+1-2 x+y^2+1+2 y=25\) \(x^2+y^2-2 x+2 y+2-25=0\) \(x^2+y^2-2 x+2 y-23=0\)
Jamia Millia Islamia-2007
Conic Section
119770
The equation of the circle passing through (1, 1) and the points of intersection of \(x^2+y^2+13 x-3 y=0\) and \(11 x+\frac{1}{2} y+\frac{25}{2}=0\) is
1 \(4 x^2+4 y^2-30 x-10 y=25\)
2 \(4 x^2+4 y^2-30 x-13 y-25=0\)
3 \(4 x^2+4 y^2-17 x-10 y+25=0\)
4 None of the above
Explanation:
D The required equation of circle is \(\left(x^2+y^2+13 x-3 y\right)+\lambda\left(11 x+\frac{1}{2} y+\frac{25}{2}\right)=0\) This circle passes through \((1,1)\). \(\left(1^2+1^2+13-3\right)+\lambda\left(11+\frac{1}{2}+\frac{25}{2}\right)=0\) \(\therefore \quad 12+\lambda(24)=0\) \(\Rightarrow \quad \lambda=-\frac{1}{2}\) On putting this value of \(\lambda\) in Eq. (i), we get \(x^2+y^2+13 x-3 y-\frac{11}{2} x-\frac{1}{4} y-\frac{25}{4}=0\) \(\Rightarrow 4 x^2+4 y^2+52 x-12 y-22 x-y-25=0\) \(\Rightarrow 4 x^2+4 y^2+30 x-13 y-25=0\)
Manipal UGET-2012
Conic Section
119771
The \(\triangle \mathrm{PQR}\) is inscribed in the circle \(\mathrm{x}^2+\mathrm{y}^2=25\) If \(Q\) and \(R\) have coordinates \((3,4)\) and \((-4,3)\), respectively. Then, \(\angle \mathrm{QPR}\) is equal to
1 \(\frac{\pi}{2}\)
2 \(\frac{\pi}{3}\)
3 \(\frac{\pi}{4}\)
4 \(\frac{\pi}{6}\)
Explanation:
C Let \(\mathrm{O}\) is the point on centre and \(\mathrm{P}\) is the point on circumference. Therefore, angle QOR is double the angle QPR. So, it is sufficient to find the angle QOR. Now, slope of OQ, \(\mathrm{m}_1=\frac{4-0}{3-0}=\frac{4}{3}\) Slope of OR, \(\mathrm{m}_2=\frac{3-0}{-4-0}=-\frac{3}{4}\) Again, \(\mathrm{m}_1 \mathrm{~m}_2=-1\) Therefore, \(\quad \angle \mathrm{QPR}=\frac{90^{\circ}}{2}=45^{\circ}\)
Manipal UGET-2012
Conic Section
119772
One diagonal of a square is along the line \(8 x-\) \(15 y=0\) and one of its vertex is \((1,2)\). Then, the equation of the sides of the square passing though this vertex are
B Slope of \(\mathrm{BD}=\frac{- \text { Coeffcient of } \mathrm{x}}{\text { Coffcient } \mathrm{y}}=\frac{8}{15} \text { angle }\) \(\text { made by } \mathrm{BD} \text { with } \mathrm{AD} \text { and } \mathrm{DC} \text { is } 45^{\circ} \text {. }\) \(\tan 45^{\circ}= \pm \frac{\mathrm{m}-8 / 15}{1+\mathrm{m} \cdot 8 / 15}\) \(\text { or }(15+8 m)= \pm(15 m-8)\) \(\mathrm{m}=\frac{23}{7} \text { and } \frac{-7}{23}\) \(\mathrm{y}-2=\frac{23}{7}(\mathrm{x}-1)\) \(\text { or } 23 x-7 y-9=0\) \(\mathrm{y}-2=-\frac{7}{23}(\mathrm{x}-1)\) \(7 \mathrm{x}+23 \mathrm{y}-53=0\) Let slope of DC be \(\mathrm{m}\), then- \(\therefore\) Equation of line DC and equation of line \(\mathrm{BC}\) is
119768
The co-ordinate of a point on the auxiliary circle of the ellipse \(x^2+2 y^2=4\) corresponding to the point on the ellipse whose eccentric angle is \(60^{\circ}\) will be
1 \((\sqrt{3}, 1)\)
2 \((1, \sqrt{3})\)
3 \((1,1)\)
4 \((1,2)\)
Explanation:
B \(\mathrm{x}=\mathrm{a} \cos \theta, \mathrm{y}=\mathrm{b} \sin \theta\). By the formula of \(\frac{\mathrm{x}^2}{\mathrm{a}^2}+\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1\) \(x^2+2 y^2=4\) \(\frac{\mathrm{x}^2}{4}+\frac{\mathrm{y}^2}{2}=1\) Compare eq \({ }^{\mathrm{n}}\). (i) \& (ii) \(a^2=4 \Rightarrow a=2,\) \(b^2=2 \Rightarrow b=\sqrt{2}\) \(x=a \cos \theta, y=a \sin \theta\) \(x=2 \cos 60^{\circ}, y=2 \sin \theta\) \(x=2 \times \frac{1}{2}=1, y=2 \times \frac{\sqrt{3}}{2}\) \(x=1, y=\sqrt{3}\) Parametric form \(\mathrm{x}=1, \quad \mathrm{y}=\sqrt{3}\) Coordinate of the point \((1, \sqrt{3})\) Hence option (b) is correct.
WB JEE-2021
Conic Section
119769
If the lines \(2 x+3 y+1=0\) and \(3 x-y-4=0\) lie along diameters of a circle of circumference \(10 \pi\), then the equation of the circle is
1 \(x^2+y^2-2 x+2 y-23=0\)
2 \(x^2+y^2-2 x-2 y-23=0\)
3 \(x^2+y^2+2 x+2 y-23=0\)
4 \(x^2+y^2+2 x-2 y-23=0\)
Explanation:
A Let centre of circle be \((\mathrm{h}, \mathrm{k})\) and radius \(=\mathrm{r}\) Circumference of circle \(=2 \pi \mathrm{r}\) \(2 \pi \mathrm{r}=10 \pi\) \(\mathrm{r}=5 \text { units }\) Also point \((\mathrm{h}, \mathrm{k})\) lies on both lines \(2 \mathrm{~h}+3 \mathrm{k}+1=0\) And \(3 \mathrm{~h}-\mathrm{k}-4=0\) Solving these two equations \(\mathrm{h}=1 \text { and } \mathrm{k}=-1\) Equation of circle becomes \((x-1)^2+(y+1)^2=(5)^2\) \(x^2+1-2 x+y^2+1+2 y=25\) \(x^2+y^2-2 x+2 y+2-25=0\) \(x^2+y^2-2 x+2 y-23=0\)
Jamia Millia Islamia-2007
Conic Section
119770
The equation of the circle passing through (1, 1) and the points of intersection of \(x^2+y^2+13 x-3 y=0\) and \(11 x+\frac{1}{2} y+\frac{25}{2}=0\) is
1 \(4 x^2+4 y^2-30 x-10 y=25\)
2 \(4 x^2+4 y^2-30 x-13 y-25=0\)
3 \(4 x^2+4 y^2-17 x-10 y+25=0\)
4 None of the above
Explanation:
D The required equation of circle is \(\left(x^2+y^2+13 x-3 y\right)+\lambda\left(11 x+\frac{1}{2} y+\frac{25}{2}\right)=0\) This circle passes through \((1,1)\). \(\left(1^2+1^2+13-3\right)+\lambda\left(11+\frac{1}{2}+\frac{25}{2}\right)=0\) \(\therefore \quad 12+\lambda(24)=0\) \(\Rightarrow \quad \lambda=-\frac{1}{2}\) On putting this value of \(\lambda\) in Eq. (i), we get \(x^2+y^2+13 x-3 y-\frac{11}{2} x-\frac{1}{4} y-\frac{25}{4}=0\) \(\Rightarrow 4 x^2+4 y^2+52 x-12 y-22 x-y-25=0\) \(\Rightarrow 4 x^2+4 y^2+30 x-13 y-25=0\)
Manipal UGET-2012
Conic Section
119771
The \(\triangle \mathrm{PQR}\) is inscribed in the circle \(\mathrm{x}^2+\mathrm{y}^2=25\) If \(Q\) and \(R\) have coordinates \((3,4)\) and \((-4,3)\), respectively. Then, \(\angle \mathrm{QPR}\) is equal to
1 \(\frac{\pi}{2}\)
2 \(\frac{\pi}{3}\)
3 \(\frac{\pi}{4}\)
4 \(\frac{\pi}{6}\)
Explanation:
C Let \(\mathrm{O}\) is the point on centre and \(\mathrm{P}\) is the point on circumference. Therefore, angle QOR is double the angle QPR. So, it is sufficient to find the angle QOR. Now, slope of OQ, \(\mathrm{m}_1=\frac{4-0}{3-0}=\frac{4}{3}\) Slope of OR, \(\mathrm{m}_2=\frac{3-0}{-4-0}=-\frac{3}{4}\) Again, \(\mathrm{m}_1 \mathrm{~m}_2=-1\) Therefore, \(\quad \angle \mathrm{QPR}=\frac{90^{\circ}}{2}=45^{\circ}\)
Manipal UGET-2012
Conic Section
119772
One diagonal of a square is along the line \(8 x-\) \(15 y=0\) and one of its vertex is \((1,2)\). Then, the equation of the sides of the square passing though this vertex are
B Slope of \(\mathrm{BD}=\frac{- \text { Coeffcient of } \mathrm{x}}{\text { Coffcient } \mathrm{y}}=\frac{8}{15} \text { angle }\) \(\text { made by } \mathrm{BD} \text { with } \mathrm{AD} \text { and } \mathrm{DC} \text { is } 45^{\circ} \text {. }\) \(\tan 45^{\circ}= \pm \frac{\mathrm{m}-8 / 15}{1+\mathrm{m} \cdot 8 / 15}\) \(\text { or }(15+8 m)= \pm(15 m-8)\) \(\mathrm{m}=\frac{23}{7} \text { and } \frac{-7}{23}\) \(\mathrm{y}-2=\frac{23}{7}(\mathrm{x}-1)\) \(\text { or } 23 x-7 y-9=0\) \(\mathrm{y}-2=-\frac{7}{23}(\mathrm{x}-1)\) \(7 \mathrm{x}+23 \mathrm{y}-53=0\) Let slope of DC be \(\mathrm{m}\), then- \(\therefore\) Equation of line DC and equation of line \(\mathrm{BC}\) is
