D Given, equation is
\(2 x^2+5 y-6 x+1=0\)
\(2 x^2-6 x+\frac{9}{2}+5 y+1-\frac{9}{2}=0\)
\(2\left(x-\frac{3}{2}\right)^2=-5\left(y-\frac{7}{10}\right)\)
\(\left(x-\frac{3}{2}\right)^2=\frac{-5}{2}\left(y-\frac{7}{10}\right)\)
\(X^2=\frac{-5}{2} Y\)
Where \(\mathrm{X}=\mathrm{x}-\frac{3}{2}\) and \(\mathrm{Y}=\mathrm{y}-\frac{7}{10}\)
For vertex, on putting \(\mathrm{X}=0, \mathrm{Y}=0\)
\(\mathrm{x}=\frac{3}{2} \text { and } \mathrm{y}=\frac{7}{10}\)
\(\therefore \quad \text { Vertex }=\left(\frac{3}{2}, \frac{7}{10}\right)\)
\(\mathrm{x}=\frac{3}{2} \text { and } \mathrm{y}-\frac{7}{10}=\frac{-5}{8}\)
\(\mathrm{y}=\frac{3}{40}\)
\(\therefore \quad \text { Focus }=\left(\frac{3}{2}, \frac{3}{40}\right)\)
For focus, \(\mathrm{X}=0\) and \(\mathrm{Y}=-\mathrm{a}\)
TS EAMCET-18.07.2022
Parabola
120974
If all the vertices of an equilateral triangle lie on the parabola \(y^2=16 x\) and one of them coincides with the vertex of that parabola, then the length of the side of that triangle is
1 \(32 \sqrt{3}\)
2 \(16 \sqrt{3}\)
3 \(8 \sqrt{3}\)
4 32
Explanation:
A Given equation of parabola, \(y^2=16 x\) and all the vertices are of an equilateral triangle with one of them coincides with the vertex of parabola
Since, given parabola is symmetrical about \(\mathrm{X}\)-axis So, \(\angle \mathrm{AOM}=30^{\circ}\)
In \(\angle \mathrm{AOM}, \tan 30^{\circ}=\frac{\mathrm{AM}}{\mathrm{OM}}=\frac{8 \mathrm{t}}{4 \mathrm{t}^2} \Rightarrow \frac{1}{\sqrt{3}}=\frac{2}{\mathrm{t}} \Rightarrow \mathrm{t}=2 \sqrt{3}\)
\(\therefore\) Distance \(\mathrm{OA}=\sqrt{\left(\mathrm{x}_{\mathrm{A}}-\mathrm{x}_{\mathrm{o}}\right)^2+\left(\mathrm{y}_{\mathrm{A}}-\mathrm{y}_{\mathrm{O}}\right)^2}\)
\(=\sqrt{(48-0)^2+(16 \sqrt{3}-0)^2}\)
\(\mathrm{OA} =32 \sqrt{3}\)\(\therefore\) Length of the side of equilateral triangle \(=32 \sqrt{3}\)
TS EAMCET-11.09.2020
Parabola
120975
If a circle with its centre at the focus of the parabola \(\mathrm{y}^2=2 \mathrm{px}\) is such that it touches the directrix of the parabola, then a point of intersection of the circle and the parabola is
C Equation of given parabola is \(\mathrm{y}^2=2 \mathrm{px}\) having focus \(\mathrm{F}\left(\frac{\mathrm{P}}{2}, 0\right)\) and equation of directrix is \(x+\frac{P}{2}=0\), so equation of circle having centre \(\left(\frac{\mathrm{p}}{2}, 0\right)\) and radius \(\mathrm{r}=\mathrm{p}\), because the circle touches the directrix of the given parabola, is
\((\mathrm{x}-\mathrm{p} / 2)^2+\mathrm{y}^2=\mathrm{p}^2\)
Now, on solving the equation of the given parabola and circle (i), we get
\(\left(\mathrm{x}^2-\mathrm{px}+\frac{\mathrm{p}^2}{4}\right)+2 \mathrm{px}=\mathrm{p}^2 \Rightarrow\left(\mathrm{x}^2+\mathrm{px}+\frac{\mathrm{p}^2}{4}\right)=\mathrm{p}^2\)
\(\left(\mathrm{x}+\frac{\mathrm{p}}{2}\right)^2=\mathrm{p}^2 \Rightarrow \mathrm{x}+\frac{\mathrm{p}}{2}= \pm \mathrm{p}\)
\(\mathrm{x}=\frac{\mathrm{p}}{2} \quad[\because \mathrm{xp}>0]\)
\(y= \pm p\)
\(\left(\frac{\mathrm{p}}{2}, \mathrm{p}\right),\left(\frac{\mathrm{p}}{2},-\mathrm{p}\right)\)
and
\(\therefore\) Point of intersection of circle and parabola are
\(\left(\frac{\mathrm{p}}{2}, \mathrm{p}\right),\left(\frac{\mathrm{p}}{2},-\mathrm{p}\right)\)
TS EAMCET-11.09.2020
Parabola
120976
A point on the parabola whose focus and vertex are respectively at \(\left(\frac{5}{4},-2\right)\) and \((1,-2)\) is
1 \((4,0)\)
2 \((15,2)\)
3 \((3,-1)\)
4 \((10,1)\)
Explanation:
D :
Given,
\(\text { Focus }=\left(\frac{5}{4},-2\right)\)
\(\text { Vertex }=(1,-2)\)
\(\mathrm{a}=\sqrt{\left(1-\frac{5}{4}\right)^2+(-2+2)^2}\)
\(=\sqrt{\left(\frac{-1}{4}\right)^2}\)
\(\mathrm{a}=\frac{1}{4}\)
Equation of parabola,
\(\quad(y+2)^2 =4 \times \frac{1}{4}(x-1)\)
\(\Rightarrow \quad(y+2)^2 =(x-1)\)
\(\therefore \quad(10,1)\) satisfies equation (i)
Therefore \((10,1)\) is a point on the parabola.
D Given, equation is
\(2 x^2+5 y-6 x+1=0\)
\(2 x^2-6 x+\frac{9}{2}+5 y+1-\frac{9}{2}=0\)
\(2\left(x-\frac{3}{2}\right)^2=-5\left(y-\frac{7}{10}\right)\)
\(\left(x-\frac{3}{2}\right)^2=\frac{-5}{2}\left(y-\frac{7}{10}\right)\)
\(X^2=\frac{-5}{2} Y\)
Where \(\mathrm{X}=\mathrm{x}-\frac{3}{2}\) and \(\mathrm{Y}=\mathrm{y}-\frac{7}{10}\)
For vertex, on putting \(\mathrm{X}=0, \mathrm{Y}=0\)
\(\mathrm{x}=\frac{3}{2} \text { and } \mathrm{y}=\frac{7}{10}\)
\(\therefore \quad \text { Vertex }=\left(\frac{3}{2}, \frac{7}{10}\right)\)
\(\mathrm{x}=\frac{3}{2} \text { and } \mathrm{y}-\frac{7}{10}=\frac{-5}{8}\)
\(\mathrm{y}=\frac{3}{40}\)
\(\therefore \quad \text { Focus }=\left(\frac{3}{2}, \frac{3}{40}\right)\)
For focus, \(\mathrm{X}=0\) and \(\mathrm{Y}=-\mathrm{a}\)
TS EAMCET-18.07.2022
Parabola
120974
If all the vertices of an equilateral triangle lie on the parabola \(y^2=16 x\) and one of them coincides with the vertex of that parabola, then the length of the side of that triangle is
1 \(32 \sqrt{3}\)
2 \(16 \sqrt{3}\)
3 \(8 \sqrt{3}\)
4 32
Explanation:
A Given equation of parabola, \(y^2=16 x\) and all the vertices are of an equilateral triangle with one of them coincides with the vertex of parabola
Since, given parabola is symmetrical about \(\mathrm{X}\)-axis So, \(\angle \mathrm{AOM}=30^{\circ}\)
In \(\angle \mathrm{AOM}, \tan 30^{\circ}=\frac{\mathrm{AM}}{\mathrm{OM}}=\frac{8 \mathrm{t}}{4 \mathrm{t}^2} \Rightarrow \frac{1}{\sqrt{3}}=\frac{2}{\mathrm{t}} \Rightarrow \mathrm{t}=2 \sqrt{3}\)
\(\therefore\) Distance \(\mathrm{OA}=\sqrt{\left(\mathrm{x}_{\mathrm{A}}-\mathrm{x}_{\mathrm{o}}\right)^2+\left(\mathrm{y}_{\mathrm{A}}-\mathrm{y}_{\mathrm{O}}\right)^2}\)
\(=\sqrt{(48-0)^2+(16 \sqrt{3}-0)^2}\)
\(\mathrm{OA} =32 \sqrt{3}\)\(\therefore\) Length of the side of equilateral triangle \(=32 \sqrt{3}\)
TS EAMCET-11.09.2020
Parabola
120975
If a circle with its centre at the focus of the parabola \(\mathrm{y}^2=2 \mathrm{px}\) is such that it touches the directrix of the parabola, then a point of intersection of the circle and the parabola is
C Equation of given parabola is \(\mathrm{y}^2=2 \mathrm{px}\) having focus \(\mathrm{F}\left(\frac{\mathrm{P}}{2}, 0\right)\) and equation of directrix is \(x+\frac{P}{2}=0\), so equation of circle having centre \(\left(\frac{\mathrm{p}}{2}, 0\right)\) and radius \(\mathrm{r}=\mathrm{p}\), because the circle touches the directrix of the given parabola, is
\((\mathrm{x}-\mathrm{p} / 2)^2+\mathrm{y}^2=\mathrm{p}^2\)
Now, on solving the equation of the given parabola and circle (i), we get
\(\left(\mathrm{x}^2-\mathrm{px}+\frac{\mathrm{p}^2}{4}\right)+2 \mathrm{px}=\mathrm{p}^2 \Rightarrow\left(\mathrm{x}^2+\mathrm{px}+\frac{\mathrm{p}^2}{4}\right)=\mathrm{p}^2\)
\(\left(\mathrm{x}+\frac{\mathrm{p}}{2}\right)^2=\mathrm{p}^2 \Rightarrow \mathrm{x}+\frac{\mathrm{p}}{2}= \pm \mathrm{p}\)
\(\mathrm{x}=\frac{\mathrm{p}}{2} \quad[\because \mathrm{xp}>0]\)
\(y= \pm p\)
\(\left(\frac{\mathrm{p}}{2}, \mathrm{p}\right),\left(\frac{\mathrm{p}}{2},-\mathrm{p}\right)\)
and
\(\therefore\) Point of intersection of circle and parabola are
\(\left(\frac{\mathrm{p}}{2}, \mathrm{p}\right),\left(\frac{\mathrm{p}}{2},-\mathrm{p}\right)\)
TS EAMCET-11.09.2020
Parabola
120976
A point on the parabola whose focus and vertex are respectively at \(\left(\frac{5}{4},-2\right)\) and \((1,-2)\) is
1 \((4,0)\)
2 \((15,2)\)
3 \((3,-1)\)
4 \((10,1)\)
Explanation:
D :
Given,
\(\text { Focus }=\left(\frac{5}{4},-2\right)\)
\(\text { Vertex }=(1,-2)\)
\(\mathrm{a}=\sqrt{\left(1-\frac{5}{4}\right)^2+(-2+2)^2}\)
\(=\sqrt{\left(\frac{-1}{4}\right)^2}\)
\(\mathrm{a}=\frac{1}{4}\)
Equation of parabola,
\(\quad(y+2)^2 =4 \times \frac{1}{4}(x-1)\)
\(\Rightarrow \quad(y+2)^2 =(x-1)\)
\(\therefore \quad(10,1)\) satisfies equation (i)
Therefore \((10,1)\) is a point on the parabola.
D Given, equation is
\(2 x^2+5 y-6 x+1=0\)
\(2 x^2-6 x+\frac{9}{2}+5 y+1-\frac{9}{2}=0\)
\(2\left(x-\frac{3}{2}\right)^2=-5\left(y-\frac{7}{10}\right)\)
\(\left(x-\frac{3}{2}\right)^2=\frac{-5}{2}\left(y-\frac{7}{10}\right)\)
\(X^2=\frac{-5}{2} Y\)
Where \(\mathrm{X}=\mathrm{x}-\frac{3}{2}\) and \(\mathrm{Y}=\mathrm{y}-\frac{7}{10}\)
For vertex, on putting \(\mathrm{X}=0, \mathrm{Y}=0\)
\(\mathrm{x}=\frac{3}{2} \text { and } \mathrm{y}=\frac{7}{10}\)
\(\therefore \quad \text { Vertex }=\left(\frac{3}{2}, \frac{7}{10}\right)\)
\(\mathrm{x}=\frac{3}{2} \text { and } \mathrm{y}-\frac{7}{10}=\frac{-5}{8}\)
\(\mathrm{y}=\frac{3}{40}\)
\(\therefore \quad \text { Focus }=\left(\frac{3}{2}, \frac{3}{40}\right)\)
For focus, \(\mathrm{X}=0\) and \(\mathrm{Y}=-\mathrm{a}\)
TS EAMCET-18.07.2022
Parabola
120974
If all the vertices of an equilateral triangle lie on the parabola \(y^2=16 x\) and one of them coincides with the vertex of that parabola, then the length of the side of that triangle is
1 \(32 \sqrt{3}\)
2 \(16 \sqrt{3}\)
3 \(8 \sqrt{3}\)
4 32
Explanation:
A Given equation of parabola, \(y^2=16 x\) and all the vertices are of an equilateral triangle with one of them coincides with the vertex of parabola
Since, given parabola is symmetrical about \(\mathrm{X}\)-axis So, \(\angle \mathrm{AOM}=30^{\circ}\)
In \(\angle \mathrm{AOM}, \tan 30^{\circ}=\frac{\mathrm{AM}}{\mathrm{OM}}=\frac{8 \mathrm{t}}{4 \mathrm{t}^2} \Rightarrow \frac{1}{\sqrt{3}}=\frac{2}{\mathrm{t}} \Rightarrow \mathrm{t}=2 \sqrt{3}\)
\(\therefore\) Distance \(\mathrm{OA}=\sqrt{\left(\mathrm{x}_{\mathrm{A}}-\mathrm{x}_{\mathrm{o}}\right)^2+\left(\mathrm{y}_{\mathrm{A}}-\mathrm{y}_{\mathrm{O}}\right)^2}\)
\(=\sqrt{(48-0)^2+(16 \sqrt{3}-0)^2}\)
\(\mathrm{OA} =32 \sqrt{3}\)\(\therefore\) Length of the side of equilateral triangle \(=32 \sqrt{3}\)
TS EAMCET-11.09.2020
Parabola
120975
If a circle with its centre at the focus of the parabola \(\mathrm{y}^2=2 \mathrm{px}\) is such that it touches the directrix of the parabola, then a point of intersection of the circle and the parabola is
C Equation of given parabola is \(\mathrm{y}^2=2 \mathrm{px}\) having focus \(\mathrm{F}\left(\frac{\mathrm{P}}{2}, 0\right)\) and equation of directrix is \(x+\frac{P}{2}=0\), so equation of circle having centre \(\left(\frac{\mathrm{p}}{2}, 0\right)\) and radius \(\mathrm{r}=\mathrm{p}\), because the circle touches the directrix of the given parabola, is
\((\mathrm{x}-\mathrm{p} / 2)^2+\mathrm{y}^2=\mathrm{p}^2\)
Now, on solving the equation of the given parabola and circle (i), we get
\(\left(\mathrm{x}^2-\mathrm{px}+\frac{\mathrm{p}^2}{4}\right)+2 \mathrm{px}=\mathrm{p}^2 \Rightarrow\left(\mathrm{x}^2+\mathrm{px}+\frac{\mathrm{p}^2}{4}\right)=\mathrm{p}^2\)
\(\left(\mathrm{x}+\frac{\mathrm{p}}{2}\right)^2=\mathrm{p}^2 \Rightarrow \mathrm{x}+\frac{\mathrm{p}}{2}= \pm \mathrm{p}\)
\(\mathrm{x}=\frac{\mathrm{p}}{2} \quad[\because \mathrm{xp}>0]\)
\(y= \pm p\)
\(\left(\frac{\mathrm{p}}{2}, \mathrm{p}\right),\left(\frac{\mathrm{p}}{2},-\mathrm{p}\right)\)
and
\(\therefore\) Point of intersection of circle and parabola are
\(\left(\frac{\mathrm{p}}{2}, \mathrm{p}\right),\left(\frac{\mathrm{p}}{2},-\mathrm{p}\right)\)
TS EAMCET-11.09.2020
Parabola
120976
A point on the parabola whose focus and vertex are respectively at \(\left(\frac{5}{4},-2\right)\) and \((1,-2)\) is
1 \((4,0)\)
2 \((15,2)\)
3 \((3,-1)\)
4 \((10,1)\)
Explanation:
D :
Given,
\(\text { Focus }=\left(\frac{5}{4},-2\right)\)
\(\text { Vertex }=(1,-2)\)
\(\mathrm{a}=\sqrt{\left(1-\frac{5}{4}\right)^2+(-2+2)^2}\)
\(=\sqrt{\left(\frac{-1}{4}\right)^2}\)
\(\mathrm{a}=\frac{1}{4}\)
Equation of parabola,
\(\quad(y+2)^2 =4 \times \frac{1}{4}(x-1)\)
\(\Rightarrow \quad(y+2)^2 =(x-1)\)
\(\therefore \quad(10,1)\) satisfies equation (i)
Therefore \((10,1)\) is a point on the parabola.
D Given, equation is
\(2 x^2+5 y-6 x+1=0\)
\(2 x^2-6 x+\frac{9}{2}+5 y+1-\frac{9}{2}=0\)
\(2\left(x-\frac{3}{2}\right)^2=-5\left(y-\frac{7}{10}\right)\)
\(\left(x-\frac{3}{2}\right)^2=\frac{-5}{2}\left(y-\frac{7}{10}\right)\)
\(X^2=\frac{-5}{2} Y\)
Where \(\mathrm{X}=\mathrm{x}-\frac{3}{2}\) and \(\mathrm{Y}=\mathrm{y}-\frac{7}{10}\)
For vertex, on putting \(\mathrm{X}=0, \mathrm{Y}=0\)
\(\mathrm{x}=\frac{3}{2} \text { and } \mathrm{y}=\frac{7}{10}\)
\(\therefore \quad \text { Vertex }=\left(\frac{3}{2}, \frac{7}{10}\right)\)
\(\mathrm{x}=\frac{3}{2} \text { and } \mathrm{y}-\frac{7}{10}=\frac{-5}{8}\)
\(\mathrm{y}=\frac{3}{40}\)
\(\therefore \quad \text { Focus }=\left(\frac{3}{2}, \frac{3}{40}\right)\)
For focus, \(\mathrm{X}=0\) and \(\mathrm{Y}=-\mathrm{a}\)
TS EAMCET-18.07.2022
Parabola
120974
If all the vertices of an equilateral triangle lie on the parabola \(y^2=16 x\) and one of them coincides with the vertex of that parabola, then the length of the side of that triangle is
1 \(32 \sqrt{3}\)
2 \(16 \sqrt{3}\)
3 \(8 \sqrt{3}\)
4 32
Explanation:
A Given equation of parabola, \(y^2=16 x\) and all the vertices are of an equilateral triangle with one of them coincides with the vertex of parabola
Since, given parabola is symmetrical about \(\mathrm{X}\)-axis So, \(\angle \mathrm{AOM}=30^{\circ}\)
In \(\angle \mathrm{AOM}, \tan 30^{\circ}=\frac{\mathrm{AM}}{\mathrm{OM}}=\frac{8 \mathrm{t}}{4 \mathrm{t}^2} \Rightarrow \frac{1}{\sqrt{3}}=\frac{2}{\mathrm{t}} \Rightarrow \mathrm{t}=2 \sqrt{3}\)
\(\therefore\) Distance \(\mathrm{OA}=\sqrt{\left(\mathrm{x}_{\mathrm{A}}-\mathrm{x}_{\mathrm{o}}\right)^2+\left(\mathrm{y}_{\mathrm{A}}-\mathrm{y}_{\mathrm{O}}\right)^2}\)
\(=\sqrt{(48-0)^2+(16 \sqrt{3}-0)^2}\)
\(\mathrm{OA} =32 \sqrt{3}\)\(\therefore\) Length of the side of equilateral triangle \(=32 \sqrt{3}\)
TS EAMCET-11.09.2020
Parabola
120975
If a circle with its centre at the focus of the parabola \(\mathrm{y}^2=2 \mathrm{px}\) is such that it touches the directrix of the parabola, then a point of intersection of the circle and the parabola is
C Equation of given parabola is \(\mathrm{y}^2=2 \mathrm{px}\) having focus \(\mathrm{F}\left(\frac{\mathrm{P}}{2}, 0\right)\) and equation of directrix is \(x+\frac{P}{2}=0\), so equation of circle having centre \(\left(\frac{\mathrm{p}}{2}, 0\right)\) and radius \(\mathrm{r}=\mathrm{p}\), because the circle touches the directrix of the given parabola, is
\((\mathrm{x}-\mathrm{p} / 2)^2+\mathrm{y}^2=\mathrm{p}^2\)
Now, on solving the equation of the given parabola and circle (i), we get
\(\left(\mathrm{x}^2-\mathrm{px}+\frac{\mathrm{p}^2}{4}\right)+2 \mathrm{px}=\mathrm{p}^2 \Rightarrow\left(\mathrm{x}^2+\mathrm{px}+\frac{\mathrm{p}^2}{4}\right)=\mathrm{p}^2\)
\(\left(\mathrm{x}+\frac{\mathrm{p}}{2}\right)^2=\mathrm{p}^2 \Rightarrow \mathrm{x}+\frac{\mathrm{p}}{2}= \pm \mathrm{p}\)
\(\mathrm{x}=\frac{\mathrm{p}}{2} \quad[\because \mathrm{xp}>0]\)
\(y= \pm p\)
\(\left(\frac{\mathrm{p}}{2}, \mathrm{p}\right),\left(\frac{\mathrm{p}}{2},-\mathrm{p}\right)\)
and
\(\therefore\) Point of intersection of circle and parabola are
\(\left(\frac{\mathrm{p}}{2}, \mathrm{p}\right),\left(\frac{\mathrm{p}}{2},-\mathrm{p}\right)\)
TS EAMCET-11.09.2020
Parabola
120976
A point on the parabola whose focus and vertex are respectively at \(\left(\frac{5}{4},-2\right)\) and \((1,-2)\) is
1 \((4,0)\)
2 \((15,2)\)
3 \((3,-1)\)
4 \((10,1)\)
Explanation:
D :
Given,
\(\text { Focus }=\left(\frac{5}{4},-2\right)\)
\(\text { Vertex }=(1,-2)\)
\(\mathrm{a}=\sqrt{\left(1-\frac{5}{4}\right)^2+(-2+2)^2}\)
\(=\sqrt{\left(\frac{-1}{4}\right)^2}\)
\(\mathrm{a}=\frac{1}{4}\)
Equation of parabola,
\(\quad(y+2)^2 =4 \times \frac{1}{4}(x-1)\)
\(\Rightarrow \quad(y+2)^2 =(x-1)\)
\(\therefore \quad(10,1)\) satisfies equation (i)
Therefore \((10,1)\) is a point on the parabola.
