Standard Equation of Parabola (parametric form)
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Parabola

120094 An equilateral triangle is inscribed in the parabola \(y^2=8 x\), with one of its vertices is the vertex of the parabola. Then length of the side of the triangle is

1 \(24 \sqrt{3}\) units
2 \(16 \sqrt{3}\) units
3 \(8 \sqrt{3}\) units
4 \(4 \sqrt{3}\) units
AP EAMCET-2015
Parabola

120095 The line which is parallel to \(x\)-axis and crosses the curve \(y=\sqrt{x}\) at an angle of \(45^{\circ}\) is \(\qquad\)

1 \(\mathrm{y}=\frac{1}{4}\)
2 \(\mathrm{y}=\frac{1}{2}\)
3 \(\mathrm{y}=1\)
4 \(y=4\)
AP EAMCET-19.08.2021
Parabola

120096 For the parabola \(y^2+2 x+2 y-3=0\), match the items in List-I with those from List-II.
| List-I | List-II |
| :--- | :--- |
| A. Vertex | I. $2 x-5=0$ |
| B. Focus | II. $\left(\frac{3}{2},-1\right)$ |
| C. Equation of the \lt br> Directrix | III. $x-2=0$ |
| D. Equation of the Axis | IV. $y+1=0$ |
| | V. $(2,-1)$ |
| | VI. $\left(2, \frac{3}{2}\right)$ |
The correct match is

1 A-V, B-VI, C-I, D-III
2 A-V, B-II, C-I, D-IV
3 A-VI, B-V, C-VI, D-I
4 A-II, B-VI, C-III, D-IV
AP EAMCET-22.04.2019
Parabola

120099 Let \(P\) be the mid- point of a chord joining the vertex of the parabola \(y^2=8 x\) to another point on it. Then, the locus of \(P\) is

1 \(y^2=2 x\)
2 \(y^2=4 x\)
3 \(\frac{x^2}{4}+y^2=1\)
4 \(x^2+\frac{y^2}{4}=1\)
WB JEE-2012
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NEET Test Series from KOTA - 10 Papers In MS WORD WhatsApp Here
Parabola

120094 An equilateral triangle is inscribed in the parabola \(y^2=8 x\), with one of its vertices is the vertex of the parabola. Then length of the side of the triangle is

1 \(24 \sqrt{3}\) units
2 \(16 \sqrt{3}\) units
3 \(8 \sqrt{3}\) units
4 \(4 \sqrt{3}\) units
AP EAMCET-2015
Parabola

120095 The line which is parallel to \(x\)-axis and crosses the curve \(y=\sqrt{x}\) at an angle of \(45^{\circ}\) is \(\qquad\)

1 \(\mathrm{y}=\frac{1}{4}\)
2 \(\mathrm{y}=\frac{1}{2}\)
3 \(\mathrm{y}=1\)
4 \(y=4\)
AP EAMCET-19.08.2021
Parabola

120096 For the parabola \(y^2+2 x+2 y-3=0\), match the items in List-I with those from List-II.
| List-I | List-II |
| :--- | :--- |
| A. Vertex | I. $2 x-5=0$ |
| B. Focus | II. $\left(\frac{3}{2},-1\right)$ |
| C. Equation of the \lt br> Directrix | III. $x-2=0$ |
| D. Equation of the Axis | IV. $y+1=0$ |
| | V. $(2,-1)$ |
| | VI. $\left(2, \frac{3}{2}\right)$ |
The correct match is

1 A-V, B-VI, C-I, D-III
2 A-V, B-II, C-I, D-IV
3 A-VI, B-V, C-VI, D-I
4 A-II, B-VI, C-III, D-IV
AP EAMCET-22.04.2019
Parabola

120099 Let \(P\) be the mid- point of a chord joining the vertex of the parabola \(y^2=8 x\) to another point on it. Then, the locus of \(P\) is

1 \(y^2=2 x\)
2 \(y^2=4 x\)
3 \(\frac{x^2}{4}+y^2=1\)
4 \(x^2+\frac{y^2}{4}=1\)
WB JEE-2012
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Parabola

120094 An equilateral triangle is inscribed in the parabola \(y^2=8 x\), with one of its vertices is the vertex of the parabola. Then length of the side of the triangle is

1 \(24 \sqrt{3}\) units
2 \(16 \sqrt{3}\) units
3 \(8 \sqrt{3}\) units
4 \(4 \sqrt{3}\) units
AP EAMCET-2015
Parabola

120095 The line which is parallel to \(x\)-axis and crosses the curve \(y=\sqrt{x}\) at an angle of \(45^{\circ}\) is \(\qquad\)

1 \(\mathrm{y}=\frac{1}{4}\)
2 \(\mathrm{y}=\frac{1}{2}\)
3 \(\mathrm{y}=1\)
4 \(y=4\)
AP EAMCET-19.08.2021
Parabola

120096 For the parabola \(y^2+2 x+2 y-3=0\), match the items in List-I with those from List-II.
| List-I | List-II |
| :--- | :--- |
| A. Vertex | I. $2 x-5=0$ |
| B. Focus | II. $\left(\frac{3}{2},-1\right)$ |
| C. Equation of the \lt br> Directrix | III. $x-2=0$ |
| D. Equation of the Axis | IV. $y+1=0$ |
| | V. $(2,-1)$ |
| | VI. $\left(2, \frac{3}{2}\right)$ |
The correct match is

1 A-V, B-VI, C-I, D-III
2 A-V, B-II, C-I, D-IV
3 A-VI, B-V, C-VI, D-I
4 A-II, B-VI, C-III, D-IV
AP EAMCET-22.04.2019
Parabola

120099 Let \(P\) be the mid- point of a chord joining the vertex of the parabola \(y^2=8 x\) to another point on it. Then, the locus of \(P\) is

1 \(y^2=2 x\)
2 \(y^2=4 x\)
3 \(\frac{x^2}{4}+y^2=1\)
4 \(x^2+\frac{y^2}{4}=1\)
WB JEE-2012
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Parabola

120094 An equilateral triangle is inscribed in the parabola \(y^2=8 x\), with one of its vertices is the vertex of the parabola. Then length of the side of the triangle is

1 \(24 \sqrt{3}\) units
2 \(16 \sqrt{3}\) units
3 \(8 \sqrt{3}\) units
4 \(4 \sqrt{3}\) units
AP EAMCET-2015
Parabola

120095 The line which is parallel to \(x\)-axis and crosses the curve \(y=\sqrt{x}\) at an angle of \(45^{\circ}\) is \(\qquad\)

1 \(\mathrm{y}=\frac{1}{4}\)
2 \(\mathrm{y}=\frac{1}{2}\)
3 \(\mathrm{y}=1\)
4 \(y=4\)
AP EAMCET-19.08.2021
Parabola

120096 For the parabola \(y^2+2 x+2 y-3=0\), match the items in List-I with those from List-II.
| List-I | List-II |
| :--- | :--- |
| A. Vertex | I. $2 x-5=0$ |
| B. Focus | II. $\left(\frac{3}{2},-1\right)$ |
| C. Equation of the \lt br> Directrix | III. $x-2=0$ |
| D. Equation of the Axis | IV. $y+1=0$ |
| | V. $(2,-1)$ |
| | VI. $\left(2, \frac{3}{2}\right)$ |
The correct match is

1 A-V, B-VI, C-I, D-III
2 A-V, B-II, C-I, D-IV
3 A-VI, B-V, C-VI, D-I
4 A-II, B-VI, C-III, D-IV
AP EAMCET-22.04.2019
Parabola

120099 Let \(P\) be the mid- point of a chord joining the vertex of the parabola \(y^2=8 x\) to another point on it. Then, the locus of \(P\) is

1 \(y^2=2 x\)
2 \(y^2=4 x\)
3 \(\frac{x^2}{4}+y^2=1\)
4 \(x^2+\frac{y^2}{4}=1\)
WB JEE-2012
For More Updates join with Us