120944
The focus of the parabola \(y^2-x-2 y+2=0\) is
1 \((1 / 4,0)\)
2 \((1,2)\)
3 \((5 / 4,1)\)
4 \((3 / 4,5 / 2)\)
Explanation:
C Given, parabola is
\(\mathrm{y}^2-\mathrm{x}-2 \mathrm{y}+2=0\)
\(\mathrm{y}^2-2 \mathrm{y}+1-1-\mathrm{x}+2=0\)
\((\mathrm{y}-1)^2=\mathrm{x}-1\)
\(\mathrm{Y}^2=\mathrm{X}\)
Where, \(\mathrm{Y}=\mathrm{y}-1, \mathrm{X}=\mathrm{x}-1, \mathrm{a}=\frac{1}{4}\)
\(\therefore\) Focus is \((a, 0)\)
\(\mathrm{X}=\mathrm{a}\)
\(x-1=\frac{1}{4}\)
\(\mathrm{x}=\frac{1}{4}+1=\frac{5}{4}\)
\(\mathrm{Y}=0\)
\(\mathrm{y}-1=0\)
\(\mathrm{y}=1\)\(\therefore\) Required focus is \(\left(\frac{5}{4}, 1\right)\)
BCECE-2010
Parabola
120945
If parabola is passing through \((2,3)\), vertex \((0\), \(0)\) and axis is along \(X\)-axis, then the equation of parabola is
1 \(\mathrm{y}^2=\frac{9}{2} \mathrm{x}\)
2 \(\mathrm{x}^2=\frac{9}{2} \mathrm{y}\)
3 \(\mathrm{y}^2=\frac{3}{2} \mathrm{x}\)
4 \(x^2=\frac{3}{2} y\)
Explanation:
A Given,
Vertex \((0,0)\)
Point \(=(2,3)\)
We have to find out the equation of the parabola with vertex at the origin
So, the equation of the parabola can be \(y^2=4 \mathrm{ax}\) or \(\mathrm{y}^2=-4 \mathrm{ax}\)
\(\because\) Point given \((2,3)\) which lies in the \(1^{\text {st }}\) quadrant So, the required formed of parabola
\(y^2=4 a x\)
\(x=2, y=3 \text { putting in parabola }\)
\((3)^2=4 \times a \times 2\)
\(9=8 a\)
\(a=9 / 8\)
\(\therefore\) Equation of parabola is -
\(y^2=4 a x\)
\(y^2=4 \times \frac{9}{8} \times x\)
\(y^2=\frac{9}{2} x\)
BCECE-2018
Parabola
120946
If the parabolas \(y^2=4 x\) and \(x^2=32 y\) intersect at \((16,8)\) at an angle \(\theta\), then \(\theta=\)
1 \(\tan ^{-1} 5 / 3\)
2 \(\tan ^{-1} 4 / 5\)
3 \(\tan ^{-1} 3 / 5\)
4 \(\pi / 2\)
Explanation:
C Given, the parabola \(y^2=4 x\) and \(x^2=32 y\) intersect at \((16,8)\) at an angle \(\theta\)
Slope of the tangent \(y^2=4 x\)
\(\mathrm{m}_1=\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{4}{2 \mathrm{y}}\)
\(\mathrm{m}_1=\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{4}{2 \times 8}=\frac{1}{4}\)
Again slope of tangent \(x^2=32 y\)
\(2 \mathrm{x}=\frac{32 \mathrm{dy}}{\mathrm{dx}}\)
\(\mathrm{m}_2=\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{2 \mathrm{x}}{32}=\frac{2 \times 16}{32}=1\)
We know that, \(\tan \theta=\frac{\mathrm{m}_2-\mathrm{m}_1}{1+\mathrm{m}_1 \mathrm{~m}_2}\)
\(\Rightarrow \frac{1-\frac{1}{4}}{1+\frac{1}{4} \times 1}=\frac{3}{5}\)
\(\theta=\tan ^{-1}\left(\frac{3}{5}\right)\)
AMU-2011
Parabola
120947
If \(a x^2+4 x y+y^2+a x+3 y+2=0\) represents a parabola, then a is
1 -4
2 4
3 0
4 6
Explanation:
B Given,
\(a x^2+4 x y+y^2+a x+3 y+2=0\)
General form of the parabola -
\(a x^2+2 h x y+b y^2+2 g x+2 f y+c=0\)
Equation (ii) represents the parabola if \(\mathrm{h}^2=\mathrm{ab}\) Compare (i) and (ii), we get \(\mathrm{h}=2, \mathrm{a}=\mathrm{ab}=1\),
Then, \(\mathrm{h}^2=\mathrm{ab}\)
\((2)^2=a \times 1\)
\(a=4\)
AMU-2010
Parabola
120948
If the vertex is \((3,0)\) and the extremities of the latusrectum are \((4,3)\) and \((4,-3)\), then the equation of the parabola is
1 \(y^2=4(x-3)\)
2 \(x^2=4(y-3)\)
3 \(y^2=-4(x+3)\)
4 \(x^2=-4(y+3)\)
Explanation:
A :
Vertex \(=(3,0)\)
Focus will be the midpoint of extremities of latus rectum
\(\therefore\) focus \(=(4,0)\)
And, distance between focus and vertex is \(\mathrm{a}=1\)
Since, axis of the parabola is \(\mathrm{x}\)-axis.
\(\therefore\) Its equation will be \(\left(\mathrm{y}-\mathrm{y}_1\right)^2=4 . \mathrm{a}\left(\mathrm{x}-\mathrm{x}_1\right)\)
Here, \(\left(\mathrm{x}_1, \mathrm{y}_1\right)=(3,0)\)
\(\mathrm{a}=1\)
So, \((\mathrm{y}-0)^2=4 \times 1 \times(\mathrm{x}-3)\)
\(\mathrm{y}^2=4(\mathrm{x}-3)\)
120944
The focus of the parabola \(y^2-x-2 y+2=0\) is
1 \((1 / 4,0)\)
2 \((1,2)\)
3 \((5 / 4,1)\)
4 \((3 / 4,5 / 2)\)
Explanation:
C Given, parabola is
\(\mathrm{y}^2-\mathrm{x}-2 \mathrm{y}+2=0\)
\(\mathrm{y}^2-2 \mathrm{y}+1-1-\mathrm{x}+2=0\)
\((\mathrm{y}-1)^2=\mathrm{x}-1\)
\(\mathrm{Y}^2=\mathrm{X}\)
Where, \(\mathrm{Y}=\mathrm{y}-1, \mathrm{X}=\mathrm{x}-1, \mathrm{a}=\frac{1}{4}\)
\(\therefore\) Focus is \((a, 0)\)
\(\mathrm{X}=\mathrm{a}\)
\(x-1=\frac{1}{4}\)
\(\mathrm{x}=\frac{1}{4}+1=\frac{5}{4}\)
\(\mathrm{Y}=0\)
\(\mathrm{y}-1=0\)
\(\mathrm{y}=1\)\(\therefore\) Required focus is \(\left(\frac{5}{4}, 1\right)\)
BCECE-2010
Parabola
120945
If parabola is passing through \((2,3)\), vertex \((0\), \(0)\) and axis is along \(X\)-axis, then the equation of parabola is
1 \(\mathrm{y}^2=\frac{9}{2} \mathrm{x}\)
2 \(\mathrm{x}^2=\frac{9}{2} \mathrm{y}\)
3 \(\mathrm{y}^2=\frac{3}{2} \mathrm{x}\)
4 \(x^2=\frac{3}{2} y\)
Explanation:
A Given,
Vertex \((0,0)\)
Point \(=(2,3)\)
We have to find out the equation of the parabola with vertex at the origin
So, the equation of the parabola can be \(y^2=4 \mathrm{ax}\) or \(\mathrm{y}^2=-4 \mathrm{ax}\)
\(\because\) Point given \((2,3)\) which lies in the \(1^{\text {st }}\) quadrant So, the required formed of parabola
\(y^2=4 a x\)
\(x=2, y=3 \text { putting in parabola }\)
\((3)^2=4 \times a \times 2\)
\(9=8 a\)
\(a=9 / 8\)
\(\therefore\) Equation of parabola is -
\(y^2=4 a x\)
\(y^2=4 \times \frac{9}{8} \times x\)
\(y^2=\frac{9}{2} x\)
BCECE-2018
Parabola
120946
If the parabolas \(y^2=4 x\) and \(x^2=32 y\) intersect at \((16,8)\) at an angle \(\theta\), then \(\theta=\)
1 \(\tan ^{-1} 5 / 3\)
2 \(\tan ^{-1} 4 / 5\)
3 \(\tan ^{-1} 3 / 5\)
4 \(\pi / 2\)
Explanation:
C Given, the parabola \(y^2=4 x\) and \(x^2=32 y\) intersect at \((16,8)\) at an angle \(\theta\)
Slope of the tangent \(y^2=4 x\)
\(\mathrm{m}_1=\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{4}{2 \mathrm{y}}\)
\(\mathrm{m}_1=\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{4}{2 \times 8}=\frac{1}{4}\)
Again slope of tangent \(x^2=32 y\)
\(2 \mathrm{x}=\frac{32 \mathrm{dy}}{\mathrm{dx}}\)
\(\mathrm{m}_2=\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{2 \mathrm{x}}{32}=\frac{2 \times 16}{32}=1\)
We know that, \(\tan \theta=\frac{\mathrm{m}_2-\mathrm{m}_1}{1+\mathrm{m}_1 \mathrm{~m}_2}\)
\(\Rightarrow \frac{1-\frac{1}{4}}{1+\frac{1}{4} \times 1}=\frac{3}{5}\)
\(\theta=\tan ^{-1}\left(\frac{3}{5}\right)\)
AMU-2011
Parabola
120947
If \(a x^2+4 x y+y^2+a x+3 y+2=0\) represents a parabola, then a is
1 -4
2 4
3 0
4 6
Explanation:
B Given,
\(a x^2+4 x y+y^2+a x+3 y+2=0\)
General form of the parabola -
\(a x^2+2 h x y+b y^2+2 g x+2 f y+c=0\)
Equation (ii) represents the parabola if \(\mathrm{h}^2=\mathrm{ab}\) Compare (i) and (ii), we get \(\mathrm{h}=2, \mathrm{a}=\mathrm{ab}=1\),
Then, \(\mathrm{h}^2=\mathrm{ab}\)
\((2)^2=a \times 1\)
\(a=4\)
AMU-2010
Parabola
120948
If the vertex is \((3,0)\) and the extremities of the latusrectum are \((4,3)\) and \((4,-3)\), then the equation of the parabola is
1 \(y^2=4(x-3)\)
2 \(x^2=4(y-3)\)
3 \(y^2=-4(x+3)\)
4 \(x^2=-4(y+3)\)
Explanation:
A :
Vertex \(=(3,0)\)
Focus will be the midpoint of extremities of latus rectum
\(\therefore\) focus \(=(4,0)\)
And, distance between focus and vertex is \(\mathrm{a}=1\)
Since, axis of the parabola is \(\mathrm{x}\)-axis.
\(\therefore\) Its equation will be \(\left(\mathrm{y}-\mathrm{y}_1\right)^2=4 . \mathrm{a}\left(\mathrm{x}-\mathrm{x}_1\right)\)
Here, \(\left(\mathrm{x}_1, \mathrm{y}_1\right)=(3,0)\)
\(\mathrm{a}=1\)
So, \((\mathrm{y}-0)^2=4 \times 1 \times(\mathrm{x}-3)\)
\(\mathrm{y}^2=4(\mathrm{x}-3)\)
120944
The focus of the parabola \(y^2-x-2 y+2=0\) is
1 \((1 / 4,0)\)
2 \((1,2)\)
3 \((5 / 4,1)\)
4 \((3 / 4,5 / 2)\)
Explanation:
C Given, parabola is
\(\mathrm{y}^2-\mathrm{x}-2 \mathrm{y}+2=0\)
\(\mathrm{y}^2-2 \mathrm{y}+1-1-\mathrm{x}+2=0\)
\((\mathrm{y}-1)^2=\mathrm{x}-1\)
\(\mathrm{Y}^2=\mathrm{X}\)
Where, \(\mathrm{Y}=\mathrm{y}-1, \mathrm{X}=\mathrm{x}-1, \mathrm{a}=\frac{1}{4}\)
\(\therefore\) Focus is \((a, 0)\)
\(\mathrm{X}=\mathrm{a}\)
\(x-1=\frac{1}{4}\)
\(\mathrm{x}=\frac{1}{4}+1=\frac{5}{4}\)
\(\mathrm{Y}=0\)
\(\mathrm{y}-1=0\)
\(\mathrm{y}=1\)\(\therefore\) Required focus is \(\left(\frac{5}{4}, 1\right)\)
BCECE-2010
Parabola
120945
If parabola is passing through \((2,3)\), vertex \((0\), \(0)\) and axis is along \(X\)-axis, then the equation of parabola is
1 \(\mathrm{y}^2=\frac{9}{2} \mathrm{x}\)
2 \(\mathrm{x}^2=\frac{9}{2} \mathrm{y}\)
3 \(\mathrm{y}^2=\frac{3}{2} \mathrm{x}\)
4 \(x^2=\frac{3}{2} y\)
Explanation:
A Given,
Vertex \((0,0)\)
Point \(=(2,3)\)
We have to find out the equation of the parabola with vertex at the origin
So, the equation of the parabola can be \(y^2=4 \mathrm{ax}\) or \(\mathrm{y}^2=-4 \mathrm{ax}\)
\(\because\) Point given \((2,3)\) which lies in the \(1^{\text {st }}\) quadrant So, the required formed of parabola
\(y^2=4 a x\)
\(x=2, y=3 \text { putting in parabola }\)
\((3)^2=4 \times a \times 2\)
\(9=8 a\)
\(a=9 / 8\)
\(\therefore\) Equation of parabola is -
\(y^2=4 a x\)
\(y^2=4 \times \frac{9}{8} \times x\)
\(y^2=\frac{9}{2} x\)
BCECE-2018
Parabola
120946
If the parabolas \(y^2=4 x\) and \(x^2=32 y\) intersect at \((16,8)\) at an angle \(\theta\), then \(\theta=\)
1 \(\tan ^{-1} 5 / 3\)
2 \(\tan ^{-1} 4 / 5\)
3 \(\tan ^{-1} 3 / 5\)
4 \(\pi / 2\)
Explanation:
C Given, the parabola \(y^2=4 x\) and \(x^2=32 y\) intersect at \((16,8)\) at an angle \(\theta\)
Slope of the tangent \(y^2=4 x\)
\(\mathrm{m}_1=\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{4}{2 \mathrm{y}}\)
\(\mathrm{m}_1=\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{4}{2 \times 8}=\frac{1}{4}\)
Again slope of tangent \(x^2=32 y\)
\(2 \mathrm{x}=\frac{32 \mathrm{dy}}{\mathrm{dx}}\)
\(\mathrm{m}_2=\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{2 \mathrm{x}}{32}=\frac{2 \times 16}{32}=1\)
We know that, \(\tan \theta=\frac{\mathrm{m}_2-\mathrm{m}_1}{1+\mathrm{m}_1 \mathrm{~m}_2}\)
\(\Rightarrow \frac{1-\frac{1}{4}}{1+\frac{1}{4} \times 1}=\frac{3}{5}\)
\(\theta=\tan ^{-1}\left(\frac{3}{5}\right)\)
AMU-2011
Parabola
120947
If \(a x^2+4 x y+y^2+a x+3 y+2=0\) represents a parabola, then a is
1 -4
2 4
3 0
4 6
Explanation:
B Given,
\(a x^2+4 x y+y^2+a x+3 y+2=0\)
General form of the parabola -
\(a x^2+2 h x y+b y^2+2 g x+2 f y+c=0\)
Equation (ii) represents the parabola if \(\mathrm{h}^2=\mathrm{ab}\) Compare (i) and (ii), we get \(\mathrm{h}=2, \mathrm{a}=\mathrm{ab}=1\),
Then, \(\mathrm{h}^2=\mathrm{ab}\)
\((2)^2=a \times 1\)
\(a=4\)
AMU-2010
Parabola
120948
If the vertex is \((3,0)\) and the extremities of the latusrectum are \((4,3)\) and \((4,-3)\), then the equation of the parabola is
1 \(y^2=4(x-3)\)
2 \(x^2=4(y-3)\)
3 \(y^2=-4(x+3)\)
4 \(x^2=-4(y+3)\)
Explanation:
A :
Vertex \(=(3,0)\)
Focus will be the midpoint of extremities of latus rectum
\(\therefore\) focus \(=(4,0)\)
And, distance between focus and vertex is \(\mathrm{a}=1\)
Since, axis of the parabola is \(\mathrm{x}\)-axis.
\(\therefore\) Its equation will be \(\left(\mathrm{y}-\mathrm{y}_1\right)^2=4 . \mathrm{a}\left(\mathrm{x}-\mathrm{x}_1\right)\)
Here, \(\left(\mathrm{x}_1, \mathrm{y}_1\right)=(3,0)\)
\(\mathrm{a}=1\)
So, \((\mathrm{y}-0)^2=4 \times 1 \times(\mathrm{x}-3)\)
\(\mathrm{y}^2=4(\mathrm{x}-3)\)
120944
The focus of the parabola \(y^2-x-2 y+2=0\) is
1 \((1 / 4,0)\)
2 \((1,2)\)
3 \((5 / 4,1)\)
4 \((3 / 4,5 / 2)\)
Explanation:
C Given, parabola is
\(\mathrm{y}^2-\mathrm{x}-2 \mathrm{y}+2=0\)
\(\mathrm{y}^2-2 \mathrm{y}+1-1-\mathrm{x}+2=0\)
\((\mathrm{y}-1)^2=\mathrm{x}-1\)
\(\mathrm{Y}^2=\mathrm{X}\)
Where, \(\mathrm{Y}=\mathrm{y}-1, \mathrm{X}=\mathrm{x}-1, \mathrm{a}=\frac{1}{4}\)
\(\therefore\) Focus is \((a, 0)\)
\(\mathrm{X}=\mathrm{a}\)
\(x-1=\frac{1}{4}\)
\(\mathrm{x}=\frac{1}{4}+1=\frac{5}{4}\)
\(\mathrm{Y}=0\)
\(\mathrm{y}-1=0\)
\(\mathrm{y}=1\)\(\therefore\) Required focus is \(\left(\frac{5}{4}, 1\right)\)
BCECE-2010
Parabola
120945
If parabola is passing through \((2,3)\), vertex \((0\), \(0)\) and axis is along \(X\)-axis, then the equation of parabola is
1 \(\mathrm{y}^2=\frac{9}{2} \mathrm{x}\)
2 \(\mathrm{x}^2=\frac{9}{2} \mathrm{y}\)
3 \(\mathrm{y}^2=\frac{3}{2} \mathrm{x}\)
4 \(x^2=\frac{3}{2} y\)
Explanation:
A Given,
Vertex \((0,0)\)
Point \(=(2,3)\)
We have to find out the equation of the parabola with vertex at the origin
So, the equation of the parabola can be \(y^2=4 \mathrm{ax}\) or \(\mathrm{y}^2=-4 \mathrm{ax}\)
\(\because\) Point given \((2,3)\) which lies in the \(1^{\text {st }}\) quadrant So, the required formed of parabola
\(y^2=4 a x\)
\(x=2, y=3 \text { putting in parabola }\)
\((3)^2=4 \times a \times 2\)
\(9=8 a\)
\(a=9 / 8\)
\(\therefore\) Equation of parabola is -
\(y^2=4 a x\)
\(y^2=4 \times \frac{9}{8} \times x\)
\(y^2=\frac{9}{2} x\)
BCECE-2018
Parabola
120946
If the parabolas \(y^2=4 x\) and \(x^2=32 y\) intersect at \((16,8)\) at an angle \(\theta\), then \(\theta=\)
1 \(\tan ^{-1} 5 / 3\)
2 \(\tan ^{-1} 4 / 5\)
3 \(\tan ^{-1} 3 / 5\)
4 \(\pi / 2\)
Explanation:
C Given, the parabola \(y^2=4 x\) and \(x^2=32 y\) intersect at \((16,8)\) at an angle \(\theta\)
Slope of the tangent \(y^2=4 x\)
\(\mathrm{m}_1=\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{4}{2 \mathrm{y}}\)
\(\mathrm{m}_1=\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{4}{2 \times 8}=\frac{1}{4}\)
Again slope of tangent \(x^2=32 y\)
\(2 \mathrm{x}=\frac{32 \mathrm{dy}}{\mathrm{dx}}\)
\(\mathrm{m}_2=\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{2 \mathrm{x}}{32}=\frac{2 \times 16}{32}=1\)
We know that, \(\tan \theta=\frac{\mathrm{m}_2-\mathrm{m}_1}{1+\mathrm{m}_1 \mathrm{~m}_2}\)
\(\Rightarrow \frac{1-\frac{1}{4}}{1+\frac{1}{4} \times 1}=\frac{3}{5}\)
\(\theta=\tan ^{-1}\left(\frac{3}{5}\right)\)
AMU-2011
Parabola
120947
If \(a x^2+4 x y+y^2+a x+3 y+2=0\) represents a parabola, then a is
1 -4
2 4
3 0
4 6
Explanation:
B Given,
\(a x^2+4 x y+y^2+a x+3 y+2=0\)
General form of the parabola -
\(a x^2+2 h x y+b y^2+2 g x+2 f y+c=0\)
Equation (ii) represents the parabola if \(\mathrm{h}^2=\mathrm{ab}\) Compare (i) and (ii), we get \(\mathrm{h}=2, \mathrm{a}=\mathrm{ab}=1\),
Then, \(\mathrm{h}^2=\mathrm{ab}\)
\((2)^2=a \times 1\)
\(a=4\)
AMU-2010
Parabola
120948
If the vertex is \((3,0)\) and the extremities of the latusrectum are \((4,3)\) and \((4,-3)\), then the equation of the parabola is
1 \(y^2=4(x-3)\)
2 \(x^2=4(y-3)\)
3 \(y^2=-4(x+3)\)
4 \(x^2=-4(y+3)\)
Explanation:
A :
Vertex \(=(3,0)\)
Focus will be the midpoint of extremities of latus rectum
\(\therefore\) focus \(=(4,0)\)
And, distance between focus and vertex is \(\mathrm{a}=1\)
Since, axis of the parabola is \(\mathrm{x}\)-axis.
\(\therefore\) Its equation will be \(\left(\mathrm{y}-\mathrm{y}_1\right)^2=4 . \mathrm{a}\left(\mathrm{x}-\mathrm{x}_1\right)\)
Here, \(\left(\mathrm{x}_1, \mathrm{y}_1\right)=(3,0)\)
\(\mathrm{a}=1\)
So, \((\mathrm{y}-0)^2=4 \times 1 \times(\mathrm{x}-3)\)
\(\mathrm{y}^2=4(\mathrm{x}-3)\)
120944
The focus of the parabola \(y^2-x-2 y+2=0\) is
1 \((1 / 4,0)\)
2 \((1,2)\)
3 \((5 / 4,1)\)
4 \((3 / 4,5 / 2)\)
Explanation:
C Given, parabola is
\(\mathrm{y}^2-\mathrm{x}-2 \mathrm{y}+2=0\)
\(\mathrm{y}^2-2 \mathrm{y}+1-1-\mathrm{x}+2=0\)
\((\mathrm{y}-1)^2=\mathrm{x}-1\)
\(\mathrm{Y}^2=\mathrm{X}\)
Where, \(\mathrm{Y}=\mathrm{y}-1, \mathrm{X}=\mathrm{x}-1, \mathrm{a}=\frac{1}{4}\)
\(\therefore\) Focus is \((a, 0)\)
\(\mathrm{X}=\mathrm{a}\)
\(x-1=\frac{1}{4}\)
\(\mathrm{x}=\frac{1}{4}+1=\frac{5}{4}\)
\(\mathrm{Y}=0\)
\(\mathrm{y}-1=0\)
\(\mathrm{y}=1\)\(\therefore\) Required focus is \(\left(\frac{5}{4}, 1\right)\)
BCECE-2010
Parabola
120945
If parabola is passing through \((2,3)\), vertex \((0\), \(0)\) and axis is along \(X\)-axis, then the equation of parabola is
1 \(\mathrm{y}^2=\frac{9}{2} \mathrm{x}\)
2 \(\mathrm{x}^2=\frac{9}{2} \mathrm{y}\)
3 \(\mathrm{y}^2=\frac{3}{2} \mathrm{x}\)
4 \(x^2=\frac{3}{2} y\)
Explanation:
A Given,
Vertex \((0,0)\)
Point \(=(2,3)\)
We have to find out the equation of the parabola with vertex at the origin
So, the equation of the parabola can be \(y^2=4 \mathrm{ax}\) or \(\mathrm{y}^2=-4 \mathrm{ax}\)
\(\because\) Point given \((2,3)\) which lies in the \(1^{\text {st }}\) quadrant So, the required formed of parabola
\(y^2=4 a x\)
\(x=2, y=3 \text { putting in parabola }\)
\((3)^2=4 \times a \times 2\)
\(9=8 a\)
\(a=9 / 8\)
\(\therefore\) Equation of parabola is -
\(y^2=4 a x\)
\(y^2=4 \times \frac{9}{8} \times x\)
\(y^2=\frac{9}{2} x\)
BCECE-2018
Parabola
120946
If the parabolas \(y^2=4 x\) and \(x^2=32 y\) intersect at \((16,8)\) at an angle \(\theta\), then \(\theta=\)
1 \(\tan ^{-1} 5 / 3\)
2 \(\tan ^{-1} 4 / 5\)
3 \(\tan ^{-1} 3 / 5\)
4 \(\pi / 2\)
Explanation:
C Given, the parabola \(y^2=4 x\) and \(x^2=32 y\) intersect at \((16,8)\) at an angle \(\theta\)
Slope of the tangent \(y^2=4 x\)
\(\mathrm{m}_1=\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{4}{2 \mathrm{y}}\)
\(\mathrm{m}_1=\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{4}{2 \times 8}=\frac{1}{4}\)
Again slope of tangent \(x^2=32 y\)
\(2 \mathrm{x}=\frac{32 \mathrm{dy}}{\mathrm{dx}}\)
\(\mathrm{m}_2=\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{2 \mathrm{x}}{32}=\frac{2 \times 16}{32}=1\)
We know that, \(\tan \theta=\frac{\mathrm{m}_2-\mathrm{m}_1}{1+\mathrm{m}_1 \mathrm{~m}_2}\)
\(\Rightarrow \frac{1-\frac{1}{4}}{1+\frac{1}{4} \times 1}=\frac{3}{5}\)
\(\theta=\tan ^{-1}\left(\frac{3}{5}\right)\)
AMU-2011
Parabola
120947
If \(a x^2+4 x y+y^2+a x+3 y+2=0\) represents a parabola, then a is
1 -4
2 4
3 0
4 6
Explanation:
B Given,
\(a x^2+4 x y+y^2+a x+3 y+2=0\)
General form of the parabola -
\(a x^2+2 h x y+b y^2+2 g x+2 f y+c=0\)
Equation (ii) represents the parabola if \(\mathrm{h}^2=\mathrm{ab}\) Compare (i) and (ii), we get \(\mathrm{h}=2, \mathrm{a}=\mathrm{ab}=1\),
Then, \(\mathrm{h}^2=\mathrm{ab}\)
\((2)^2=a \times 1\)
\(a=4\)
AMU-2010
Parabola
120948
If the vertex is \((3,0)\) and the extremities of the latusrectum are \((4,3)\) and \((4,-3)\), then the equation of the parabola is
1 \(y^2=4(x-3)\)
2 \(x^2=4(y-3)\)
3 \(y^2=-4(x+3)\)
4 \(x^2=-4(y+3)\)
Explanation:
A :
Vertex \(=(3,0)\)
Focus will be the midpoint of extremities of latus rectum
\(\therefore\) focus \(=(4,0)\)
And, distance between focus and vertex is \(\mathrm{a}=1\)
Since, axis of the parabola is \(\mathrm{x}\)-axis.
\(\therefore\) Its equation will be \(\left(\mathrm{y}-\mathrm{y}_1\right)^2=4 . \mathrm{a}\left(\mathrm{x}-\mathrm{x}_1\right)\)
Here, \(\left(\mathrm{x}_1, \mathrm{y}_1\right)=(3,0)\)
\(\mathrm{a}=1\)
So, \((\mathrm{y}-0)^2=4 \times 1 \times(\mathrm{x}-3)\)
\(\mathrm{y}^2=4(\mathrm{x}-3)\)
