120986
Equation of the directrix of the conic \(x^2+4 y+\) \(4=0\) is
1 \(y=1\)
2 \(y=-1\)
3 \(y=0\)
4 \(x=0\)
5 \(x=1\)
Explanation:
C We have the equation of conic, \(\mathrm{x}^2+4 \mathrm{y}+4=0\)
\(x^2=-4 y-4=-4(y+1)\)
\(\mathrm{x}^2=-4 \mathrm{Y}\)
Where \(\mathrm{X}=\mathrm{x}\) and \(\mathrm{Y}=\mathrm{y}+1\)
Equation of directrix is \(\mathrm{Y}=1\)
\(\quad \mathrm{y}+1=1\)
\(\therefore \quad \mathrm{y}=0\)
Kerala CEE-2008
Parabola
120987
If the vertex of the parabola \(y=x^2-16 x+k\) lies on \(x\)-axis, then the value of \(k\) is :
1 16
2 8
3 64
4 -64
5 -8
Explanation:
C Given, Vertex of the parabola
\(y=x^2-16 x+k\)
\(=x^2-2 x(8)+k\)
\(=x^2-2 x(8)+64-64+k\)
\(=(x-8)^2-64+k\)
Since, vertex lies on \(\mathrm{x}\)-axis
\(\therefore \mathrm{y}=0\)So, it can be possible only when, \(\mathrm{x}=8, \mathrm{k}=64\)
Kerala CEE-2006
Parabola
120988
An equilateral triangle \(\mathrm{SAB}\) is inscribed in the parabola \(y^2=4 a x\) having its focus at \(S\). It chord \(A B\) lies towards the left of \(S\), then side length of this triangle is
1 \(2 \mathrm{a}(2-\sqrt{3})\)
2 \(4 \mathrm{a}(2-\sqrt{3})\)
3 \(\mathrm{a}(2-\sqrt{3})\)
4 \(8 \mathrm{a}(2-\sqrt{3})\)
Explanation:
B Let A \(\left(\mathrm{at}_1^2, 2 \mathrm{at}_2\right), \mathrm{B}\left(\mathrm{at}_2^2,-2 \mathrm{at}_2\right)\)
We have,
\(\mathrm{m}_{\mathrm{AS}}=\tan \left(\frac{5 \pi}{6}\right)\)
\(\frac{2 \mathrm{at}_1}{\mathrm{at}_1^2-\mathrm{a}}=-\frac{1}{\sqrt{3}}\)
\(\mathrm{t}_1^2+2 \sqrt{3} \mathrm{t}_1-1=0\)
\(\mathrm{t}_1=\frac{-2 \sqrt{3} \pm \sqrt{12+4}}{2}\)
\(\mathrm{t}_1=\frac{+2(-\sqrt{3} \pm 2)}{2}\)
\(\mathrm{t}_1=(-\sqrt{3} \pm 2)\)
Clearly, \(t_1=-\sqrt{3}-2\) is rejected.
thus, \(\quad t_1=(2-\sqrt{3})\)
Hence, \(\mathrm{AB}=4 \mathrm{at}_1\)
\(=4 \mathrm{a}(2-\sqrt{3})\)
Manipal UGET-2013
Parabola
120931
The two ends of a latus rectum of a parabola are \((5,8)\) and \((-7,8)\) Then its focus is
1 \((-1,8)\)
2 \((-2,8)\)
3 \((-2,16)\)
4 \((-1,-8)\)
Explanation:
A Ends of the lotus rectum are \((5,8)\) and \((-7,8)\) We know focus is the midpoint of two end points of latus rectum.
\(\text { So }, \text { focus } =\left(\frac{5-7}{2}, \frac{8+8}{2}\right)\)
\(=(-1,8)\)
120986
Equation of the directrix of the conic \(x^2+4 y+\) \(4=0\) is
1 \(y=1\)
2 \(y=-1\)
3 \(y=0\)
4 \(x=0\)
5 \(x=1\)
Explanation:
C We have the equation of conic, \(\mathrm{x}^2+4 \mathrm{y}+4=0\)
\(x^2=-4 y-4=-4(y+1)\)
\(\mathrm{x}^2=-4 \mathrm{Y}\)
Where \(\mathrm{X}=\mathrm{x}\) and \(\mathrm{Y}=\mathrm{y}+1\)
Equation of directrix is \(\mathrm{Y}=1\)
\(\quad \mathrm{y}+1=1\)
\(\therefore \quad \mathrm{y}=0\)
Kerala CEE-2008
Parabola
120987
If the vertex of the parabola \(y=x^2-16 x+k\) lies on \(x\)-axis, then the value of \(k\) is :
1 16
2 8
3 64
4 -64
5 -8
Explanation:
C Given, Vertex of the parabola
\(y=x^2-16 x+k\)
\(=x^2-2 x(8)+k\)
\(=x^2-2 x(8)+64-64+k\)
\(=(x-8)^2-64+k\)
Since, vertex lies on \(\mathrm{x}\)-axis
\(\therefore \mathrm{y}=0\)So, it can be possible only when, \(\mathrm{x}=8, \mathrm{k}=64\)
Kerala CEE-2006
Parabola
120988
An equilateral triangle \(\mathrm{SAB}\) is inscribed in the parabola \(y^2=4 a x\) having its focus at \(S\). It chord \(A B\) lies towards the left of \(S\), then side length of this triangle is
1 \(2 \mathrm{a}(2-\sqrt{3})\)
2 \(4 \mathrm{a}(2-\sqrt{3})\)
3 \(\mathrm{a}(2-\sqrt{3})\)
4 \(8 \mathrm{a}(2-\sqrt{3})\)
Explanation:
B Let A \(\left(\mathrm{at}_1^2, 2 \mathrm{at}_2\right), \mathrm{B}\left(\mathrm{at}_2^2,-2 \mathrm{at}_2\right)\)
We have,
\(\mathrm{m}_{\mathrm{AS}}=\tan \left(\frac{5 \pi}{6}\right)\)
\(\frac{2 \mathrm{at}_1}{\mathrm{at}_1^2-\mathrm{a}}=-\frac{1}{\sqrt{3}}\)
\(\mathrm{t}_1^2+2 \sqrt{3} \mathrm{t}_1-1=0\)
\(\mathrm{t}_1=\frac{-2 \sqrt{3} \pm \sqrt{12+4}}{2}\)
\(\mathrm{t}_1=\frac{+2(-\sqrt{3} \pm 2)}{2}\)
\(\mathrm{t}_1=(-\sqrt{3} \pm 2)\)
Clearly, \(t_1=-\sqrt{3}-2\) is rejected.
thus, \(\quad t_1=(2-\sqrt{3})\)
Hence, \(\mathrm{AB}=4 \mathrm{at}_1\)
\(=4 \mathrm{a}(2-\sqrt{3})\)
Manipal UGET-2013
Parabola
120931
The two ends of a latus rectum of a parabola are \((5,8)\) and \((-7,8)\) Then its focus is
1 \((-1,8)\)
2 \((-2,8)\)
3 \((-2,16)\)
4 \((-1,-8)\)
Explanation:
A Ends of the lotus rectum are \((5,8)\) and \((-7,8)\) We know focus is the midpoint of two end points of latus rectum.
\(\text { So }, \text { focus } =\left(\frac{5-7}{2}, \frac{8+8}{2}\right)\)
\(=(-1,8)\)
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Parabola
120986
Equation of the directrix of the conic \(x^2+4 y+\) \(4=0\) is
1 \(y=1\)
2 \(y=-1\)
3 \(y=0\)
4 \(x=0\)
5 \(x=1\)
Explanation:
C We have the equation of conic, \(\mathrm{x}^2+4 \mathrm{y}+4=0\)
\(x^2=-4 y-4=-4(y+1)\)
\(\mathrm{x}^2=-4 \mathrm{Y}\)
Where \(\mathrm{X}=\mathrm{x}\) and \(\mathrm{Y}=\mathrm{y}+1\)
Equation of directrix is \(\mathrm{Y}=1\)
\(\quad \mathrm{y}+1=1\)
\(\therefore \quad \mathrm{y}=0\)
Kerala CEE-2008
Parabola
120987
If the vertex of the parabola \(y=x^2-16 x+k\) lies on \(x\)-axis, then the value of \(k\) is :
1 16
2 8
3 64
4 -64
5 -8
Explanation:
C Given, Vertex of the parabola
\(y=x^2-16 x+k\)
\(=x^2-2 x(8)+k\)
\(=x^2-2 x(8)+64-64+k\)
\(=(x-8)^2-64+k\)
Since, vertex lies on \(\mathrm{x}\)-axis
\(\therefore \mathrm{y}=0\)So, it can be possible only when, \(\mathrm{x}=8, \mathrm{k}=64\)
Kerala CEE-2006
Parabola
120988
An equilateral triangle \(\mathrm{SAB}\) is inscribed in the parabola \(y^2=4 a x\) having its focus at \(S\). It chord \(A B\) lies towards the left of \(S\), then side length of this triangle is
1 \(2 \mathrm{a}(2-\sqrt{3})\)
2 \(4 \mathrm{a}(2-\sqrt{3})\)
3 \(\mathrm{a}(2-\sqrt{3})\)
4 \(8 \mathrm{a}(2-\sqrt{3})\)
Explanation:
B Let A \(\left(\mathrm{at}_1^2, 2 \mathrm{at}_2\right), \mathrm{B}\left(\mathrm{at}_2^2,-2 \mathrm{at}_2\right)\)
We have,
\(\mathrm{m}_{\mathrm{AS}}=\tan \left(\frac{5 \pi}{6}\right)\)
\(\frac{2 \mathrm{at}_1}{\mathrm{at}_1^2-\mathrm{a}}=-\frac{1}{\sqrt{3}}\)
\(\mathrm{t}_1^2+2 \sqrt{3} \mathrm{t}_1-1=0\)
\(\mathrm{t}_1=\frac{-2 \sqrt{3} \pm \sqrt{12+4}}{2}\)
\(\mathrm{t}_1=\frac{+2(-\sqrt{3} \pm 2)}{2}\)
\(\mathrm{t}_1=(-\sqrt{3} \pm 2)\)
Clearly, \(t_1=-\sqrt{3}-2\) is rejected.
thus, \(\quad t_1=(2-\sqrt{3})\)
Hence, \(\mathrm{AB}=4 \mathrm{at}_1\)
\(=4 \mathrm{a}(2-\sqrt{3})\)
Manipal UGET-2013
Parabola
120931
The two ends of a latus rectum of a parabola are \((5,8)\) and \((-7,8)\) Then its focus is
1 \((-1,8)\)
2 \((-2,8)\)
3 \((-2,16)\)
4 \((-1,-8)\)
Explanation:
A Ends of the lotus rectum are \((5,8)\) and \((-7,8)\) We know focus is the midpoint of two end points of latus rectum.
\(\text { So }, \text { focus } =\left(\frac{5-7}{2}, \frac{8+8}{2}\right)\)
\(=(-1,8)\)
120986
Equation of the directrix of the conic \(x^2+4 y+\) \(4=0\) is
1 \(y=1\)
2 \(y=-1\)
3 \(y=0\)
4 \(x=0\)
5 \(x=1\)
Explanation:
C We have the equation of conic, \(\mathrm{x}^2+4 \mathrm{y}+4=0\)
\(x^2=-4 y-4=-4(y+1)\)
\(\mathrm{x}^2=-4 \mathrm{Y}\)
Where \(\mathrm{X}=\mathrm{x}\) and \(\mathrm{Y}=\mathrm{y}+1\)
Equation of directrix is \(\mathrm{Y}=1\)
\(\quad \mathrm{y}+1=1\)
\(\therefore \quad \mathrm{y}=0\)
Kerala CEE-2008
Parabola
120987
If the vertex of the parabola \(y=x^2-16 x+k\) lies on \(x\)-axis, then the value of \(k\) is :
1 16
2 8
3 64
4 -64
5 -8
Explanation:
C Given, Vertex of the parabola
\(y=x^2-16 x+k\)
\(=x^2-2 x(8)+k\)
\(=x^2-2 x(8)+64-64+k\)
\(=(x-8)^2-64+k\)
Since, vertex lies on \(\mathrm{x}\)-axis
\(\therefore \mathrm{y}=0\)So, it can be possible only when, \(\mathrm{x}=8, \mathrm{k}=64\)
Kerala CEE-2006
Parabola
120988
An equilateral triangle \(\mathrm{SAB}\) is inscribed in the parabola \(y^2=4 a x\) having its focus at \(S\). It chord \(A B\) lies towards the left of \(S\), then side length of this triangle is
1 \(2 \mathrm{a}(2-\sqrt{3})\)
2 \(4 \mathrm{a}(2-\sqrt{3})\)
3 \(\mathrm{a}(2-\sqrt{3})\)
4 \(8 \mathrm{a}(2-\sqrt{3})\)
Explanation:
B Let A \(\left(\mathrm{at}_1^2, 2 \mathrm{at}_2\right), \mathrm{B}\left(\mathrm{at}_2^2,-2 \mathrm{at}_2\right)\)
We have,
\(\mathrm{m}_{\mathrm{AS}}=\tan \left(\frac{5 \pi}{6}\right)\)
\(\frac{2 \mathrm{at}_1}{\mathrm{at}_1^2-\mathrm{a}}=-\frac{1}{\sqrt{3}}\)
\(\mathrm{t}_1^2+2 \sqrt{3} \mathrm{t}_1-1=0\)
\(\mathrm{t}_1=\frac{-2 \sqrt{3} \pm \sqrt{12+4}}{2}\)
\(\mathrm{t}_1=\frac{+2(-\sqrt{3} \pm 2)}{2}\)
\(\mathrm{t}_1=(-\sqrt{3} \pm 2)\)
Clearly, \(t_1=-\sqrt{3}-2\) is rejected.
thus, \(\quad t_1=(2-\sqrt{3})\)
Hence, \(\mathrm{AB}=4 \mathrm{at}_1\)
\(=4 \mathrm{a}(2-\sqrt{3})\)
Manipal UGET-2013
Parabola
120931
The two ends of a latus rectum of a parabola are \((5,8)\) and \((-7,8)\) Then its focus is
1 \((-1,8)\)
2 \((-2,8)\)
3 \((-2,16)\)
4 \((-1,-8)\)
Explanation:
A Ends of the lotus rectum are \((5,8)\) and \((-7,8)\) We know focus is the midpoint of two end points of latus rectum.
\(\text { So }, \text { focus } =\left(\frac{5-7}{2}, \frac{8+8}{2}\right)\)
\(=(-1,8)\)