120073
The distance between the vertex of the parabola \(y=x^2-4 x+3\) and the centre of the circle \(x^2=9-(y-3)^2\) is
1 \(2 \sqrt{3}\)
2 \(3 \sqrt{2}\)
3 \(2 \sqrt{2}\)
4 \(2 \sqrt{5}\)
Explanation:
D Given,
\(y=x^2-4 x+3\)
\(y+1=x^2-4 x+4\)
\(y+1=(x-2)^2\)
So, the vertex is \((2,-1)\)
And for the circle, \(x^2+(y-3)^2=9\), the centre is \((0,3)\)
\(\therefore\) Distance between vertex and centre,
\(=\sqrt{2^2+(-4)^2}=2 \sqrt{5}\)
COMEDK-2019
Parabola
120074
If the parabola \(y^2=4 a x\) passes through the point \((1,-2)\), then the tangent at this point is
1 \(x+y-1=0\)
2 \(x-y-1=0\)
3 \(x+y+1=0\)
4 \(\mathrm{x}-\mathrm{y}-1=0\)
Explanation:
C Given,
parabola \(\mathrm{y}^2=4 \mathrm{ax}\) passes through the point \((1,-2)\).
\(\therefore(-2)^2=4 \times \mathrm{a} \times 1\)
\(4=4 \mathrm{a}\)
\(\mathrm{a}=1\)
Equation of tangent to the parabola at \((1,-2)\) is
\(y_1=2 a\left(x+x_1\right)\)
\(y(-2)=2(1)(x+1)\)
\(-2 y=2 x+2\)
\(2 x+2 y+2=0\)
Dividing by 2 on both side, we get -
\(\mathrm{x}+\mathrm{y}+1=0\)
VITEEE-2018
Parabola
120075
The locus of a point which moves such that its distance from the point \((0,0)\) is twice its distance from the \(\mathbf{Y}\)-axis, is
1 \(x^2-y^2=0\)
2 \(x^2-3 y^2=0\)
3 \(3 x^2-y^2=0\)
4 None of these
Explanation:
C By hypothesis,
\(|\mathrm{OP}|=2|\mathrm{PM}|\)
\(\sqrt{\mathrm{x}_1^2+\mathrm{y}_1^2}=2\left|\mathrm{x}_1\right|\)
\(|\mathrm{OP}|=2|\mathrm{PM}|\)
On squaring both sides, we get,
\(\mathrm{x}_1^2+\mathrm{y}_1^2=4 \mathrm{x}_1^2\)
\(3 \mathrm{x}_1^2-\mathrm{y}_1^2=0\)\(\therefore\) Locus of the points is \(3 \mathrm{x}^2-\mathrm{y}^2=0\)
UPSEE-2014
Parabola
120076
Locus of the middle points of all chords of \(\frac{x^2}{4}+\frac{y^2}{9}=1\), which are at a distance of 2 units from the vertex of parabola \(y^2=-8 a x\), is
B Let, (h, k) be the midpoint of a chord of the ellipse
\(\frac{\mathrm{x}^2}{4}+\frac{\mathrm{y}^2}{9}=1\)
\(\frac{\mathrm{hx}}{4}+\frac{\mathrm{ky}}{9}-1=\frac{\mathrm{h}^2}{4}+\frac{\mathrm{k}^2}{9}-1\)
\(\frac{\mathrm{hx}}{4}+\frac{\mathrm{ky}}{9}=\frac{\mathrm{h}^2}{4}+\frac{\mathrm{k}^2}{9}\)
Since, it is a distance of 2 units from the vertex \((0,0)\) of the parabola \(y^2=-8 a x\)
\(\therefore \quad \quad\left|\frac{\frac{\mathrm{h}^2}{4}+\frac{\mathrm{k}^2}{9}}{\sqrt{\frac{\mathrm{h}^2}{16}+\frac{\mathrm{k}^2}{81}}}\right| =2\)
\(\left(\frac{\mathrm{h}^2}{4}+\frac{\mathrm{k}^2}{9}\right)^2 =4\left(\frac{\mathrm{h}^2}{16}+\frac{\mathrm{k}^2}{81}\right)\)
Hence, the locus of \((h, k)\) is
\(\left(\frac{x^2}{4}+\frac{y^2}{9}\right)=4\left(\frac{x^2}{16}+\frac{y^2}{81}\right)\)
120073
The distance between the vertex of the parabola \(y=x^2-4 x+3\) and the centre of the circle \(x^2=9-(y-3)^2\) is
1 \(2 \sqrt{3}\)
2 \(3 \sqrt{2}\)
3 \(2 \sqrt{2}\)
4 \(2 \sqrt{5}\)
Explanation:
D Given,
\(y=x^2-4 x+3\)
\(y+1=x^2-4 x+4\)
\(y+1=(x-2)^2\)
So, the vertex is \((2,-1)\)
And for the circle, \(x^2+(y-3)^2=9\), the centre is \((0,3)\)
\(\therefore\) Distance between vertex and centre,
\(=\sqrt{2^2+(-4)^2}=2 \sqrt{5}\)
COMEDK-2019
Parabola
120074
If the parabola \(y^2=4 a x\) passes through the point \((1,-2)\), then the tangent at this point is
1 \(x+y-1=0\)
2 \(x-y-1=0\)
3 \(x+y+1=0\)
4 \(\mathrm{x}-\mathrm{y}-1=0\)
Explanation:
C Given,
parabola \(\mathrm{y}^2=4 \mathrm{ax}\) passes through the point \((1,-2)\).
\(\therefore(-2)^2=4 \times \mathrm{a} \times 1\)
\(4=4 \mathrm{a}\)
\(\mathrm{a}=1\)
Equation of tangent to the parabola at \((1,-2)\) is
\(y_1=2 a\left(x+x_1\right)\)
\(y(-2)=2(1)(x+1)\)
\(-2 y=2 x+2\)
\(2 x+2 y+2=0\)
Dividing by 2 on both side, we get -
\(\mathrm{x}+\mathrm{y}+1=0\)
VITEEE-2018
Parabola
120075
The locus of a point which moves such that its distance from the point \((0,0)\) is twice its distance from the \(\mathbf{Y}\)-axis, is
1 \(x^2-y^2=0\)
2 \(x^2-3 y^2=0\)
3 \(3 x^2-y^2=0\)
4 None of these
Explanation:
C By hypothesis,
\(|\mathrm{OP}|=2|\mathrm{PM}|\)
\(\sqrt{\mathrm{x}_1^2+\mathrm{y}_1^2}=2\left|\mathrm{x}_1\right|\)
\(|\mathrm{OP}|=2|\mathrm{PM}|\)
On squaring both sides, we get,
\(\mathrm{x}_1^2+\mathrm{y}_1^2=4 \mathrm{x}_1^2\)
\(3 \mathrm{x}_1^2-\mathrm{y}_1^2=0\)\(\therefore\) Locus of the points is \(3 \mathrm{x}^2-\mathrm{y}^2=0\)
UPSEE-2014
Parabola
120076
Locus of the middle points of all chords of \(\frac{x^2}{4}+\frac{y^2}{9}=1\), which are at a distance of 2 units from the vertex of parabola \(y^2=-8 a x\), is
B Let, (h, k) be the midpoint of a chord of the ellipse
\(\frac{\mathrm{x}^2}{4}+\frac{\mathrm{y}^2}{9}=1\)
\(\frac{\mathrm{hx}}{4}+\frac{\mathrm{ky}}{9}-1=\frac{\mathrm{h}^2}{4}+\frac{\mathrm{k}^2}{9}-1\)
\(\frac{\mathrm{hx}}{4}+\frac{\mathrm{ky}}{9}=\frac{\mathrm{h}^2}{4}+\frac{\mathrm{k}^2}{9}\)
Since, it is a distance of 2 units from the vertex \((0,0)\) of the parabola \(y^2=-8 a x\)
\(\therefore \quad \quad\left|\frac{\frac{\mathrm{h}^2}{4}+\frac{\mathrm{k}^2}{9}}{\sqrt{\frac{\mathrm{h}^2}{16}+\frac{\mathrm{k}^2}{81}}}\right| =2\)
\(\left(\frac{\mathrm{h}^2}{4}+\frac{\mathrm{k}^2}{9}\right)^2 =4\left(\frac{\mathrm{h}^2}{16}+\frac{\mathrm{k}^2}{81}\right)\)
Hence, the locus of \((h, k)\) is
\(\left(\frac{x^2}{4}+\frac{y^2}{9}\right)=4\left(\frac{x^2}{16}+\frac{y^2}{81}\right)\)
120073
The distance between the vertex of the parabola \(y=x^2-4 x+3\) and the centre of the circle \(x^2=9-(y-3)^2\) is
1 \(2 \sqrt{3}\)
2 \(3 \sqrt{2}\)
3 \(2 \sqrt{2}\)
4 \(2 \sqrt{5}\)
Explanation:
D Given,
\(y=x^2-4 x+3\)
\(y+1=x^2-4 x+4\)
\(y+1=(x-2)^2\)
So, the vertex is \((2,-1)\)
And for the circle, \(x^2+(y-3)^2=9\), the centre is \((0,3)\)
\(\therefore\) Distance between vertex and centre,
\(=\sqrt{2^2+(-4)^2}=2 \sqrt{5}\)
COMEDK-2019
Parabola
120074
If the parabola \(y^2=4 a x\) passes through the point \((1,-2)\), then the tangent at this point is
1 \(x+y-1=0\)
2 \(x-y-1=0\)
3 \(x+y+1=0\)
4 \(\mathrm{x}-\mathrm{y}-1=0\)
Explanation:
C Given,
parabola \(\mathrm{y}^2=4 \mathrm{ax}\) passes through the point \((1,-2)\).
\(\therefore(-2)^2=4 \times \mathrm{a} \times 1\)
\(4=4 \mathrm{a}\)
\(\mathrm{a}=1\)
Equation of tangent to the parabola at \((1,-2)\) is
\(y_1=2 a\left(x+x_1\right)\)
\(y(-2)=2(1)(x+1)\)
\(-2 y=2 x+2\)
\(2 x+2 y+2=0\)
Dividing by 2 on both side, we get -
\(\mathrm{x}+\mathrm{y}+1=0\)
VITEEE-2018
Parabola
120075
The locus of a point which moves such that its distance from the point \((0,0)\) is twice its distance from the \(\mathbf{Y}\)-axis, is
1 \(x^2-y^2=0\)
2 \(x^2-3 y^2=0\)
3 \(3 x^2-y^2=0\)
4 None of these
Explanation:
C By hypothesis,
\(|\mathrm{OP}|=2|\mathrm{PM}|\)
\(\sqrt{\mathrm{x}_1^2+\mathrm{y}_1^2}=2\left|\mathrm{x}_1\right|\)
\(|\mathrm{OP}|=2|\mathrm{PM}|\)
On squaring both sides, we get,
\(\mathrm{x}_1^2+\mathrm{y}_1^2=4 \mathrm{x}_1^2\)
\(3 \mathrm{x}_1^2-\mathrm{y}_1^2=0\)\(\therefore\) Locus of the points is \(3 \mathrm{x}^2-\mathrm{y}^2=0\)
UPSEE-2014
Parabola
120076
Locus of the middle points of all chords of \(\frac{x^2}{4}+\frac{y^2}{9}=1\), which are at a distance of 2 units from the vertex of parabola \(y^2=-8 a x\), is
B Let, (h, k) be the midpoint of a chord of the ellipse
\(\frac{\mathrm{x}^2}{4}+\frac{\mathrm{y}^2}{9}=1\)
\(\frac{\mathrm{hx}}{4}+\frac{\mathrm{ky}}{9}-1=\frac{\mathrm{h}^2}{4}+\frac{\mathrm{k}^2}{9}-1\)
\(\frac{\mathrm{hx}}{4}+\frac{\mathrm{ky}}{9}=\frac{\mathrm{h}^2}{4}+\frac{\mathrm{k}^2}{9}\)
Since, it is a distance of 2 units from the vertex \((0,0)\) of the parabola \(y^2=-8 a x\)
\(\therefore \quad \quad\left|\frac{\frac{\mathrm{h}^2}{4}+\frac{\mathrm{k}^2}{9}}{\sqrt{\frac{\mathrm{h}^2}{16}+\frac{\mathrm{k}^2}{81}}}\right| =2\)
\(\left(\frac{\mathrm{h}^2}{4}+\frac{\mathrm{k}^2}{9}\right)^2 =4\left(\frac{\mathrm{h}^2}{16}+\frac{\mathrm{k}^2}{81}\right)\)
Hence, the locus of \((h, k)\) is
\(\left(\frac{x^2}{4}+\frac{y^2}{9}\right)=4\left(\frac{x^2}{16}+\frac{y^2}{81}\right)\)
120073
The distance between the vertex of the parabola \(y=x^2-4 x+3\) and the centre of the circle \(x^2=9-(y-3)^2\) is
1 \(2 \sqrt{3}\)
2 \(3 \sqrt{2}\)
3 \(2 \sqrt{2}\)
4 \(2 \sqrt{5}\)
Explanation:
D Given,
\(y=x^2-4 x+3\)
\(y+1=x^2-4 x+4\)
\(y+1=(x-2)^2\)
So, the vertex is \((2,-1)\)
And for the circle, \(x^2+(y-3)^2=9\), the centre is \((0,3)\)
\(\therefore\) Distance between vertex and centre,
\(=\sqrt{2^2+(-4)^2}=2 \sqrt{5}\)
COMEDK-2019
Parabola
120074
If the parabola \(y^2=4 a x\) passes through the point \((1,-2)\), then the tangent at this point is
1 \(x+y-1=0\)
2 \(x-y-1=0\)
3 \(x+y+1=0\)
4 \(\mathrm{x}-\mathrm{y}-1=0\)
Explanation:
C Given,
parabola \(\mathrm{y}^2=4 \mathrm{ax}\) passes through the point \((1,-2)\).
\(\therefore(-2)^2=4 \times \mathrm{a} \times 1\)
\(4=4 \mathrm{a}\)
\(\mathrm{a}=1\)
Equation of tangent to the parabola at \((1,-2)\) is
\(y_1=2 a\left(x+x_1\right)\)
\(y(-2)=2(1)(x+1)\)
\(-2 y=2 x+2\)
\(2 x+2 y+2=0\)
Dividing by 2 on both side, we get -
\(\mathrm{x}+\mathrm{y}+1=0\)
VITEEE-2018
Parabola
120075
The locus of a point which moves such that its distance from the point \((0,0)\) is twice its distance from the \(\mathbf{Y}\)-axis, is
1 \(x^2-y^2=0\)
2 \(x^2-3 y^2=0\)
3 \(3 x^2-y^2=0\)
4 None of these
Explanation:
C By hypothesis,
\(|\mathrm{OP}|=2|\mathrm{PM}|\)
\(\sqrt{\mathrm{x}_1^2+\mathrm{y}_1^2}=2\left|\mathrm{x}_1\right|\)
\(|\mathrm{OP}|=2|\mathrm{PM}|\)
On squaring both sides, we get,
\(\mathrm{x}_1^2+\mathrm{y}_1^2=4 \mathrm{x}_1^2\)
\(3 \mathrm{x}_1^2-\mathrm{y}_1^2=0\)\(\therefore\) Locus of the points is \(3 \mathrm{x}^2-\mathrm{y}^2=0\)
UPSEE-2014
Parabola
120076
Locus of the middle points of all chords of \(\frac{x^2}{4}+\frac{y^2}{9}=1\), which are at a distance of 2 units from the vertex of parabola \(y^2=-8 a x\), is
B Let, (h, k) be the midpoint of a chord of the ellipse
\(\frac{\mathrm{x}^2}{4}+\frac{\mathrm{y}^2}{9}=1\)
\(\frac{\mathrm{hx}}{4}+\frac{\mathrm{ky}}{9}-1=\frac{\mathrm{h}^2}{4}+\frac{\mathrm{k}^2}{9}-1\)
\(\frac{\mathrm{hx}}{4}+\frac{\mathrm{ky}}{9}=\frac{\mathrm{h}^2}{4}+\frac{\mathrm{k}^2}{9}\)
Since, it is a distance of 2 units from the vertex \((0,0)\) of the parabola \(y^2=-8 a x\)
\(\therefore \quad \quad\left|\frac{\frac{\mathrm{h}^2}{4}+\frac{\mathrm{k}^2}{9}}{\sqrt{\frac{\mathrm{h}^2}{16}+\frac{\mathrm{k}^2}{81}}}\right| =2\)
\(\left(\frac{\mathrm{h}^2}{4}+\frac{\mathrm{k}^2}{9}\right)^2 =4\left(\frac{\mathrm{h}^2}{16}+\frac{\mathrm{k}^2}{81}\right)\)
Hence, the locus of \((h, k)\) is
\(\left(\frac{x^2}{4}+\frac{y^2}{9}\right)=4\left(\frac{x^2}{16}+\frac{y^2}{81}\right)\)
