120962
A circle of radius 2 units lies in the first quadrant and touches both the axis. The equation of the circle with centre at \((6,5)\) and touching the circle externally is
1 \(x^2+y^2-12 x-10 y+12=0\)
2 \(x^2+y^2-12 x-10 y-20=0\)
3 \(x^2+y^2-12 x-10 y+25=0\)
4 \(x^2+y^2-12 x-10 y+52=0\)
Explanation:
D :
Let, radius of circle is \(\mathrm{r}\).
If two circle touch externally then,
\(c_1 c_2=r_1+r_2\)
\(\sqrt{(6-2)^2+(5-2)^2}=2+r\)
\(5=2+r\)
\(r=3\)
\(\therefore \text { The equation of circle } c_2\)
\((x-6)^2+(y-5)^2=3^2\)
\(x^2+y^2-12 x-10 y+52=0\)
AP EAMCET-23.04.2019
Parabola
120963
The parabola with directrix \(x+2 y-1=0\) and focus \((1,0)\) is.....
1 \(4 x^2-4 x y+y^2-8 x+4 y+4=0\)
2 \(4 x^2+4 x y+y^2-8 x+4 y+4=0\)
3 \(4 x^2+4 x y+y^2+8 x-4 y+4=0\)
4 \(4 x^2-4 x y+y^2-8 x-4 y+4=0\)
Explanation:
A Given,
\(\text { Directrix of parabola } \mathrm{x}+2 \mathrm{y}-1=0\)
\(\text { Let, a point on parabola } \mathrm{P}(\mathrm{h}, \mathrm{k})\)
\(\text { Focus }=(1,0)\)
\(\sqrt{(\mathrm{h}-1)^2+\mathrm{k}^2}=\frac{\mathrm{h}+2 \mathrm{k}-1}{\sqrt{1^2+2^2}}\)
\(\left(\sqrt{(\mathrm{h}-1)^2+\mathrm{k}^2}\right)^2=\left(\frac{\mathrm{h}+2 \mathrm{k}-1}{\sqrt{5}}\right)^2\)
\((\mathrm{~h}-1)^2+\mathrm{k}^2=\frac{(\mathrm{h}+2 \mathrm{k}-1)^2}{5}\)
\(\quad 5\left(\mathrm{~h}^2-2 \mathrm{~h}+1+\mathrm{k}^2\right)=\mathrm{h}^2+4 \mathrm{k}^2+1-2 \mathrm{~h}-4 \mathrm{k}+4 \mathrm{kh}\)
\(5 \mathrm{~h}^2-10 \mathrm{~h}+5+5 \mathrm{k}^2=\mathrm{h}^2+4 \mathrm{k}^2-2 \mathrm{~h}-4 \mathrm{k}+4 \mathrm{kh}+1\)
\(\quad \mathrm{h}^2-4 \mathrm{kh}+\mathrm{k}^2-8 \mathrm{~h}+4 \mathrm{k}+4=0\)
\(\text { Put } \mathrm{h}=\mathrm{x} \text { and } \mathrm{k}=\mathrm{y}\)
\(\text { Then, } 4 \mathrm{x}^2-4 \mathrm{xy}+\mathrm{y}^2-8 \mathrm{x}+4 \mathrm{y}+4=0\)
AP EAMCET-21.09.2020
Parabola
120965
The value of ' \(a\) ' for which the function
\(f(x)=a \sin x+\left(\frac{1}{3}\right) \sin 3 x\) has an extremum at \(x=\frac{\pi}{3}\), is
1 1
2 -1
3 2
4 0
Explanation:
C Given, \(\mathrm{f}(\mathrm{x})=\mathrm{a} \sin \mathrm{x}+\frac{1}{3} \sin 3 \mathrm{x}\)
On differentiating w.r.t.x, we get
\(\mathrm{f}^{\prime}(\mathrm{x})=\mathrm{a} \cos \mathrm{x}+\frac{3}{3} \cos 3 \mathrm{x}=0\)
\(\text { at } \mathrm{x}=\frac{\pi}{3}\)
\(\text { a } \cos \frac{\pi}{3}+\cos \left(\frac{\pi}{3} \cdot 3\right)=0\)
\(\text { a. } \frac{1}{2}+(-1)=0\)
\(\frac{\mathrm{a}}{2}=1\)
\(\mathrm{a}=2\)
Manipal UGET-2018
Parabola
120966
The point on the curve \(x^2=4 y\), which is nearest to the point \((1,2)\) is
1 \((0,0)\)
2 \((-2,1)\)
3 \((2,1)\)
4 \((2,-1)\)
Explanation:
C Given,
\(x^2=4 y\)
Using distance formula for point \((1,2)\)
Let, \((x, y)\) be the point on \(x^2=4 y\) which is nearest to the point \((+1,2)\)
\(\because(\mathrm{x}, \mathrm{y})\) lies on curve \(\mathrm{y}=\frac{\mathrm{x}^2}{4}\)
\(\therefore d=\sqrt{(x-1)^2+\left(\frac{x^2}{4}-2\right)^2}\)
\(d^2=(x-1)^2+\left(\frac{x^2}{4}-2\right)^2\)
Put \(d^2=\) D
\(\therefore D=(x-1)^2+\left(\frac{x^2}{4}-2\right)^2\)
\(\frac{d D}{d x}=2(x-1)+2\left(\frac{x^2}{4}-2\right) \cdot \frac{2 x}{4}\)
\(=2 x-2+\frac{x^3}{4}-2 x\)
\(=-2+\frac{x^3}{4}\)
\(\frac{d D}{d x}=0\)
\(=-8+x^3=0\)
\(\therefore x^3=8\)
\(x=2\)
\(\frac{d^2 D}{d x^2}=\frac{3 x^2}{4}\)
\(A t x=2\)
\(\frac{d^2 D}{d x^2}=\frac{3}{4} \times 4=3>0\)
\(\mathrm{D}\) is minimum when \(\mathrm{x}=2, \mathrm{y}=\frac{4}{4}=1\)
\(\therefore\) Required point \((2,1)\)
Assam CEE-2022
Parabola
120967
The equation of parabola whose focus is \((5,3)\) and directrix is \(3 x-4 y+1=0\) is
120962
A circle of radius 2 units lies in the first quadrant and touches both the axis. The equation of the circle with centre at \((6,5)\) and touching the circle externally is
1 \(x^2+y^2-12 x-10 y+12=0\)
2 \(x^2+y^2-12 x-10 y-20=0\)
3 \(x^2+y^2-12 x-10 y+25=0\)
4 \(x^2+y^2-12 x-10 y+52=0\)
Explanation:
D :
Let, radius of circle is \(\mathrm{r}\).
If two circle touch externally then,
\(c_1 c_2=r_1+r_2\)
\(\sqrt{(6-2)^2+(5-2)^2}=2+r\)
\(5=2+r\)
\(r=3\)
\(\therefore \text { The equation of circle } c_2\)
\((x-6)^2+(y-5)^2=3^2\)
\(x^2+y^2-12 x-10 y+52=0\)
AP EAMCET-23.04.2019
Parabola
120963
The parabola with directrix \(x+2 y-1=0\) and focus \((1,0)\) is.....
1 \(4 x^2-4 x y+y^2-8 x+4 y+4=0\)
2 \(4 x^2+4 x y+y^2-8 x+4 y+4=0\)
3 \(4 x^2+4 x y+y^2+8 x-4 y+4=0\)
4 \(4 x^2-4 x y+y^2-8 x-4 y+4=0\)
Explanation:
A Given,
\(\text { Directrix of parabola } \mathrm{x}+2 \mathrm{y}-1=0\)
\(\text { Let, a point on parabola } \mathrm{P}(\mathrm{h}, \mathrm{k})\)
\(\text { Focus }=(1,0)\)
\(\sqrt{(\mathrm{h}-1)^2+\mathrm{k}^2}=\frac{\mathrm{h}+2 \mathrm{k}-1}{\sqrt{1^2+2^2}}\)
\(\left(\sqrt{(\mathrm{h}-1)^2+\mathrm{k}^2}\right)^2=\left(\frac{\mathrm{h}+2 \mathrm{k}-1}{\sqrt{5}}\right)^2\)
\((\mathrm{~h}-1)^2+\mathrm{k}^2=\frac{(\mathrm{h}+2 \mathrm{k}-1)^2}{5}\)
\(\quad 5\left(\mathrm{~h}^2-2 \mathrm{~h}+1+\mathrm{k}^2\right)=\mathrm{h}^2+4 \mathrm{k}^2+1-2 \mathrm{~h}-4 \mathrm{k}+4 \mathrm{kh}\)
\(5 \mathrm{~h}^2-10 \mathrm{~h}+5+5 \mathrm{k}^2=\mathrm{h}^2+4 \mathrm{k}^2-2 \mathrm{~h}-4 \mathrm{k}+4 \mathrm{kh}+1\)
\(\quad \mathrm{h}^2-4 \mathrm{kh}+\mathrm{k}^2-8 \mathrm{~h}+4 \mathrm{k}+4=0\)
\(\text { Put } \mathrm{h}=\mathrm{x} \text { and } \mathrm{k}=\mathrm{y}\)
\(\text { Then, } 4 \mathrm{x}^2-4 \mathrm{xy}+\mathrm{y}^2-8 \mathrm{x}+4 \mathrm{y}+4=0\)
AP EAMCET-21.09.2020
Parabola
120965
The value of ' \(a\) ' for which the function
\(f(x)=a \sin x+\left(\frac{1}{3}\right) \sin 3 x\) has an extremum at \(x=\frac{\pi}{3}\), is
1 1
2 -1
3 2
4 0
Explanation:
C Given, \(\mathrm{f}(\mathrm{x})=\mathrm{a} \sin \mathrm{x}+\frac{1}{3} \sin 3 \mathrm{x}\)
On differentiating w.r.t.x, we get
\(\mathrm{f}^{\prime}(\mathrm{x})=\mathrm{a} \cos \mathrm{x}+\frac{3}{3} \cos 3 \mathrm{x}=0\)
\(\text { at } \mathrm{x}=\frac{\pi}{3}\)
\(\text { a } \cos \frac{\pi}{3}+\cos \left(\frac{\pi}{3} \cdot 3\right)=0\)
\(\text { a. } \frac{1}{2}+(-1)=0\)
\(\frac{\mathrm{a}}{2}=1\)
\(\mathrm{a}=2\)
Manipal UGET-2018
Parabola
120966
The point on the curve \(x^2=4 y\), which is nearest to the point \((1,2)\) is
1 \((0,0)\)
2 \((-2,1)\)
3 \((2,1)\)
4 \((2,-1)\)
Explanation:
C Given,
\(x^2=4 y\)
Using distance formula for point \((1,2)\)
Let, \((x, y)\) be the point on \(x^2=4 y\) which is nearest to the point \((+1,2)\)
\(\because(\mathrm{x}, \mathrm{y})\) lies on curve \(\mathrm{y}=\frac{\mathrm{x}^2}{4}\)
\(\therefore d=\sqrt{(x-1)^2+\left(\frac{x^2}{4}-2\right)^2}\)
\(d^2=(x-1)^2+\left(\frac{x^2}{4}-2\right)^2\)
Put \(d^2=\) D
\(\therefore D=(x-1)^2+\left(\frac{x^2}{4}-2\right)^2\)
\(\frac{d D}{d x}=2(x-1)+2\left(\frac{x^2}{4}-2\right) \cdot \frac{2 x}{4}\)
\(=2 x-2+\frac{x^3}{4}-2 x\)
\(=-2+\frac{x^3}{4}\)
\(\frac{d D}{d x}=0\)
\(=-8+x^3=0\)
\(\therefore x^3=8\)
\(x=2\)
\(\frac{d^2 D}{d x^2}=\frac{3 x^2}{4}\)
\(A t x=2\)
\(\frac{d^2 D}{d x^2}=\frac{3}{4} \times 4=3>0\)
\(\mathrm{D}\) is minimum when \(\mathrm{x}=2, \mathrm{y}=\frac{4}{4}=1\)
\(\therefore\) Required point \((2,1)\)
Assam CEE-2022
Parabola
120967
The equation of parabola whose focus is \((5,3)\) and directrix is \(3 x-4 y+1=0\) is
120962
A circle of radius 2 units lies in the first quadrant and touches both the axis. The equation of the circle with centre at \((6,5)\) and touching the circle externally is
1 \(x^2+y^2-12 x-10 y+12=0\)
2 \(x^2+y^2-12 x-10 y-20=0\)
3 \(x^2+y^2-12 x-10 y+25=0\)
4 \(x^2+y^2-12 x-10 y+52=0\)
Explanation:
D :
Let, radius of circle is \(\mathrm{r}\).
If two circle touch externally then,
\(c_1 c_2=r_1+r_2\)
\(\sqrt{(6-2)^2+(5-2)^2}=2+r\)
\(5=2+r\)
\(r=3\)
\(\therefore \text { The equation of circle } c_2\)
\((x-6)^2+(y-5)^2=3^2\)
\(x^2+y^2-12 x-10 y+52=0\)
AP EAMCET-23.04.2019
Parabola
120963
The parabola with directrix \(x+2 y-1=0\) and focus \((1,0)\) is.....
1 \(4 x^2-4 x y+y^2-8 x+4 y+4=0\)
2 \(4 x^2+4 x y+y^2-8 x+4 y+4=0\)
3 \(4 x^2+4 x y+y^2+8 x-4 y+4=0\)
4 \(4 x^2-4 x y+y^2-8 x-4 y+4=0\)
Explanation:
A Given,
\(\text { Directrix of parabola } \mathrm{x}+2 \mathrm{y}-1=0\)
\(\text { Let, a point on parabola } \mathrm{P}(\mathrm{h}, \mathrm{k})\)
\(\text { Focus }=(1,0)\)
\(\sqrt{(\mathrm{h}-1)^2+\mathrm{k}^2}=\frac{\mathrm{h}+2 \mathrm{k}-1}{\sqrt{1^2+2^2}}\)
\(\left(\sqrt{(\mathrm{h}-1)^2+\mathrm{k}^2}\right)^2=\left(\frac{\mathrm{h}+2 \mathrm{k}-1}{\sqrt{5}}\right)^2\)
\((\mathrm{~h}-1)^2+\mathrm{k}^2=\frac{(\mathrm{h}+2 \mathrm{k}-1)^2}{5}\)
\(\quad 5\left(\mathrm{~h}^2-2 \mathrm{~h}+1+\mathrm{k}^2\right)=\mathrm{h}^2+4 \mathrm{k}^2+1-2 \mathrm{~h}-4 \mathrm{k}+4 \mathrm{kh}\)
\(5 \mathrm{~h}^2-10 \mathrm{~h}+5+5 \mathrm{k}^2=\mathrm{h}^2+4 \mathrm{k}^2-2 \mathrm{~h}-4 \mathrm{k}+4 \mathrm{kh}+1\)
\(\quad \mathrm{h}^2-4 \mathrm{kh}+\mathrm{k}^2-8 \mathrm{~h}+4 \mathrm{k}+4=0\)
\(\text { Put } \mathrm{h}=\mathrm{x} \text { and } \mathrm{k}=\mathrm{y}\)
\(\text { Then, } 4 \mathrm{x}^2-4 \mathrm{xy}+\mathrm{y}^2-8 \mathrm{x}+4 \mathrm{y}+4=0\)
AP EAMCET-21.09.2020
Parabola
120965
The value of ' \(a\) ' for which the function
\(f(x)=a \sin x+\left(\frac{1}{3}\right) \sin 3 x\) has an extremum at \(x=\frac{\pi}{3}\), is
1 1
2 -1
3 2
4 0
Explanation:
C Given, \(\mathrm{f}(\mathrm{x})=\mathrm{a} \sin \mathrm{x}+\frac{1}{3} \sin 3 \mathrm{x}\)
On differentiating w.r.t.x, we get
\(\mathrm{f}^{\prime}(\mathrm{x})=\mathrm{a} \cos \mathrm{x}+\frac{3}{3} \cos 3 \mathrm{x}=0\)
\(\text { at } \mathrm{x}=\frac{\pi}{3}\)
\(\text { a } \cos \frac{\pi}{3}+\cos \left(\frac{\pi}{3} \cdot 3\right)=0\)
\(\text { a. } \frac{1}{2}+(-1)=0\)
\(\frac{\mathrm{a}}{2}=1\)
\(\mathrm{a}=2\)
Manipal UGET-2018
Parabola
120966
The point on the curve \(x^2=4 y\), which is nearest to the point \((1,2)\) is
1 \((0,0)\)
2 \((-2,1)\)
3 \((2,1)\)
4 \((2,-1)\)
Explanation:
C Given,
\(x^2=4 y\)
Using distance formula for point \((1,2)\)
Let, \((x, y)\) be the point on \(x^2=4 y\) which is nearest to the point \((+1,2)\)
\(\because(\mathrm{x}, \mathrm{y})\) lies on curve \(\mathrm{y}=\frac{\mathrm{x}^2}{4}\)
\(\therefore d=\sqrt{(x-1)^2+\left(\frac{x^2}{4}-2\right)^2}\)
\(d^2=(x-1)^2+\left(\frac{x^2}{4}-2\right)^2\)
Put \(d^2=\) D
\(\therefore D=(x-1)^2+\left(\frac{x^2}{4}-2\right)^2\)
\(\frac{d D}{d x}=2(x-1)+2\left(\frac{x^2}{4}-2\right) \cdot \frac{2 x}{4}\)
\(=2 x-2+\frac{x^3}{4}-2 x\)
\(=-2+\frac{x^3}{4}\)
\(\frac{d D}{d x}=0\)
\(=-8+x^3=0\)
\(\therefore x^3=8\)
\(x=2\)
\(\frac{d^2 D}{d x^2}=\frac{3 x^2}{4}\)
\(A t x=2\)
\(\frac{d^2 D}{d x^2}=\frac{3}{4} \times 4=3>0\)
\(\mathrm{D}\) is minimum when \(\mathrm{x}=2, \mathrm{y}=\frac{4}{4}=1\)
\(\therefore\) Required point \((2,1)\)
Assam CEE-2022
Parabola
120967
The equation of parabola whose focus is \((5,3)\) and directrix is \(3 x-4 y+1=0\) is
NEET Test Series from KOTA - 10 Papers In MS WORD
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Parabola
120962
A circle of radius 2 units lies in the first quadrant and touches both the axis. The equation of the circle with centre at \((6,5)\) and touching the circle externally is
1 \(x^2+y^2-12 x-10 y+12=0\)
2 \(x^2+y^2-12 x-10 y-20=0\)
3 \(x^2+y^2-12 x-10 y+25=0\)
4 \(x^2+y^2-12 x-10 y+52=0\)
Explanation:
D :
Let, radius of circle is \(\mathrm{r}\).
If two circle touch externally then,
\(c_1 c_2=r_1+r_2\)
\(\sqrt{(6-2)^2+(5-2)^2}=2+r\)
\(5=2+r\)
\(r=3\)
\(\therefore \text { The equation of circle } c_2\)
\((x-6)^2+(y-5)^2=3^2\)
\(x^2+y^2-12 x-10 y+52=0\)
AP EAMCET-23.04.2019
Parabola
120963
The parabola with directrix \(x+2 y-1=0\) and focus \((1,0)\) is.....
1 \(4 x^2-4 x y+y^2-8 x+4 y+4=0\)
2 \(4 x^2+4 x y+y^2-8 x+4 y+4=0\)
3 \(4 x^2+4 x y+y^2+8 x-4 y+4=0\)
4 \(4 x^2-4 x y+y^2-8 x-4 y+4=0\)
Explanation:
A Given,
\(\text { Directrix of parabola } \mathrm{x}+2 \mathrm{y}-1=0\)
\(\text { Let, a point on parabola } \mathrm{P}(\mathrm{h}, \mathrm{k})\)
\(\text { Focus }=(1,0)\)
\(\sqrt{(\mathrm{h}-1)^2+\mathrm{k}^2}=\frac{\mathrm{h}+2 \mathrm{k}-1}{\sqrt{1^2+2^2}}\)
\(\left(\sqrt{(\mathrm{h}-1)^2+\mathrm{k}^2}\right)^2=\left(\frac{\mathrm{h}+2 \mathrm{k}-1}{\sqrt{5}}\right)^2\)
\((\mathrm{~h}-1)^2+\mathrm{k}^2=\frac{(\mathrm{h}+2 \mathrm{k}-1)^2}{5}\)
\(\quad 5\left(\mathrm{~h}^2-2 \mathrm{~h}+1+\mathrm{k}^2\right)=\mathrm{h}^2+4 \mathrm{k}^2+1-2 \mathrm{~h}-4 \mathrm{k}+4 \mathrm{kh}\)
\(5 \mathrm{~h}^2-10 \mathrm{~h}+5+5 \mathrm{k}^2=\mathrm{h}^2+4 \mathrm{k}^2-2 \mathrm{~h}-4 \mathrm{k}+4 \mathrm{kh}+1\)
\(\quad \mathrm{h}^2-4 \mathrm{kh}+\mathrm{k}^2-8 \mathrm{~h}+4 \mathrm{k}+4=0\)
\(\text { Put } \mathrm{h}=\mathrm{x} \text { and } \mathrm{k}=\mathrm{y}\)
\(\text { Then, } 4 \mathrm{x}^2-4 \mathrm{xy}+\mathrm{y}^2-8 \mathrm{x}+4 \mathrm{y}+4=0\)
AP EAMCET-21.09.2020
Parabola
120965
The value of ' \(a\) ' for which the function
\(f(x)=a \sin x+\left(\frac{1}{3}\right) \sin 3 x\) has an extremum at \(x=\frac{\pi}{3}\), is
1 1
2 -1
3 2
4 0
Explanation:
C Given, \(\mathrm{f}(\mathrm{x})=\mathrm{a} \sin \mathrm{x}+\frac{1}{3} \sin 3 \mathrm{x}\)
On differentiating w.r.t.x, we get
\(\mathrm{f}^{\prime}(\mathrm{x})=\mathrm{a} \cos \mathrm{x}+\frac{3}{3} \cos 3 \mathrm{x}=0\)
\(\text { at } \mathrm{x}=\frac{\pi}{3}\)
\(\text { a } \cos \frac{\pi}{3}+\cos \left(\frac{\pi}{3} \cdot 3\right)=0\)
\(\text { a. } \frac{1}{2}+(-1)=0\)
\(\frac{\mathrm{a}}{2}=1\)
\(\mathrm{a}=2\)
Manipal UGET-2018
Parabola
120966
The point on the curve \(x^2=4 y\), which is nearest to the point \((1,2)\) is
1 \((0,0)\)
2 \((-2,1)\)
3 \((2,1)\)
4 \((2,-1)\)
Explanation:
C Given,
\(x^2=4 y\)
Using distance formula for point \((1,2)\)
Let, \((x, y)\) be the point on \(x^2=4 y\) which is nearest to the point \((+1,2)\)
\(\because(\mathrm{x}, \mathrm{y})\) lies on curve \(\mathrm{y}=\frac{\mathrm{x}^2}{4}\)
\(\therefore d=\sqrt{(x-1)^2+\left(\frac{x^2}{4}-2\right)^2}\)
\(d^2=(x-1)^2+\left(\frac{x^2}{4}-2\right)^2\)
Put \(d^2=\) D
\(\therefore D=(x-1)^2+\left(\frac{x^2}{4}-2\right)^2\)
\(\frac{d D}{d x}=2(x-1)+2\left(\frac{x^2}{4}-2\right) \cdot \frac{2 x}{4}\)
\(=2 x-2+\frac{x^3}{4}-2 x\)
\(=-2+\frac{x^3}{4}\)
\(\frac{d D}{d x}=0\)
\(=-8+x^3=0\)
\(\therefore x^3=8\)
\(x=2\)
\(\frac{d^2 D}{d x^2}=\frac{3 x^2}{4}\)
\(A t x=2\)
\(\frac{d^2 D}{d x^2}=\frac{3}{4} \times 4=3>0\)
\(\mathrm{D}\) is minimum when \(\mathrm{x}=2, \mathrm{y}=\frac{4}{4}=1\)
\(\therefore\) Required point \((2,1)\)
Assam CEE-2022
Parabola
120967
The equation of parabola whose focus is \((5,3)\) and directrix is \(3 x-4 y+1=0\) is
120962
A circle of radius 2 units lies in the first quadrant and touches both the axis. The equation of the circle with centre at \((6,5)\) and touching the circle externally is
1 \(x^2+y^2-12 x-10 y+12=0\)
2 \(x^2+y^2-12 x-10 y-20=0\)
3 \(x^2+y^2-12 x-10 y+25=0\)
4 \(x^2+y^2-12 x-10 y+52=0\)
Explanation:
D :
Let, radius of circle is \(\mathrm{r}\).
If two circle touch externally then,
\(c_1 c_2=r_1+r_2\)
\(\sqrt{(6-2)^2+(5-2)^2}=2+r\)
\(5=2+r\)
\(r=3\)
\(\therefore \text { The equation of circle } c_2\)
\((x-6)^2+(y-5)^2=3^2\)
\(x^2+y^2-12 x-10 y+52=0\)
AP EAMCET-23.04.2019
Parabola
120963
The parabola with directrix \(x+2 y-1=0\) and focus \((1,0)\) is.....
1 \(4 x^2-4 x y+y^2-8 x+4 y+4=0\)
2 \(4 x^2+4 x y+y^2-8 x+4 y+4=0\)
3 \(4 x^2+4 x y+y^2+8 x-4 y+4=0\)
4 \(4 x^2-4 x y+y^2-8 x-4 y+4=0\)
Explanation:
A Given,
\(\text { Directrix of parabola } \mathrm{x}+2 \mathrm{y}-1=0\)
\(\text { Let, a point on parabola } \mathrm{P}(\mathrm{h}, \mathrm{k})\)
\(\text { Focus }=(1,0)\)
\(\sqrt{(\mathrm{h}-1)^2+\mathrm{k}^2}=\frac{\mathrm{h}+2 \mathrm{k}-1}{\sqrt{1^2+2^2}}\)
\(\left(\sqrt{(\mathrm{h}-1)^2+\mathrm{k}^2}\right)^2=\left(\frac{\mathrm{h}+2 \mathrm{k}-1}{\sqrt{5}}\right)^2\)
\((\mathrm{~h}-1)^2+\mathrm{k}^2=\frac{(\mathrm{h}+2 \mathrm{k}-1)^2}{5}\)
\(\quad 5\left(\mathrm{~h}^2-2 \mathrm{~h}+1+\mathrm{k}^2\right)=\mathrm{h}^2+4 \mathrm{k}^2+1-2 \mathrm{~h}-4 \mathrm{k}+4 \mathrm{kh}\)
\(5 \mathrm{~h}^2-10 \mathrm{~h}+5+5 \mathrm{k}^2=\mathrm{h}^2+4 \mathrm{k}^2-2 \mathrm{~h}-4 \mathrm{k}+4 \mathrm{kh}+1\)
\(\quad \mathrm{h}^2-4 \mathrm{kh}+\mathrm{k}^2-8 \mathrm{~h}+4 \mathrm{k}+4=0\)
\(\text { Put } \mathrm{h}=\mathrm{x} \text { and } \mathrm{k}=\mathrm{y}\)
\(\text { Then, } 4 \mathrm{x}^2-4 \mathrm{xy}+\mathrm{y}^2-8 \mathrm{x}+4 \mathrm{y}+4=0\)
AP EAMCET-21.09.2020
Parabola
120965
The value of ' \(a\) ' for which the function
\(f(x)=a \sin x+\left(\frac{1}{3}\right) \sin 3 x\) has an extremum at \(x=\frac{\pi}{3}\), is
1 1
2 -1
3 2
4 0
Explanation:
C Given, \(\mathrm{f}(\mathrm{x})=\mathrm{a} \sin \mathrm{x}+\frac{1}{3} \sin 3 \mathrm{x}\)
On differentiating w.r.t.x, we get
\(\mathrm{f}^{\prime}(\mathrm{x})=\mathrm{a} \cos \mathrm{x}+\frac{3}{3} \cos 3 \mathrm{x}=0\)
\(\text { at } \mathrm{x}=\frac{\pi}{3}\)
\(\text { a } \cos \frac{\pi}{3}+\cos \left(\frac{\pi}{3} \cdot 3\right)=0\)
\(\text { a. } \frac{1}{2}+(-1)=0\)
\(\frac{\mathrm{a}}{2}=1\)
\(\mathrm{a}=2\)
Manipal UGET-2018
Parabola
120966
The point on the curve \(x^2=4 y\), which is nearest to the point \((1,2)\) is
1 \((0,0)\)
2 \((-2,1)\)
3 \((2,1)\)
4 \((2,-1)\)
Explanation:
C Given,
\(x^2=4 y\)
Using distance formula for point \((1,2)\)
Let, \((x, y)\) be the point on \(x^2=4 y\) which is nearest to the point \((+1,2)\)
\(\because(\mathrm{x}, \mathrm{y})\) lies on curve \(\mathrm{y}=\frac{\mathrm{x}^2}{4}\)
\(\therefore d=\sqrt{(x-1)^2+\left(\frac{x^2}{4}-2\right)^2}\)
\(d^2=(x-1)^2+\left(\frac{x^2}{4}-2\right)^2\)
Put \(d^2=\) D
\(\therefore D=(x-1)^2+\left(\frac{x^2}{4}-2\right)^2\)
\(\frac{d D}{d x}=2(x-1)+2\left(\frac{x^2}{4}-2\right) \cdot \frac{2 x}{4}\)
\(=2 x-2+\frac{x^3}{4}-2 x\)
\(=-2+\frac{x^3}{4}\)
\(\frac{d D}{d x}=0\)
\(=-8+x^3=0\)
\(\therefore x^3=8\)
\(x=2\)
\(\frac{d^2 D}{d x^2}=\frac{3 x^2}{4}\)
\(A t x=2\)
\(\frac{d^2 D}{d x^2}=\frac{3}{4} \times 4=3>0\)
\(\mathrm{D}\) is minimum when \(\mathrm{x}=2, \mathrm{y}=\frac{4}{4}=1\)
\(\therefore\) Required point \((2,1)\)
Assam CEE-2022
Parabola
120967
The equation of parabola whose focus is \((5,3)\) and directrix is \(3 x-4 y+1=0\) is
