120253
The line \(x=m y+c\) is normal to the parabola \(x^2=-4 a y\) if \(c=\)
1 \(-2 a m-a m^3\)
2 \(-2 \mathrm{am}+2 \mathrm{am}^3\)
3 \(2 \mathrm{am}-\mathrm{am}^3\)
4 \(2 a \mathrm{am}+\mathrm{am}^3\)
Explanation:
D Line \(\mathrm{x}=\mathrm{my}+\mathrm{c}\) is normal to the parabola \(\mathrm{x}^2=\) -4 ay
We know that equation of normal is
\(x=m y+2 a m+a m^3\)
Where \(1 / \mathrm{m}\) is the slope of normal
Comparing the line \(\mathrm{x}=\mathrm{my}+\mathrm{c}\)
Then, \(\mathrm{c}=2 \mathrm{am}+\mathrm{am}^3\)
AMU-2013
Parabola
120254
The equation of the common tangent with positive slope to the parabola \(y^2=8 \sqrt{3} x\) and the hyperbola \(4 x^2-y^2=4\) is
1 \(y=\sqrt{6} x+\sqrt{2}\)
2 \(y=\sqrt{6} x-\sqrt{2}\)
3 \(y=\sqrt{3} x+\sqrt{2}\)
4 \(y=\sqrt{3} x-\sqrt{2}\)
Explanation:
A Given,
Parabola: \(\mathrm{y}^2=8 \sqrt{3} \mathrm{x}=4(2 \sqrt{3}) \mathrm{x}\)
Hyperbola: \(4 x^2-y^2=4\)
Equation of tangent from parabola is
\(\mathrm{y}=\mathrm{mx}+\mathrm{c}\)
Where, \(\mathrm{c}=\frac{\mathrm{a}}{\mathrm{m}}=\frac{2 \sqrt{3}}{\mathrm{~m}}\)
Tangent from hyperbola is-
\(c^2=a^2 m^2-b^2\)
\(c^2=(1)^2 m^2-4\)
\(\frac{12}{m^2}=m^2-4 \text { \{From (ii)\} }\)
\(\mathrm{m}^4-4 m^2-12=0\)
\(\left(m^2+2\right)\left(m^2-6\right)=0\)
\(\therefore \mathrm{m}= \pm \sqrt{6}\) and \(\mathrm{m}= \pm \sqrt{-2}\)
\{Invalid\}
So,
\(\mathrm{m}=\sqrt{6} \text { (positive slope) }\)
Hence, required equation is
\(y=\sqrt{6} x+\sqrt{2}\)
WB JEE-2014
Parabola
120255
If \(y=4 x+3\) is parallel to a tangent to the parabola \(y^2=12 x\), then its distance from the normal parallel to the given line is
1 \(\frac{213}{\sqrt{17}}\)
2 \(\frac{219}{\sqrt{17}}\)
3 \(\frac{211}{\sqrt{17}}\)
4 \(\frac{210}{\sqrt{17}}\)
Explanation:
B Given, \(y=4 x+3\) is parallel to the tangent to parabola \(y^2=12 x\) \(y=4 x+3\)
Put \(x=0, y=3\)
If \(y=0, x=-3 / 4\)
\(y=m x+c\)
\(\mathrm{m}=4\)
\(\mathrm{y}^2=12 \mathrm{x}\)
\(2 \mathrm{y} \frac{\mathrm{dy}}{\mathrm{dx}}=12\)
\(\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{6}{\mathrm{y}}\)
Slope of normal \(=\frac{-1}{\frac{d y}{d x}}\)
\(4=\frac{-1}{\frac{6}{y}}\)
\(y=-24\)
\(y^2=12 x\)
\(24 \times 24=12 \times x\)
\(\mathrm{x}=48\).
\(\left(\mathrm{x}_2, \mathrm{y}_2\right)=(48,-24)\)
\(\mathrm{x} \cdot \mathrm{x}_1+\mathrm{y} \cdot \mathrm{y}_1-\mathrm{c}=0\)
\(\mathrm{d}=\frac{\mathrm{x}_1 \mathrm{x}_2+\mathrm{y}_1 \mathrm{y}_2-\mathrm{c}}{\sqrt{\mathrm{x}_1^2+\mathrm{x}_2^2}}\)
\(\mathrm{d}=\frac{4 \times 48-1 \times(-24)+3}{4^2+1^2}\)
\(\mathrm{d}=\frac{192+27}{\sqrt{17}}\)
\(\mathrm{d}=\frac{219}{\sqrt{17}}\)
WB JEE-2014
Parabola
120256
\(y=3 x-2\) is a straight line touching the parabola \((y-3)^2=12(x-2)\). If a line drawn perpendicular to this line at \(P\) on it, touches the given parabola, then the point \(P\) is
1 \((-1,-5)\)
2 \((-1,5)\)
3 \((-2,-8)\)
4 \((2,4)\)
Explanation:
A Given,
Line: \(y=3 x-2\)
Parabola \((\mathrm{y}-3)^2=12(\mathrm{x}-2)\)
According to question
Angle between tangents is \(90^{\circ}\)
The point on the directrix
There, \(x-2=-3\)
\(\mathrm{x}=-1 \text { is directrix }\)
Put the value of \(x\) in equation (i),
\(y=3(-1)-2\)
\(y=-5\)Hence, point \(\mathrm{P}(-1,-5)\)
NEET Test Series from KOTA - 10 Papers In MS WORD
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Parabola
120253
The line \(x=m y+c\) is normal to the parabola \(x^2=-4 a y\) if \(c=\)
1 \(-2 a m-a m^3\)
2 \(-2 \mathrm{am}+2 \mathrm{am}^3\)
3 \(2 \mathrm{am}-\mathrm{am}^3\)
4 \(2 a \mathrm{am}+\mathrm{am}^3\)
Explanation:
D Line \(\mathrm{x}=\mathrm{my}+\mathrm{c}\) is normal to the parabola \(\mathrm{x}^2=\) -4 ay
We know that equation of normal is
\(x=m y+2 a m+a m^3\)
Where \(1 / \mathrm{m}\) is the slope of normal
Comparing the line \(\mathrm{x}=\mathrm{my}+\mathrm{c}\)
Then, \(\mathrm{c}=2 \mathrm{am}+\mathrm{am}^3\)
AMU-2013
Parabola
120254
The equation of the common tangent with positive slope to the parabola \(y^2=8 \sqrt{3} x\) and the hyperbola \(4 x^2-y^2=4\) is
1 \(y=\sqrt{6} x+\sqrt{2}\)
2 \(y=\sqrt{6} x-\sqrt{2}\)
3 \(y=\sqrt{3} x+\sqrt{2}\)
4 \(y=\sqrt{3} x-\sqrt{2}\)
Explanation:
A Given,
Parabola: \(\mathrm{y}^2=8 \sqrt{3} \mathrm{x}=4(2 \sqrt{3}) \mathrm{x}\)
Hyperbola: \(4 x^2-y^2=4\)
Equation of tangent from parabola is
\(\mathrm{y}=\mathrm{mx}+\mathrm{c}\)
Where, \(\mathrm{c}=\frac{\mathrm{a}}{\mathrm{m}}=\frac{2 \sqrt{3}}{\mathrm{~m}}\)
Tangent from hyperbola is-
\(c^2=a^2 m^2-b^2\)
\(c^2=(1)^2 m^2-4\)
\(\frac{12}{m^2}=m^2-4 \text { \{From (ii)\} }\)
\(\mathrm{m}^4-4 m^2-12=0\)
\(\left(m^2+2\right)\left(m^2-6\right)=0\)
\(\therefore \mathrm{m}= \pm \sqrt{6}\) and \(\mathrm{m}= \pm \sqrt{-2}\)
\{Invalid\}
So,
\(\mathrm{m}=\sqrt{6} \text { (positive slope) }\)
Hence, required equation is
\(y=\sqrt{6} x+\sqrt{2}\)
WB JEE-2014
Parabola
120255
If \(y=4 x+3\) is parallel to a tangent to the parabola \(y^2=12 x\), then its distance from the normal parallel to the given line is
1 \(\frac{213}{\sqrt{17}}\)
2 \(\frac{219}{\sqrt{17}}\)
3 \(\frac{211}{\sqrt{17}}\)
4 \(\frac{210}{\sqrt{17}}\)
Explanation:
B Given, \(y=4 x+3\) is parallel to the tangent to parabola \(y^2=12 x\) \(y=4 x+3\)
Put \(x=0, y=3\)
If \(y=0, x=-3 / 4\)
\(y=m x+c\)
\(\mathrm{m}=4\)
\(\mathrm{y}^2=12 \mathrm{x}\)
\(2 \mathrm{y} \frac{\mathrm{dy}}{\mathrm{dx}}=12\)
\(\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{6}{\mathrm{y}}\)
Slope of normal \(=\frac{-1}{\frac{d y}{d x}}\)
\(4=\frac{-1}{\frac{6}{y}}\)
\(y=-24\)
\(y^2=12 x\)
\(24 \times 24=12 \times x\)
\(\mathrm{x}=48\).
\(\left(\mathrm{x}_2, \mathrm{y}_2\right)=(48,-24)\)
\(\mathrm{x} \cdot \mathrm{x}_1+\mathrm{y} \cdot \mathrm{y}_1-\mathrm{c}=0\)
\(\mathrm{d}=\frac{\mathrm{x}_1 \mathrm{x}_2+\mathrm{y}_1 \mathrm{y}_2-\mathrm{c}}{\sqrt{\mathrm{x}_1^2+\mathrm{x}_2^2}}\)
\(\mathrm{d}=\frac{4 \times 48-1 \times(-24)+3}{4^2+1^2}\)
\(\mathrm{d}=\frac{192+27}{\sqrt{17}}\)
\(\mathrm{d}=\frac{219}{\sqrt{17}}\)
WB JEE-2014
Parabola
120256
\(y=3 x-2\) is a straight line touching the parabola \((y-3)^2=12(x-2)\). If a line drawn perpendicular to this line at \(P\) on it, touches the given parabola, then the point \(P\) is
1 \((-1,-5)\)
2 \((-1,5)\)
3 \((-2,-8)\)
4 \((2,4)\)
Explanation:
A Given,
Line: \(y=3 x-2\)
Parabola \((\mathrm{y}-3)^2=12(\mathrm{x}-2)\)
According to question
Angle between tangents is \(90^{\circ}\)
The point on the directrix
There, \(x-2=-3\)
\(\mathrm{x}=-1 \text { is directrix }\)
Put the value of \(x\) in equation (i),
\(y=3(-1)-2\)
\(y=-5\)Hence, point \(\mathrm{P}(-1,-5)\)
120253
The line \(x=m y+c\) is normal to the parabola \(x^2=-4 a y\) if \(c=\)
1 \(-2 a m-a m^3\)
2 \(-2 \mathrm{am}+2 \mathrm{am}^3\)
3 \(2 \mathrm{am}-\mathrm{am}^3\)
4 \(2 a \mathrm{am}+\mathrm{am}^3\)
Explanation:
D Line \(\mathrm{x}=\mathrm{my}+\mathrm{c}\) is normal to the parabola \(\mathrm{x}^2=\) -4 ay
We know that equation of normal is
\(x=m y+2 a m+a m^3\)
Where \(1 / \mathrm{m}\) is the slope of normal
Comparing the line \(\mathrm{x}=\mathrm{my}+\mathrm{c}\)
Then, \(\mathrm{c}=2 \mathrm{am}+\mathrm{am}^3\)
AMU-2013
Parabola
120254
The equation of the common tangent with positive slope to the parabola \(y^2=8 \sqrt{3} x\) and the hyperbola \(4 x^2-y^2=4\) is
1 \(y=\sqrt{6} x+\sqrt{2}\)
2 \(y=\sqrt{6} x-\sqrt{2}\)
3 \(y=\sqrt{3} x+\sqrt{2}\)
4 \(y=\sqrt{3} x-\sqrt{2}\)
Explanation:
A Given,
Parabola: \(\mathrm{y}^2=8 \sqrt{3} \mathrm{x}=4(2 \sqrt{3}) \mathrm{x}\)
Hyperbola: \(4 x^2-y^2=4\)
Equation of tangent from parabola is
\(\mathrm{y}=\mathrm{mx}+\mathrm{c}\)
Where, \(\mathrm{c}=\frac{\mathrm{a}}{\mathrm{m}}=\frac{2 \sqrt{3}}{\mathrm{~m}}\)
Tangent from hyperbola is-
\(c^2=a^2 m^2-b^2\)
\(c^2=(1)^2 m^2-4\)
\(\frac{12}{m^2}=m^2-4 \text { \{From (ii)\} }\)
\(\mathrm{m}^4-4 m^2-12=0\)
\(\left(m^2+2\right)\left(m^2-6\right)=0\)
\(\therefore \mathrm{m}= \pm \sqrt{6}\) and \(\mathrm{m}= \pm \sqrt{-2}\)
\{Invalid\}
So,
\(\mathrm{m}=\sqrt{6} \text { (positive slope) }\)
Hence, required equation is
\(y=\sqrt{6} x+\sqrt{2}\)
WB JEE-2014
Parabola
120255
If \(y=4 x+3\) is parallel to a tangent to the parabola \(y^2=12 x\), then its distance from the normal parallel to the given line is
1 \(\frac{213}{\sqrt{17}}\)
2 \(\frac{219}{\sqrt{17}}\)
3 \(\frac{211}{\sqrt{17}}\)
4 \(\frac{210}{\sqrt{17}}\)
Explanation:
B Given, \(y=4 x+3\) is parallel to the tangent to parabola \(y^2=12 x\) \(y=4 x+3\)
Put \(x=0, y=3\)
If \(y=0, x=-3 / 4\)
\(y=m x+c\)
\(\mathrm{m}=4\)
\(\mathrm{y}^2=12 \mathrm{x}\)
\(2 \mathrm{y} \frac{\mathrm{dy}}{\mathrm{dx}}=12\)
\(\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{6}{\mathrm{y}}\)
Slope of normal \(=\frac{-1}{\frac{d y}{d x}}\)
\(4=\frac{-1}{\frac{6}{y}}\)
\(y=-24\)
\(y^2=12 x\)
\(24 \times 24=12 \times x\)
\(\mathrm{x}=48\).
\(\left(\mathrm{x}_2, \mathrm{y}_2\right)=(48,-24)\)
\(\mathrm{x} \cdot \mathrm{x}_1+\mathrm{y} \cdot \mathrm{y}_1-\mathrm{c}=0\)
\(\mathrm{d}=\frac{\mathrm{x}_1 \mathrm{x}_2+\mathrm{y}_1 \mathrm{y}_2-\mathrm{c}}{\sqrt{\mathrm{x}_1^2+\mathrm{x}_2^2}}\)
\(\mathrm{d}=\frac{4 \times 48-1 \times(-24)+3}{4^2+1^2}\)
\(\mathrm{d}=\frac{192+27}{\sqrt{17}}\)
\(\mathrm{d}=\frac{219}{\sqrt{17}}\)
WB JEE-2014
Parabola
120256
\(y=3 x-2\) is a straight line touching the parabola \((y-3)^2=12(x-2)\). If a line drawn perpendicular to this line at \(P\) on it, touches the given parabola, then the point \(P\) is
1 \((-1,-5)\)
2 \((-1,5)\)
3 \((-2,-8)\)
4 \((2,4)\)
Explanation:
A Given,
Line: \(y=3 x-2\)
Parabola \((\mathrm{y}-3)^2=12(\mathrm{x}-2)\)
According to question
Angle between tangents is \(90^{\circ}\)
The point on the directrix
There, \(x-2=-3\)
\(\mathrm{x}=-1 \text { is directrix }\)
Put the value of \(x\) in equation (i),
\(y=3(-1)-2\)
\(y=-5\)Hence, point \(\mathrm{P}(-1,-5)\)
120253
The line \(x=m y+c\) is normal to the parabola \(x^2=-4 a y\) if \(c=\)
1 \(-2 a m-a m^3\)
2 \(-2 \mathrm{am}+2 \mathrm{am}^3\)
3 \(2 \mathrm{am}-\mathrm{am}^3\)
4 \(2 a \mathrm{am}+\mathrm{am}^3\)
Explanation:
D Line \(\mathrm{x}=\mathrm{my}+\mathrm{c}\) is normal to the parabola \(\mathrm{x}^2=\) -4 ay
We know that equation of normal is
\(x=m y+2 a m+a m^3\)
Where \(1 / \mathrm{m}\) is the slope of normal
Comparing the line \(\mathrm{x}=\mathrm{my}+\mathrm{c}\)
Then, \(\mathrm{c}=2 \mathrm{am}+\mathrm{am}^3\)
AMU-2013
Parabola
120254
The equation of the common tangent with positive slope to the parabola \(y^2=8 \sqrt{3} x\) and the hyperbola \(4 x^2-y^2=4\) is
1 \(y=\sqrt{6} x+\sqrt{2}\)
2 \(y=\sqrt{6} x-\sqrt{2}\)
3 \(y=\sqrt{3} x+\sqrt{2}\)
4 \(y=\sqrt{3} x-\sqrt{2}\)
Explanation:
A Given,
Parabola: \(\mathrm{y}^2=8 \sqrt{3} \mathrm{x}=4(2 \sqrt{3}) \mathrm{x}\)
Hyperbola: \(4 x^2-y^2=4\)
Equation of tangent from parabola is
\(\mathrm{y}=\mathrm{mx}+\mathrm{c}\)
Where, \(\mathrm{c}=\frac{\mathrm{a}}{\mathrm{m}}=\frac{2 \sqrt{3}}{\mathrm{~m}}\)
Tangent from hyperbola is-
\(c^2=a^2 m^2-b^2\)
\(c^2=(1)^2 m^2-4\)
\(\frac{12}{m^2}=m^2-4 \text { \{From (ii)\} }\)
\(\mathrm{m}^4-4 m^2-12=0\)
\(\left(m^2+2\right)\left(m^2-6\right)=0\)
\(\therefore \mathrm{m}= \pm \sqrt{6}\) and \(\mathrm{m}= \pm \sqrt{-2}\)
\{Invalid\}
So,
\(\mathrm{m}=\sqrt{6} \text { (positive slope) }\)
Hence, required equation is
\(y=\sqrt{6} x+\sqrt{2}\)
WB JEE-2014
Parabola
120255
If \(y=4 x+3\) is parallel to a tangent to the parabola \(y^2=12 x\), then its distance from the normal parallel to the given line is
1 \(\frac{213}{\sqrt{17}}\)
2 \(\frac{219}{\sqrt{17}}\)
3 \(\frac{211}{\sqrt{17}}\)
4 \(\frac{210}{\sqrt{17}}\)
Explanation:
B Given, \(y=4 x+3\) is parallel to the tangent to parabola \(y^2=12 x\) \(y=4 x+3\)
Put \(x=0, y=3\)
If \(y=0, x=-3 / 4\)
\(y=m x+c\)
\(\mathrm{m}=4\)
\(\mathrm{y}^2=12 \mathrm{x}\)
\(2 \mathrm{y} \frac{\mathrm{dy}}{\mathrm{dx}}=12\)
\(\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{6}{\mathrm{y}}\)
Slope of normal \(=\frac{-1}{\frac{d y}{d x}}\)
\(4=\frac{-1}{\frac{6}{y}}\)
\(y=-24\)
\(y^2=12 x\)
\(24 \times 24=12 \times x\)
\(\mathrm{x}=48\).
\(\left(\mathrm{x}_2, \mathrm{y}_2\right)=(48,-24)\)
\(\mathrm{x} \cdot \mathrm{x}_1+\mathrm{y} \cdot \mathrm{y}_1-\mathrm{c}=0\)
\(\mathrm{d}=\frac{\mathrm{x}_1 \mathrm{x}_2+\mathrm{y}_1 \mathrm{y}_2-\mathrm{c}}{\sqrt{\mathrm{x}_1^2+\mathrm{x}_2^2}}\)
\(\mathrm{d}=\frac{4 \times 48-1 \times(-24)+3}{4^2+1^2}\)
\(\mathrm{d}=\frac{192+27}{\sqrt{17}}\)
\(\mathrm{d}=\frac{219}{\sqrt{17}}\)
WB JEE-2014
Parabola
120256
\(y=3 x-2\) is a straight line touching the parabola \((y-3)^2=12(x-2)\). If a line drawn perpendicular to this line at \(P\) on it, touches the given parabola, then the point \(P\) is
1 \((-1,-5)\)
2 \((-1,5)\)
3 \((-2,-8)\)
4 \((2,4)\)
Explanation:
A Given,
Line: \(y=3 x-2\)
Parabola \((\mathrm{y}-3)^2=12(\mathrm{x}-2)\)
According to question
Angle between tangents is \(90^{\circ}\)
The point on the directrix
There, \(x-2=-3\)
\(\mathrm{x}=-1 \text { is directrix }\)
Put the value of \(x\) in equation (i),
\(y=3(-1)-2\)
\(y=-5\)Hence, point \(\mathrm{P}(-1,-5)\)