120280
If a line along a chord of the circle \(4 x^2+4 y^2+\) \(120 x+675=0\), passes through the point \((-30\), \(0)\) and is tangent to the parabola \(y^2=30 x\), then the length of this chord is
1 5
2 7
3 \(5 \sqrt{3}\)
4 \(3 \sqrt{5}\)
Explanation:
D Given,
Tangent to the parabola, \(\mathrm{y}^2=30 \mathrm{x}\)
\(\mathrm{y}=\mathrm{mx}+\frac{30}{4 \mathrm{~m}}\)
Passes through the point \((-30,0)\)
\(\therefore \quad 0=-30 \mathrm{~m}+\frac{30}{4 \mathrm{~m}}\)
\(4 \mathrm{~m}^2 =1\)
\(\mathrm{~m} = \pm \frac{1}{2}\)
For \(\quad \mathrm{m}=\frac{1}{2}\)
\(\mathrm{y}=\frac{\mathrm{x}}{2}+15\)
\(\mathrm{x}-2 \mathrm{y}+30=0\)
Length of \(\mathrm{OB}=\left|\frac{-15+0+30}{\sqrt{5}}\right|\)
\(\mathrm{OB}=3 \sqrt{5}\)
Radius of circle \(=\frac{15}{2}\)
Length of \(B C=\sqrt{\frac{225}{4}-45}\)
\(\mathrm{BC}=\frac{3 \sqrt{5}}{2}\)
Length of chord \(A C=2 \times \frac{3 \sqrt{5}}{2}\)
\(\mathrm{AC}=3 \sqrt{5}\)
JEE Main 26.09.2021
Parabola
120281
For any non-zero real value of \(m\), the equation of the parabola to which the line \(\mathbf{m x}-\mathbf{y}+\mathbf{1 0}+\) \(\mathrm{m}^2=0\) is a tangent, is
1 \(x^2=y-10\)
2 \(y^2=4(x-2)\)
3 \(x^2=-4(y-10)\)
4 \(x^2=-4 y\)
Explanation:
C Given, equation of tangent to the parabola
\(\mathrm{mx}-\mathrm{y}+\left(10+\mathrm{m}^2\right)=0\)
\(\mathrm{~m}^2+\mathrm{mx}+(10-\mathrm{y})=0\)
Since, above line is tangent, then
\(\mathrm{D}=0\)
Now,
\(b^2-4 a c=0\)
\(x^2-4 \times 1 \times(10-y)=0\)
\(x^2=4(10-y)\)
\(x^2=-4(y-10)\)
AP EAMCET-23.04.2019
Parabola
120282
The normal at point \((1,1)\) to the curve \(y^2=x^3\) is parallel to the line
1 \(3 x-y-2=0\)
2 \(2 \mathrm{x}+3 \mathrm{y}-7=0\)
3 \(2 \mathrm{x}-3 \mathrm{y}+1=0\)
4 \(2 \mathrm{y}-3 \mathrm{x}+1=0\)
Explanation:
B Given,
\(y^2=x^3\)
Differentiating with respect to \(\mathrm{x}\),
\(2 \mathrm{y} \frac{\mathrm{dy}}{\mathrm{dx}}=3 \mathrm{x}^2\)
The normal of point \((1,1)\),
\(2(1) \frac{d y}{d x}=3(1)^2\)
\(2 \cdot \frac{d y}{d x}=3\)
\(\frac{d y}{d x}=m_t=\frac{3}{2}\)
So, \(\mathrm{m}_{\mathrm{t}} \cdot \mathrm{m}_{\mathrm{n}}=-1\)
Where, \(\mathrm{m}_{\mathrm{t}}, \mathrm{m}_{\mathrm{n}}\) are slopes of tangent and normal respectively
\(\mathrm{m}_{\mathrm{n}}=\frac{-2}{3}\)
The slope of the line \(2 x+3 y-7=0\) is given as
\(\mathrm{m}=\frac{-2}{3}\)So, as the slopes are equal the line \(2 x+3 y-7=0\) is parallel to the normal
Rajasthan PET-2009
Parabola
120283
The equation of the normal to the curve \(y^2=4\) ax at point ( \(\left.\mathrm{at}^2, 2 \mathrm{at}\right)\) is
1 \(y-t x=a^3+2 a t\)
2 \(y+t x=a t^3+2 a t\)
3 \(y-t x=a t^3-2 a t\)
4 None of these
Explanation:
B The condition for \(\mathrm{y}=\mathrm{mx}+\mathrm{c}\) to be the tangent to \(\mathrm{y}^2=4 \mathrm{ax}\) is \(\mathrm{c}=\frac{\mathrm{a}}{\mathrm{m}}\)
\(\text { So, } y= \left.m x+\frac{a}{m} \text { passes through (at } t^2, 2 a t\right)\)
\(2 a t=m \cdot a t^2+\frac{a}{m}\)
\(m^2 t^2-2 m t+1=0\)
\(m=\frac{1}{t}\)
\(\therefore\) Slope of tangent \(=\frac{1}{\mathrm{t}}\)
\(\text { Also slope of normal }=-\frac{1}{\text { slope of tangent }}=-\frac{1}{\frac{1}{t}}\)
\(=-t\)
\(\therefore\) Equation of normal is
\(y-y_1=\text { slope of normal }\left(x-x_1\right)\)
\(y-2 a t=-t\left(x-a t^2\right)\)
\(y-2 a t=-t x+a t^3\)
\(y+t x=a t^3+2 a t\)
Rajasthan PET-2004
Parabola
120284
Let \(x+y=k\) be a normal to the parabola \(y^2=\) \(12 x\). If \(p\) is length of the perpendicular from the focus of the parabola onto this normal, then \(4 \mathrm{k}\) \(-2 p^2\) is equal to
1 1
2 0
3 -1
4 2
Explanation:
B Given,
\(y^2=12 x\)
Here, \(\quad \mathrm{a}=3\)
\(\therefore\) Equation of normal
\(\mathrm{y}=\mathrm{mx}-2 \mathrm{am}-\mathrm{am}^3 \quad(\because \mathrm{m}=-1)\)
\(y=-x-2(3)(-1)-3(-1)^3\)
\(\mathrm{y}=-\mathrm{x}+6+3\)
\(\mathrm{x}+\mathrm{y}=9\)
But given normal,
\(\mathrm{x}+\mathrm{y}=\mathrm{k}\)
\(\therefore \quad \mathrm{k} =9\)
Focus of a given parabola is \((3,0)\)
Now, perpendicular distance from \((3,0)\) to the equation
(i) is
\(\mathrm{p} =\frac{|3(1)+0-9|}{\sqrt{1^2+1^2}}\)
\(\mathrm{p} =\frac{|3-9|}{\sqrt{2}}\)
\(\mathrm{p} =\frac{6}{\sqrt{2}}\)
\(\therefore 4 \mathrm{k}-2 \mathrm{p}^2 =4(9)-2\left(\frac{6}{\sqrt{2}}\right)^2=36-2 \times \frac{36}{2}\)
\(=36-36=0\)
120280
If a line along a chord of the circle \(4 x^2+4 y^2+\) \(120 x+675=0\), passes through the point \((-30\), \(0)\) and is tangent to the parabola \(y^2=30 x\), then the length of this chord is
1 5
2 7
3 \(5 \sqrt{3}\)
4 \(3 \sqrt{5}\)
Explanation:
D Given,
Tangent to the parabola, \(\mathrm{y}^2=30 \mathrm{x}\)
\(\mathrm{y}=\mathrm{mx}+\frac{30}{4 \mathrm{~m}}\)
Passes through the point \((-30,0)\)
\(\therefore \quad 0=-30 \mathrm{~m}+\frac{30}{4 \mathrm{~m}}\)
\(4 \mathrm{~m}^2 =1\)
\(\mathrm{~m} = \pm \frac{1}{2}\)
For \(\quad \mathrm{m}=\frac{1}{2}\)
\(\mathrm{y}=\frac{\mathrm{x}}{2}+15\)
\(\mathrm{x}-2 \mathrm{y}+30=0\)
Length of \(\mathrm{OB}=\left|\frac{-15+0+30}{\sqrt{5}}\right|\)
\(\mathrm{OB}=3 \sqrt{5}\)
Radius of circle \(=\frac{15}{2}\)
Length of \(B C=\sqrt{\frac{225}{4}-45}\)
\(\mathrm{BC}=\frac{3 \sqrt{5}}{2}\)
Length of chord \(A C=2 \times \frac{3 \sqrt{5}}{2}\)
\(\mathrm{AC}=3 \sqrt{5}\)
JEE Main 26.09.2021
Parabola
120281
For any non-zero real value of \(m\), the equation of the parabola to which the line \(\mathbf{m x}-\mathbf{y}+\mathbf{1 0}+\) \(\mathrm{m}^2=0\) is a tangent, is
1 \(x^2=y-10\)
2 \(y^2=4(x-2)\)
3 \(x^2=-4(y-10)\)
4 \(x^2=-4 y\)
Explanation:
C Given, equation of tangent to the parabola
\(\mathrm{mx}-\mathrm{y}+\left(10+\mathrm{m}^2\right)=0\)
\(\mathrm{~m}^2+\mathrm{mx}+(10-\mathrm{y})=0\)
Since, above line is tangent, then
\(\mathrm{D}=0\)
Now,
\(b^2-4 a c=0\)
\(x^2-4 \times 1 \times(10-y)=0\)
\(x^2=4(10-y)\)
\(x^2=-4(y-10)\)
AP EAMCET-23.04.2019
Parabola
120282
The normal at point \((1,1)\) to the curve \(y^2=x^3\) is parallel to the line
1 \(3 x-y-2=0\)
2 \(2 \mathrm{x}+3 \mathrm{y}-7=0\)
3 \(2 \mathrm{x}-3 \mathrm{y}+1=0\)
4 \(2 \mathrm{y}-3 \mathrm{x}+1=0\)
Explanation:
B Given,
\(y^2=x^3\)
Differentiating with respect to \(\mathrm{x}\),
\(2 \mathrm{y} \frac{\mathrm{dy}}{\mathrm{dx}}=3 \mathrm{x}^2\)
The normal of point \((1,1)\),
\(2(1) \frac{d y}{d x}=3(1)^2\)
\(2 \cdot \frac{d y}{d x}=3\)
\(\frac{d y}{d x}=m_t=\frac{3}{2}\)
So, \(\mathrm{m}_{\mathrm{t}} \cdot \mathrm{m}_{\mathrm{n}}=-1\)
Where, \(\mathrm{m}_{\mathrm{t}}, \mathrm{m}_{\mathrm{n}}\) are slopes of tangent and normal respectively
\(\mathrm{m}_{\mathrm{n}}=\frac{-2}{3}\)
The slope of the line \(2 x+3 y-7=0\) is given as
\(\mathrm{m}=\frac{-2}{3}\)So, as the slopes are equal the line \(2 x+3 y-7=0\) is parallel to the normal
Rajasthan PET-2009
Parabola
120283
The equation of the normal to the curve \(y^2=4\) ax at point ( \(\left.\mathrm{at}^2, 2 \mathrm{at}\right)\) is
1 \(y-t x=a^3+2 a t\)
2 \(y+t x=a t^3+2 a t\)
3 \(y-t x=a t^3-2 a t\)
4 None of these
Explanation:
B The condition for \(\mathrm{y}=\mathrm{mx}+\mathrm{c}\) to be the tangent to \(\mathrm{y}^2=4 \mathrm{ax}\) is \(\mathrm{c}=\frac{\mathrm{a}}{\mathrm{m}}\)
\(\text { So, } y= \left.m x+\frac{a}{m} \text { passes through (at } t^2, 2 a t\right)\)
\(2 a t=m \cdot a t^2+\frac{a}{m}\)
\(m^2 t^2-2 m t+1=0\)
\(m=\frac{1}{t}\)
\(\therefore\) Slope of tangent \(=\frac{1}{\mathrm{t}}\)
\(\text { Also slope of normal }=-\frac{1}{\text { slope of tangent }}=-\frac{1}{\frac{1}{t}}\)
\(=-t\)
\(\therefore\) Equation of normal is
\(y-y_1=\text { slope of normal }\left(x-x_1\right)\)
\(y-2 a t=-t\left(x-a t^2\right)\)
\(y-2 a t=-t x+a t^3\)
\(y+t x=a t^3+2 a t\)
Rajasthan PET-2004
Parabola
120284
Let \(x+y=k\) be a normal to the parabola \(y^2=\) \(12 x\). If \(p\) is length of the perpendicular from the focus of the parabola onto this normal, then \(4 \mathrm{k}\) \(-2 p^2\) is equal to
1 1
2 0
3 -1
4 2
Explanation:
B Given,
\(y^2=12 x\)
Here, \(\quad \mathrm{a}=3\)
\(\therefore\) Equation of normal
\(\mathrm{y}=\mathrm{mx}-2 \mathrm{am}-\mathrm{am}^3 \quad(\because \mathrm{m}=-1)\)
\(y=-x-2(3)(-1)-3(-1)^3\)
\(\mathrm{y}=-\mathrm{x}+6+3\)
\(\mathrm{x}+\mathrm{y}=9\)
But given normal,
\(\mathrm{x}+\mathrm{y}=\mathrm{k}\)
\(\therefore \quad \mathrm{k} =9\)
Focus of a given parabola is \((3,0)\)
Now, perpendicular distance from \((3,0)\) to the equation
(i) is
\(\mathrm{p} =\frac{|3(1)+0-9|}{\sqrt{1^2+1^2}}\)
\(\mathrm{p} =\frac{|3-9|}{\sqrt{2}}\)
\(\mathrm{p} =\frac{6}{\sqrt{2}}\)
\(\therefore 4 \mathrm{k}-2 \mathrm{p}^2 =4(9)-2\left(\frac{6}{\sqrt{2}}\right)^2=36-2 \times \frac{36}{2}\)
\(=36-36=0\)
120280
If a line along a chord of the circle \(4 x^2+4 y^2+\) \(120 x+675=0\), passes through the point \((-30\), \(0)\) and is tangent to the parabola \(y^2=30 x\), then the length of this chord is
1 5
2 7
3 \(5 \sqrt{3}\)
4 \(3 \sqrt{5}\)
Explanation:
D Given,
Tangent to the parabola, \(\mathrm{y}^2=30 \mathrm{x}\)
\(\mathrm{y}=\mathrm{mx}+\frac{30}{4 \mathrm{~m}}\)
Passes through the point \((-30,0)\)
\(\therefore \quad 0=-30 \mathrm{~m}+\frac{30}{4 \mathrm{~m}}\)
\(4 \mathrm{~m}^2 =1\)
\(\mathrm{~m} = \pm \frac{1}{2}\)
For \(\quad \mathrm{m}=\frac{1}{2}\)
\(\mathrm{y}=\frac{\mathrm{x}}{2}+15\)
\(\mathrm{x}-2 \mathrm{y}+30=0\)
Length of \(\mathrm{OB}=\left|\frac{-15+0+30}{\sqrt{5}}\right|\)
\(\mathrm{OB}=3 \sqrt{5}\)
Radius of circle \(=\frac{15}{2}\)
Length of \(B C=\sqrt{\frac{225}{4}-45}\)
\(\mathrm{BC}=\frac{3 \sqrt{5}}{2}\)
Length of chord \(A C=2 \times \frac{3 \sqrt{5}}{2}\)
\(\mathrm{AC}=3 \sqrt{5}\)
JEE Main 26.09.2021
Parabola
120281
For any non-zero real value of \(m\), the equation of the parabola to which the line \(\mathbf{m x}-\mathbf{y}+\mathbf{1 0}+\) \(\mathrm{m}^2=0\) is a tangent, is
1 \(x^2=y-10\)
2 \(y^2=4(x-2)\)
3 \(x^2=-4(y-10)\)
4 \(x^2=-4 y\)
Explanation:
C Given, equation of tangent to the parabola
\(\mathrm{mx}-\mathrm{y}+\left(10+\mathrm{m}^2\right)=0\)
\(\mathrm{~m}^2+\mathrm{mx}+(10-\mathrm{y})=0\)
Since, above line is tangent, then
\(\mathrm{D}=0\)
Now,
\(b^2-4 a c=0\)
\(x^2-4 \times 1 \times(10-y)=0\)
\(x^2=4(10-y)\)
\(x^2=-4(y-10)\)
AP EAMCET-23.04.2019
Parabola
120282
The normal at point \((1,1)\) to the curve \(y^2=x^3\) is parallel to the line
1 \(3 x-y-2=0\)
2 \(2 \mathrm{x}+3 \mathrm{y}-7=0\)
3 \(2 \mathrm{x}-3 \mathrm{y}+1=0\)
4 \(2 \mathrm{y}-3 \mathrm{x}+1=0\)
Explanation:
B Given,
\(y^2=x^3\)
Differentiating with respect to \(\mathrm{x}\),
\(2 \mathrm{y} \frac{\mathrm{dy}}{\mathrm{dx}}=3 \mathrm{x}^2\)
The normal of point \((1,1)\),
\(2(1) \frac{d y}{d x}=3(1)^2\)
\(2 \cdot \frac{d y}{d x}=3\)
\(\frac{d y}{d x}=m_t=\frac{3}{2}\)
So, \(\mathrm{m}_{\mathrm{t}} \cdot \mathrm{m}_{\mathrm{n}}=-1\)
Where, \(\mathrm{m}_{\mathrm{t}}, \mathrm{m}_{\mathrm{n}}\) are slopes of tangent and normal respectively
\(\mathrm{m}_{\mathrm{n}}=\frac{-2}{3}\)
The slope of the line \(2 x+3 y-7=0\) is given as
\(\mathrm{m}=\frac{-2}{3}\)So, as the slopes are equal the line \(2 x+3 y-7=0\) is parallel to the normal
Rajasthan PET-2009
Parabola
120283
The equation of the normal to the curve \(y^2=4\) ax at point ( \(\left.\mathrm{at}^2, 2 \mathrm{at}\right)\) is
1 \(y-t x=a^3+2 a t\)
2 \(y+t x=a t^3+2 a t\)
3 \(y-t x=a t^3-2 a t\)
4 None of these
Explanation:
B The condition for \(\mathrm{y}=\mathrm{mx}+\mathrm{c}\) to be the tangent to \(\mathrm{y}^2=4 \mathrm{ax}\) is \(\mathrm{c}=\frac{\mathrm{a}}{\mathrm{m}}\)
\(\text { So, } y= \left.m x+\frac{a}{m} \text { passes through (at } t^2, 2 a t\right)\)
\(2 a t=m \cdot a t^2+\frac{a}{m}\)
\(m^2 t^2-2 m t+1=0\)
\(m=\frac{1}{t}\)
\(\therefore\) Slope of tangent \(=\frac{1}{\mathrm{t}}\)
\(\text { Also slope of normal }=-\frac{1}{\text { slope of tangent }}=-\frac{1}{\frac{1}{t}}\)
\(=-t\)
\(\therefore\) Equation of normal is
\(y-y_1=\text { slope of normal }\left(x-x_1\right)\)
\(y-2 a t=-t\left(x-a t^2\right)\)
\(y-2 a t=-t x+a t^3\)
\(y+t x=a t^3+2 a t\)
Rajasthan PET-2004
Parabola
120284
Let \(x+y=k\) be a normal to the parabola \(y^2=\) \(12 x\). If \(p\) is length of the perpendicular from the focus of the parabola onto this normal, then \(4 \mathrm{k}\) \(-2 p^2\) is equal to
1 1
2 0
3 -1
4 2
Explanation:
B Given,
\(y^2=12 x\)
Here, \(\quad \mathrm{a}=3\)
\(\therefore\) Equation of normal
\(\mathrm{y}=\mathrm{mx}-2 \mathrm{am}-\mathrm{am}^3 \quad(\because \mathrm{m}=-1)\)
\(y=-x-2(3)(-1)-3(-1)^3\)
\(\mathrm{y}=-\mathrm{x}+6+3\)
\(\mathrm{x}+\mathrm{y}=9\)
But given normal,
\(\mathrm{x}+\mathrm{y}=\mathrm{k}\)
\(\therefore \quad \mathrm{k} =9\)
Focus of a given parabola is \((3,0)\)
Now, perpendicular distance from \((3,0)\) to the equation
(i) is
\(\mathrm{p} =\frac{|3(1)+0-9|}{\sqrt{1^2+1^2}}\)
\(\mathrm{p} =\frac{|3-9|}{\sqrt{2}}\)
\(\mathrm{p} =\frac{6}{\sqrt{2}}\)
\(\therefore 4 \mathrm{k}-2 \mathrm{p}^2 =4(9)-2\left(\frac{6}{\sqrt{2}}\right)^2=36-2 \times \frac{36}{2}\)
\(=36-36=0\)
NEET Test Series from KOTA - 10 Papers In MS WORD
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Parabola
120280
If a line along a chord of the circle \(4 x^2+4 y^2+\) \(120 x+675=0\), passes through the point \((-30\), \(0)\) and is tangent to the parabola \(y^2=30 x\), then the length of this chord is
1 5
2 7
3 \(5 \sqrt{3}\)
4 \(3 \sqrt{5}\)
Explanation:
D Given,
Tangent to the parabola, \(\mathrm{y}^2=30 \mathrm{x}\)
\(\mathrm{y}=\mathrm{mx}+\frac{30}{4 \mathrm{~m}}\)
Passes through the point \((-30,0)\)
\(\therefore \quad 0=-30 \mathrm{~m}+\frac{30}{4 \mathrm{~m}}\)
\(4 \mathrm{~m}^2 =1\)
\(\mathrm{~m} = \pm \frac{1}{2}\)
For \(\quad \mathrm{m}=\frac{1}{2}\)
\(\mathrm{y}=\frac{\mathrm{x}}{2}+15\)
\(\mathrm{x}-2 \mathrm{y}+30=0\)
Length of \(\mathrm{OB}=\left|\frac{-15+0+30}{\sqrt{5}}\right|\)
\(\mathrm{OB}=3 \sqrt{5}\)
Radius of circle \(=\frac{15}{2}\)
Length of \(B C=\sqrt{\frac{225}{4}-45}\)
\(\mathrm{BC}=\frac{3 \sqrt{5}}{2}\)
Length of chord \(A C=2 \times \frac{3 \sqrt{5}}{2}\)
\(\mathrm{AC}=3 \sqrt{5}\)
JEE Main 26.09.2021
Parabola
120281
For any non-zero real value of \(m\), the equation of the parabola to which the line \(\mathbf{m x}-\mathbf{y}+\mathbf{1 0}+\) \(\mathrm{m}^2=0\) is a tangent, is
1 \(x^2=y-10\)
2 \(y^2=4(x-2)\)
3 \(x^2=-4(y-10)\)
4 \(x^2=-4 y\)
Explanation:
C Given, equation of tangent to the parabola
\(\mathrm{mx}-\mathrm{y}+\left(10+\mathrm{m}^2\right)=0\)
\(\mathrm{~m}^2+\mathrm{mx}+(10-\mathrm{y})=0\)
Since, above line is tangent, then
\(\mathrm{D}=0\)
Now,
\(b^2-4 a c=0\)
\(x^2-4 \times 1 \times(10-y)=0\)
\(x^2=4(10-y)\)
\(x^2=-4(y-10)\)
AP EAMCET-23.04.2019
Parabola
120282
The normal at point \((1,1)\) to the curve \(y^2=x^3\) is parallel to the line
1 \(3 x-y-2=0\)
2 \(2 \mathrm{x}+3 \mathrm{y}-7=0\)
3 \(2 \mathrm{x}-3 \mathrm{y}+1=0\)
4 \(2 \mathrm{y}-3 \mathrm{x}+1=0\)
Explanation:
B Given,
\(y^2=x^3\)
Differentiating with respect to \(\mathrm{x}\),
\(2 \mathrm{y} \frac{\mathrm{dy}}{\mathrm{dx}}=3 \mathrm{x}^2\)
The normal of point \((1,1)\),
\(2(1) \frac{d y}{d x}=3(1)^2\)
\(2 \cdot \frac{d y}{d x}=3\)
\(\frac{d y}{d x}=m_t=\frac{3}{2}\)
So, \(\mathrm{m}_{\mathrm{t}} \cdot \mathrm{m}_{\mathrm{n}}=-1\)
Where, \(\mathrm{m}_{\mathrm{t}}, \mathrm{m}_{\mathrm{n}}\) are slopes of tangent and normal respectively
\(\mathrm{m}_{\mathrm{n}}=\frac{-2}{3}\)
The slope of the line \(2 x+3 y-7=0\) is given as
\(\mathrm{m}=\frac{-2}{3}\)So, as the slopes are equal the line \(2 x+3 y-7=0\) is parallel to the normal
Rajasthan PET-2009
Parabola
120283
The equation of the normal to the curve \(y^2=4\) ax at point ( \(\left.\mathrm{at}^2, 2 \mathrm{at}\right)\) is
1 \(y-t x=a^3+2 a t\)
2 \(y+t x=a t^3+2 a t\)
3 \(y-t x=a t^3-2 a t\)
4 None of these
Explanation:
B The condition for \(\mathrm{y}=\mathrm{mx}+\mathrm{c}\) to be the tangent to \(\mathrm{y}^2=4 \mathrm{ax}\) is \(\mathrm{c}=\frac{\mathrm{a}}{\mathrm{m}}\)
\(\text { So, } y= \left.m x+\frac{a}{m} \text { passes through (at } t^2, 2 a t\right)\)
\(2 a t=m \cdot a t^2+\frac{a}{m}\)
\(m^2 t^2-2 m t+1=0\)
\(m=\frac{1}{t}\)
\(\therefore\) Slope of tangent \(=\frac{1}{\mathrm{t}}\)
\(\text { Also slope of normal }=-\frac{1}{\text { slope of tangent }}=-\frac{1}{\frac{1}{t}}\)
\(=-t\)
\(\therefore\) Equation of normal is
\(y-y_1=\text { slope of normal }\left(x-x_1\right)\)
\(y-2 a t=-t\left(x-a t^2\right)\)
\(y-2 a t=-t x+a t^3\)
\(y+t x=a t^3+2 a t\)
Rajasthan PET-2004
Parabola
120284
Let \(x+y=k\) be a normal to the parabola \(y^2=\) \(12 x\). If \(p\) is length of the perpendicular from the focus of the parabola onto this normal, then \(4 \mathrm{k}\) \(-2 p^2\) is equal to
1 1
2 0
3 -1
4 2
Explanation:
B Given,
\(y^2=12 x\)
Here, \(\quad \mathrm{a}=3\)
\(\therefore\) Equation of normal
\(\mathrm{y}=\mathrm{mx}-2 \mathrm{am}-\mathrm{am}^3 \quad(\because \mathrm{m}=-1)\)
\(y=-x-2(3)(-1)-3(-1)^3\)
\(\mathrm{y}=-\mathrm{x}+6+3\)
\(\mathrm{x}+\mathrm{y}=9\)
But given normal,
\(\mathrm{x}+\mathrm{y}=\mathrm{k}\)
\(\therefore \quad \mathrm{k} =9\)
Focus of a given parabola is \((3,0)\)
Now, perpendicular distance from \((3,0)\) to the equation
(i) is
\(\mathrm{p} =\frac{|3(1)+0-9|}{\sqrt{1^2+1^2}}\)
\(\mathrm{p} =\frac{|3-9|}{\sqrt{2}}\)
\(\mathrm{p} =\frac{6}{\sqrt{2}}\)
\(\therefore 4 \mathrm{k}-2 \mathrm{p}^2 =4(9)-2\left(\frac{6}{\sqrt{2}}\right)^2=36-2 \times \frac{36}{2}\)
\(=36-36=0\)
120280
If a line along a chord of the circle \(4 x^2+4 y^2+\) \(120 x+675=0\), passes through the point \((-30\), \(0)\) and is tangent to the parabola \(y^2=30 x\), then the length of this chord is
1 5
2 7
3 \(5 \sqrt{3}\)
4 \(3 \sqrt{5}\)
Explanation:
D Given,
Tangent to the parabola, \(\mathrm{y}^2=30 \mathrm{x}\)
\(\mathrm{y}=\mathrm{mx}+\frac{30}{4 \mathrm{~m}}\)
Passes through the point \((-30,0)\)
\(\therefore \quad 0=-30 \mathrm{~m}+\frac{30}{4 \mathrm{~m}}\)
\(4 \mathrm{~m}^2 =1\)
\(\mathrm{~m} = \pm \frac{1}{2}\)
For \(\quad \mathrm{m}=\frac{1}{2}\)
\(\mathrm{y}=\frac{\mathrm{x}}{2}+15\)
\(\mathrm{x}-2 \mathrm{y}+30=0\)
Length of \(\mathrm{OB}=\left|\frac{-15+0+30}{\sqrt{5}}\right|\)
\(\mathrm{OB}=3 \sqrt{5}\)
Radius of circle \(=\frac{15}{2}\)
Length of \(B C=\sqrt{\frac{225}{4}-45}\)
\(\mathrm{BC}=\frac{3 \sqrt{5}}{2}\)
Length of chord \(A C=2 \times \frac{3 \sqrt{5}}{2}\)
\(\mathrm{AC}=3 \sqrt{5}\)
JEE Main 26.09.2021
Parabola
120281
For any non-zero real value of \(m\), the equation of the parabola to which the line \(\mathbf{m x}-\mathbf{y}+\mathbf{1 0}+\) \(\mathrm{m}^2=0\) is a tangent, is
1 \(x^2=y-10\)
2 \(y^2=4(x-2)\)
3 \(x^2=-4(y-10)\)
4 \(x^2=-4 y\)
Explanation:
C Given, equation of tangent to the parabola
\(\mathrm{mx}-\mathrm{y}+\left(10+\mathrm{m}^2\right)=0\)
\(\mathrm{~m}^2+\mathrm{mx}+(10-\mathrm{y})=0\)
Since, above line is tangent, then
\(\mathrm{D}=0\)
Now,
\(b^2-4 a c=0\)
\(x^2-4 \times 1 \times(10-y)=0\)
\(x^2=4(10-y)\)
\(x^2=-4(y-10)\)
AP EAMCET-23.04.2019
Parabola
120282
The normal at point \((1,1)\) to the curve \(y^2=x^3\) is parallel to the line
1 \(3 x-y-2=0\)
2 \(2 \mathrm{x}+3 \mathrm{y}-7=0\)
3 \(2 \mathrm{x}-3 \mathrm{y}+1=0\)
4 \(2 \mathrm{y}-3 \mathrm{x}+1=0\)
Explanation:
B Given,
\(y^2=x^3\)
Differentiating with respect to \(\mathrm{x}\),
\(2 \mathrm{y} \frac{\mathrm{dy}}{\mathrm{dx}}=3 \mathrm{x}^2\)
The normal of point \((1,1)\),
\(2(1) \frac{d y}{d x}=3(1)^2\)
\(2 \cdot \frac{d y}{d x}=3\)
\(\frac{d y}{d x}=m_t=\frac{3}{2}\)
So, \(\mathrm{m}_{\mathrm{t}} \cdot \mathrm{m}_{\mathrm{n}}=-1\)
Where, \(\mathrm{m}_{\mathrm{t}}, \mathrm{m}_{\mathrm{n}}\) are slopes of tangent and normal respectively
\(\mathrm{m}_{\mathrm{n}}=\frac{-2}{3}\)
The slope of the line \(2 x+3 y-7=0\) is given as
\(\mathrm{m}=\frac{-2}{3}\)So, as the slopes are equal the line \(2 x+3 y-7=0\) is parallel to the normal
Rajasthan PET-2009
Parabola
120283
The equation of the normal to the curve \(y^2=4\) ax at point ( \(\left.\mathrm{at}^2, 2 \mathrm{at}\right)\) is
1 \(y-t x=a^3+2 a t\)
2 \(y+t x=a t^3+2 a t\)
3 \(y-t x=a t^3-2 a t\)
4 None of these
Explanation:
B The condition for \(\mathrm{y}=\mathrm{mx}+\mathrm{c}\) to be the tangent to \(\mathrm{y}^2=4 \mathrm{ax}\) is \(\mathrm{c}=\frac{\mathrm{a}}{\mathrm{m}}\)
\(\text { So, } y= \left.m x+\frac{a}{m} \text { passes through (at } t^2, 2 a t\right)\)
\(2 a t=m \cdot a t^2+\frac{a}{m}\)
\(m^2 t^2-2 m t+1=0\)
\(m=\frac{1}{t}\)
\(\therefore\) Slope of tangent \(=\frac{1}{\mathrm{t}}\)
\(\text { Also slope of normal }=-\frac{1}{\text { slope of tangent }}=-\frac{1}{\frac{1}{t}}\)
\(=-t\)
\(\therefore\) Equation of normal is
\(y-y_1=\text { slope of normal }\left(x-x_1\right)\)
\(y-2 a t=-t\left(x-a t^2\right)\)
\(y-2 a t=-t x+a t^3\)
\(y+t x=a t^3+2 a t\)
Rajasthan PET-2004
Parabola
120284
Let \(x+y=k\) be a normal to the parabola \(y^2=\) \(12 x\). If \(p\) is length of the perpendicular from the focus of the parabola onto this normal, then \(4 \mathrm{k}\) \(-2 p^2\) is equal to
1 1
2 0
3 -1
4 2
Explanation:
B Given,
\(y^2=12 x\)
Here, \(\quad \mathrm{a}=3\)
\(\therefore\) Equation of normal
\(\mathrm{y}=\mathrm{mx}-2 \mathrm{am}-\mathrm{am}^3 \quad(\because \mathrm{m}=-1)\)
\(y=-x-2(3)(-1)-3(-1)^3\)
\(\mathrm{y}=-\mathrm{x}+6+3\)
\(\mathrm{x}+\mathrm{y}=9\)
But given normal,
\(\mathrm{x}+\mathrm{y}=\mathrm{k}\)
\(\therefore \quad \mathrm{k} =9\)
Focus of a given parabola is \((3,0)\)
Now, perpendicular distance from \((3,0)\) to the equation
(i) is
\(\mathrm{p} =\frac{|3(1)+0-9|}{\sqrt{1^2+1^2}}\)
\(\mathrm{p} =\frac{|3-9|}{\sqrt{2}}\)
\(\mathrm{p} =\frac{6}{\sqrt{2}}\)
\(\therefore 4 \mathrm{k}-2 \mathrm{p}^2 =4(9)-2\left(\frac{6}{\sqrt{2}}\right)^2=36-2 \times \frac{36}{2}\)
\(=36-36=0\)
