NEET Test Series from KOTA - 10 Papers In MS WORD
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Parabola
120225
If \(y+b=m_1(x+a)\) and \(y+b=m_2(x+a)\) are two tangents to \(\mathrm{y}^2=4 \mathrm{ax}\), then
1 \(\mathrm{m}_1+\mathrm{m}_2=0\)
2 \(\mathrm{m}_1 \mathrm{~m}_2=1\)
3 \(\mathrm{m}_1 \mathrm{~m}_2=-1\)
4 None of these
Explanation:
C Given,
\(y+b=m_1(x+a) \ldots\) (i)
\(y+b=m_2(x+a)\)
On subtracting equation (i) from equation (ii)
\(\left(\mathrm{m}_1-\mathrm{m}_2\right)(\mathrm{x}+\mathrm{a})=0\)
\(\mathrm{~m}_1 \neq \mathrm{m}_2 \Rightarrow(\mathrm{x}+\mathrm{a})=0\)
\(\therefore\) The two tangents intersect on the directrix.
The tangents are perpendicular and \(\mathrm{m}_1 \mathrm{~m}_2=-1\).
COMEDK-2016
Parabola
120227
The condition that the line \(\frac{x}{p}+\frac{y}{q}=1\) be a normal to the parabola \(y^2=4 \mathrm{ax}\) is
1 \(\mathrm{p}^3=2 a \mathrm{p}^2+a \mathrm{q}^2\)
A Given,
The line \(\frac{x}{p}+\frac{y}{q}=1\) will be a normal to the parabola \(y^2=\) \(4 \mathrm{ax}\)
We know that, for some value of \(m\), it is identical with
\(y=m x-2 a m-a m^3\)
\(\text { i.e. } m x-y=\left(2 a m+a m^3\right)\)
Comparing coefficients, we get
\(\frac{m}{1 / p}=\frac{-1}{1 / q}=\frac{2 a m+a m^3}{1}\)
\(m p=-q\)
\(\therefore m=-\frac{q}{p} \text { and } m p=m\left(2 a+a m^2\right)\)
\(p =2 a+a m^2\)
\(p=2 a+a\left(\frac{-q}{p}\right)^2\)
\(p=2 a+\frac{a q^2}{p^2}\)
\(p^3=2 a p^2+a q^2\)Which is the required condition.
VITEEE-2017
Parabola
120228
The normal at the point \(\left(a_1{ }^2, 2 a t_1\right)\) on the parabola, cuts the parabola again at the point whose parameter is
C Let the normal at ' \(t_1\) ' cuts the parabola again at the point ' \(t_2\) '. the equation of the normal at \(\left(a_1{ }^2, 2 a_1\right)\) is \(y\) \(+\mathrm{t}_1 \mathrm{x}=2 \mathrm{at}_1+\mathrm{at}_1{ }^3\)
Since, it passes through the point ' \(t_2\) ' i.e
\(\quad\left(\mathrm{at}_2{ }^2, 2 \mathrm{at}_2\right)\)
\(\therefore \quad 2 \mathrm{at}_2+\mathrm{at}_1 \mathrm{t}_2{ }^2=2 \mathrm{at}_1+\mathrm{at}_1^3\)
\(2 \mathrm{a}\left(\mathrm{t}_1-\mathrm{t}_2\right)+\mathrm{at}_1\left(\mathrm{t}_1{ }^2-\mathrm{t}_2{ }^2\right)=0\)
\(2+\mathrm{t}_1\left(\mathrm{t}_1+\mathrm{t}_2\right)=0\)
\(2+\mathrm{t}_1{ }^2+\mathrm{t}_1 \mathrm{t}_2=0\)
\(\mathrm{t}_1 \mathrm{t}_2=-\left(\mathrm{t}_1{ }^2+2\right)\)
\(\mathrm{t}_2=-\left(\mathrm{t}_1+\frac{2}{\mathrm{t}_1}\right)\)
VITEEE-2016
Parabola
120229
If \(2 x+y+k=0\) is a normal to the parabola \(y^2\) \(=-8 x\), then the value of \(k\), is
1 8
2 16
3 24
4 32
Explanation:
C The equation of any normal to the parabola
\(y^2=-8 x \text { is } y=M x+4 M+2 M^3\)
(Using equation of normal of parabola in slope form \(y=\) \(\mathrm{Mx}-2 \mathrm{aM}-\mathrm{aM}^3\) and \(\mathrm{a}=-2\) )
The given normal is
\(2 \mathrm{x}+\mathrm{y}+\mathrm{k}=0 \Rightarrow \mathrm{y}=-2 \mathrm{x}-\mathrm{k}\)
Comparing equation (i) and (ii), we get
\(M=-2 \text { and }-4 M-2 M^3=k \Rightarrow-4(-2)-2(-8)=k\)
\(+8+16=\mathrm{k}\)
\(\mathrm{k}=24\)
120225
If \(y+b=m_1(x+a)\) and \(y+b=m_2(x+a)\) are two tangents to \(\mathrm{y}^2=4 \mathrm{ax}\), then
1 \(\mathrm{m}_1+\mathrm{m}_2=0\)
2 \(\mathrm{m}_1 \mathrm{~m}_2=1\)
3 \(\mathrm{m}_1 \mathrm{~m}_2=-1\)
4 None of these
Explanation:
C Given,
\(y+b=m_1(x+a) \ldots\) (i)
\(y+b=m_2(x+a)\)
On subtracting equation (i) from equation (ii)
\(\left(\mathrm{m}_1-\mathrm{m}_2\right)(\mathrm{x}+\mathrm{a})=0\)
\(\mathrm{~m}_1 \neq \mathrm{m}_2 \Rightarrow(\mathrm{x}+\mathrm{a})=0\)
\(\therefore\) The two tangents intersect on the directrix.
The tangents are perpendicular and \(\mathrm{m}_1 \mathrm{~m}_2=-1\).
COMEDK-2016
Parabola
120227
The condition that the line \(\frac{x}{p}+\frac{y}{q}=1\) be a normal to the parabola \(y^2=4 \mathrm{ax}\) is
1 \(\mathrm{p}^3=2 a \mathrm{p}^2+a \mathrm{q}^2\)
A Given,
The line \(\frac{x}{p}+\frac{y}{q}=1\) will be a normal to the parabola \(y^2=\) \(4 \mathrm{ax}\)
We know that, for some value of \(m\), it is identical with
\(y=m x-2 a m-a m^3\)
\(\text { i.e. } m x-y=\left(2 a m+a m^3\right)\)
Comparing coefficients, we get
\(\frac{m}{1 / p}=\frac{-1}{1 / q}=\frac{2 a m+a m^3}{1}\)
\(m p=-q\)
\(\therefore m=-\frac{q}{p} \text { and } m p=m\left(2 a+a m^2\right)\)
\(p =2 a+a m^2\)
\(p=2 a+a\left(\frac{-q}{p}\right)^2\)
\(p=2 a+\frac{a q^2}{p^2}\)
\(p^3=2 a p^2+a q^2\)Which is the required condition.
VITEEE-2017
Parabola
120228
The normal at the point \(\left(a_1{ }^2, 2 a t_1\right)\) on the parabola, cuts the parabola again at the point whose parameter is
C Let the normal at ' \(t_1\) ' cuts the parabola again at the point ' \(t_2\) '. the equation of the normal at \(\left(a_1{ }^2, 2 a_1\right)\) is \(y\) \(+\mathrm{t}_1 \mathrm{x}=2 \mathrm{at}_1+\mathrm{at}_1{ }^3\)
Since, it passes through the point ' \(t_2\) ' i.e
\(\quad\left(\mathrm{at}_2{ }^2, 2 \mathrm{at}_2\right)\)
\(\therefore \quad 2 \mathrm{at}_2+\mathrm{at}_1 \mathrm{t}_2{ }^2=2 \mathrm{at}_1+\mathrm{at}_1^3\)
\(2 \mathrm{a}\left(\mathrm{t}_1-\mathrm{t}_2\right)+\mathrm{at}_1\left(\mathrm{t}_1{ }^2-\mathrm{t}_2{ }^2\right)=0\)
\(2+\mathrm{t}_1\left(\mathrm{t}_1+\mathrm{t}_2\right)=0\)
\(2+\mathrm{t}_1{ }^2+\mathrm{t}_1 \mathrm{t}_2=0\)
\(\mathrm{t}_1 \mathrm{t}_2=-\left(\mathrm{t}_1{ }^2+2\right)\)
\(\mathrm{t}_2=-\left(\mathrm{t}_1+\frac{2}{\mathrm{t}_1}\right)\)
VITEEE-2016
Parabola
120229
If \(2 x+y+k=0\) is a normal to the parabola \(y^2\) \(=-8 x\), then the value of \(k\), is
1 8
2 16
3 24
4 32
Explanation:
C The equation of any normal to the parabola
\(y^2=-8 x \text { is } y=M x+4 M+2 M^3\)
(Using equation of normal of parabola in slope form \(y=\) \(\mathrm{Mx}-2 \mathrm{aM}-\mathrm{aM}^3\) and \(\mathrm{a}=-2\) )
The given normal is
\(2 \mathrm{x}+\mathrm{y}+\mathrm{k}=0 \Rightarrow \mathrm{y}=-2 \mathrm{x}-\mathrm{k}\)
Comparing equation (i) and (ii), we get
\(M=-2 \text { and }-4 M-2 M^3=k \Rightarrow-4(-2)-2(-8)=k\)
\(+8+16=\mathrm{k}\)
\(\mathrm{k}=24\)
120225
If \(y+b=m_1(x+a)\) and \(y+b=m_2(x+a)\) are two tangents to \(\mathrm{y}^2=4 \mathrm{ax}\), then
1 \(\mathrm{m}_1+\mathrm{m}_2=0\)
2 \(\mathrm{m}_1 \mathrm{~m}_2=1\)
3 \(\mathrm{m}_1 \mathrm{~m}_2=-1\)
4 None of these
Explanation:
C Given,
\(y+b=m_1(x+a) \ldots\) (i)
\(y+b=m_2(x+a)\)
On subtracting equation (i) from equation (ii)
\(\left(\mathrm{m}_1-\mathrm{m}_2\right)(\mathrm{x}+\mathrm{a})=0\)
\(\mathrm{~m}_1 \neq \mathrm{m}_2 \Rightarrow(\mathrm{x}+\mathrm{a})=0\)
\(\therefore\) The two tangents intersect on the directrix.
The tangents are perpendicular and \(\mathrm{m}_1 \mathrm{~m}_2=-1\).
COMEDK-2016
Parabola
120227
The condition that the line \(\frac{x}{p}+\frac{y}{q}=1\) be a normal to the parabola \(y^2=4 \mathrm{ax}\) is
1 \(\mathrm{p}^3=2 a \mathrm{p}^2+a \mathrm{q}^2\)
A Given,
The line \(\frac{x}{p}+\frac{y}{q}=1\) will be a normal to the parabola \(y^2=\) \(4 \mathrm{ax}\)
We know that, for some value of \(m\), it is identical with
\(y=m x-2 a m-a m^3\)
\(\text { i.e. } m x-y=\left(2 a m+a m^3\right)\)
Comparing coefficients, we get
\(\frac{m}{1 / p}=\frac{-1}{1 / q}=\frac{2 a m+a m^3}{1}\)
\(m p=-q\)
\(\therefore m=-\frac{q}{p} \text { and } m p=m\left(2 a+a m^2\right)\)
\(p =2 a+a m^2\)
\(p=2 a+a\left(\frac{-q}{p}\right)^2\)
\(p=2 a+\frac{a q^2}{p^2}\)
\(p^3=2 a p^2+a q^2\)Which is the required condition.
VITEEE-2017
Parabola
120228
The normal at the point \(\left(a_1{ }^2, 2 a t_1\right)\) on the parabola, cuts the parabola again at the point whose parameter is
C Let the normal at ' \(t_1\) ' cuts the parabola again at the point ' \(t_2\) '. the equation of the normal at \(\left(a_1{ }^2, 2 a_1\right)\) is \(y\) \(+\mathrm{t}_1 \mathrm{x}=2 \mathrm{at}_1+\mathrm{at}_1{ }^3\)
Since, it passes through the point ' \(t_2\) ' i.e
\(\quad\left(\mathrm{at}_2{ }^2, 2 \mathrm{at}_2\right)\)
\(\therefore \quad 2 \mathrm{at}_2+\mathrm{at}_1 \mathrm{t}_2{ }^2=2 \mathrm{at}_1+\mathrm{at}_1^3\)
\(2 \mathrm{a}\left(\mathrm{t}_1-\mathrm{t}_2\right)+\mathrm{at}_1\left(\mathrm{t}_1{ }^2-\mathrm{t}_2{ }^2\right)=0\)
\(2+\mathrm{t}_1\left(\mathrm{t}_1+\mathrm{t}_2\right)=0\)
\(2+\mathrm{t}_1{ }^2+\mathrm{t}_1 \mathrm{t}_2=0\)
\(\mathrm{t}_1 \mathrm{t}_2=-\left(\mathrm{t}_1{ }^2+2\right)\)
\(\mathrm{t}_2=-\left(\mathrm{t}_1+\frac{2}{\mathrm{t}_1}\right)\)
VITEEE-2016
Parabola
120229
If \(2 x+y+k=0\) is a normal to the parabola \(y^2\) \(=-8 x\), then the value of \(k\), is
1 8
2 16
3 24
4 32
Explanation:
C The equation of any normal to the parabola
\(y^2=-8 x \text { is } y=M x+4 M+2 M^3\)
(Using equation of normal of parabola in slope form \(y=\) \(\mathrm{Mx}-2 \mathrm{aM}-\mathrm{aM}^3\) and \(\mathrm{a}=-2\) )
The given normal is
\(2 \mathrm{x}+\mathrm{y}+\mathrm{k}=0 \Rightarrow \mathrm{y}=-2 \mathrm{x}-\mathrm{k}\)
Comparing equation (i) and (ii), we get
\(M=-2 \text { and }-4 M-2 M^3=k \Rightarrow-4(-2)-2(-8)=k\)
\(+8+16=\mathrm{k}\)
\(\mathrm{k}=24\)
120225
If \(y+b=m_1(x+a)\) and \(y+b=m_2(x+a)\) are two tangents to \(\mathrm{y}^2=4 \mathrm{ax}\), then
1 \(\mathrm{m}_1+\mathrm{m}_2=0\)
2 \(\mathrm{m}_1 \mathrm{~m}_2=1\)
3 \(\mathrm{m}_1 \mathrm{~m}_2=-1\)
4 None of these
Explanation:
C Given,
\(y+b=m_1(x+a) \ldots\) (i)
\(y+b=m_2(x+a)\)
On subtracting equation (i) from equation (ii)
\(\left(\mathrm{m}_1-\mathrm{m}_2\right)(\mathrm{x}+\mathrm{a})=0\)
\(\mathrm{~m}_1 \neq \mathrm{m}_2 \Rightarrow(\mathrm{x}+\mathrm{a})=0\)
\(\therefore\) The two tangents intersect on the directrix.
The tangents are perpendicular and \(\mathrm{m}_1 \mathrm{~m}_2=-1\).
COMEDK-2016
Parabola
120227
The condition that the line \(\frac{x}{p}+\frac{y}{q}=1\) be a normal to the parabola \(y^2=4 \mathrm{ax}\) is
1 \(\mathrm{p}^3=2 a \mathrm{p}^2+a \mathrm{q}^2\)
A Given,
The line \(\frac{x}{p}+\frac{y}{q}=1\) will be a normal to the parabola \(y^2=\) \(4 \mathrm{ax}\)
We know that, for some value of \(m\), it is identical with
\(y=m x-2 a m-a m^3\)
\(\text { i.e. } m x-y=\left(2 a m+a m^3\right)\)
Comparing coefficients, we get
\(\frac{m}{1 / p}=\frac{-1}{1 / q}=\frac{2 a m+a m^3}{1}\)
\(m p=-q\)
\(\therefore m=-\frac{q}{p} \text { and } m p=m\left(2 a+a m^2\right)\)
\(p =2 a+a m^2\)
\(p=2 a+a\left(\frac{-q}{p}\right)^2\)
\(p=2 a+\frac{a q^2}{p^2}\)
\(p^3=2 a p^2+a q^2\)Which is the required condition.
VITEEE-2017
Parabola
120228
The normal at the point \(\left(a_1{ }^2, 2 a t_1\right)\) on the parabola, cuts the parabola again at the point whose parameter is
C Let the normal at ' \(t_1\) ' cuts the parabola again at the point ' \(t_2\) '. the equation of the normal at \(\left(a_1{ }^2, 2 a_1\right)\) is \(y\) \(+\mathrm{t}_1 \mathrm{x}=2 \mathrm{at}_1+\mathrm{at}_1{ }^3\)
Since, it passes through the point ' \(t_2\) ' i.e
\(\quad\left(\mathrm{at}_2{ }^2, 2 \mathrm{at}_2\right)\)
\(\therefore \quad 2 \mathrm{at}_2+\mathrm{at}_1 \mathrm{t}_2{ }^2=2 \mathrm{at}_1+\mathrm{at}_1^3\)
\(2 \mathrm{a}\left(\mathrm{t}_1-\mathrm{t}_2\right)+\mathrm{at}_1\left(\mathrm{t}_1{ }^2-\mathrm{t}_2{ }^2\right)=0\)
\(2+\mathrm{t}_1\left(\mathrm{t}_1+\mathrm{t}_2\right)=0\)
\(2+\mathrm{t}_1{ }^2+\mathrm{t}_1 \mathrm{t}_2=0\)
\(\mathrm{t}_1 \mathrm{t}_2=-\left(\mathrm{t}_1{ }^2+2\right)\)
\(\mathrm{t}_2=-\left(\mathrm{t}_1+\frac{2}{\mathrm{t}_1}\right)\)
VITEEE-2016
Parabola
120229
If \(2 x+y+k=0\) is a normal to the parabola \(y^2\) \(=-8 x\), then the value of \(k\), is
1 8
2 16
3 24
4 32
Explanation:
C The equation of any normal to the parabola
\(y^2=-8 x \text { is } y=M x+4 M+2 M^3\)
(Using equation of normal of parabola in slope form \(y=\) \(\mathrm{Mx}-2 \mathrm{aM}-\mathrm{aM}^3\) and \(\mathrm{a}=-2\) )
The given normal is
\(2 \mathrm{x}+\mathrm{y}+\mathrm{k}=0 \Rightarrow \mathrm{y}=-2 \mathrm{x}-\mathrm{k}\)
Comparing equation (i) and (ii), we get
\(M=-2 \text { and }-4 M-2 M^3=k \Rightarrow-4(-2)-2(-8)=k\)
\(+8+16=\mathrm{k}\)
\(\mathrm{k}=24\)
