120235
The normal chord at a point \(t\) on the parabola \(y^2=4 a x\) subtends a right angle at the vertex. Then, \(t^2\) is equal to
1 2
2 \(\sqrt{2}\)
3 4
4 None of these
Explanation:
A : Given,
The equation of any normal to \(y^2=4 \mathrm{ax}\) at (at \(\left.{ }^2, 2 a t\right)\) is
\(y+t x=2 a t+a t^3\)
The equation of the line joining the vertex (origin) to the points of intersection of the parabola and equation (i) is,
\(y^2=4 a x\left(\frac{y+t x}{2 a t+a t^3}\right) \Rightarrow\left(2 t+t^3\right) y^2=4 x(y+t x)\)
If equation (i) makes a right angle at the vertex, then coefficient of \(x^2+\) coefficient of \(y^2=0\)
\(4 \mathrm{t}-2 \mathrm{t}-\mathrm{t}^3=0\)
\(\mathrm{t}^2=2\)
BCECE-2010
Parabola
120236
If a line \(y=3 x+1\) cuts the parabola \(x^2-4 x-\) \(4 y+20=0\) at \(A\) and \(B\), then the tangent of the angle subtended by the line segment \(A B\) at origin is
1 \(\frac{8 \sqrt{3}}{205}\)
2 \(\frac{8 \sqrt{3}}{209}\)
3 \(\frac{8 \sqrt{3}}{215}\)
4 None of these
Explanation:
B Given,
Line : \(y=3 x+1\)
Parabola: \(x^2-4 x-4 y+20=0\)
Joint equation of \(\mathrm{OA}\) and \(\mathrm{OB}\) is found by making equation of parabola homogeneous using straight line :
\(x^2-4 x(y-3 x)-4 y-(y-3 x)+20(y-3 x)^2=0\)
\(x^2(1+12+180)-y^2(4-20)-x y(4-12+120)=\)
0
\(193 x^2+16 y^2-112 x y=0\)
Here,
\(\mathrm{a}=193, \mathrm{~b}=16,2 \mathrm{~h}=112\)
\(\therefore \quad \tan \theta=\frac{2 \sqrt{\mathrm{h}^2-\mathrm{ab}}}{\mathrm{a}+\mathrm{b}}\)
\(=\frac{2 \times \sqrt{(56)^2-193 \times 16}}{(193+16)}\)
\(\tan \theta=\frac{8 \sqrt{3}}{209}\)
AMU-2019
Parabola
120237
The equation of the common tangent to the parabola \(y^2=8 x\) and rectangular hyperbola \(x y\) \(=-1\) is
1 \(x-y+2=0\)
2 \(9 x-3 y+2=0\)
3 \(2 \mathrm{x}+\mathrm{y}+1=0\)
4 \(x+2 y-1=0\)
Explanation:
A The parabola \(\mathrm{y}^2=8 \mathrm{x}\)
Tangent to the curve \(\mathrm{y}=\mathrm{mx}+\frac{2}{\mathrm{~m}}\)
So, it must satisfy \(\mathrm{xy}=-1\)
\(x\left(m x+\frac{2}{m}\right)=-1\)
\(m x^2+\frac{2}{m} x+1=0\)
Now, quadratic equation then \(b^2-4 a c=0\)
\(\frac{4}{\mathrm{~m}^2}-4 \mathrm{~m}=0\)
\(\mathrm{~m}^3=1 \quad \mathrm{~m}=1\)
So, the common tangent \(\mathrm{y}=\mathrm{x}+2\)
\(x-y+2=0\)
AMU-2011
Parabola
120238
The line \(x+y=6\) is normal to the parabola \(y^2=8 x\) at the point
1 \((4,2)\)
2 \((2,4)\)
3 \((2,2)\)
4 \((3,3)\)
Explanation:
B \(y=-x+6\)
is normal to \(y^2=4 a x\) slope of line \((1)=-1\)
Now, equation of any normal
\(y^2=4 a x\) is \(y=m x-2 a m-a m^3\)
Comparing (i) and (ii), we get
\(\mathrm{mx}=-\mathrm{x} \Rightarrow \mathrm{m}=-1\)
\(6=-2 a(-1)-a(-1)^3\)
\(6=2 a+a=3 a \Rightarrow a=2\)
\(\therefore\) Point \(\left(\mathrm{am}^2,-2 \mathrm{am}\right)=\left(2(-1)^2,-2 \times 2 \times(-1)\right)=(2,4)\)
120235
The normal chord at a point \(t\) on the parabola \(y^2=4 a x\) subtends a right angle at the vertex. Then, \(t^2\) is equal to
1 2
2 \(\sqrt{2}\)
3 4
4 None of these
Explanation:
A : Given,
The equation of any normal to \(y^2=4 \mathrm{ax}\) at (at \(\left.{ }^2, 2 a t\right)\) is
\(y+t x=2 a t+a t^3\)
The equation of the line joining the vertex (origin) to the points of intersection of the parabola and equation (i) is,
\(y^2=4 a x\left(\frac{y+t x}{2 a t+a t^3}\right) \Rightarrow\left(2 t+t^3\right) y^2=4 x(y+t x)\)
If equation (i) makes a right angle at the vertex, then coefficient of \(x^2+\) coefficient of \(y^2=0\)
\(4 \mathrm{t}-2 \mathrm{t}-\mathrm{t}^3=0\)
\(\mathrm{t}^2=2\)
BCECE-2010
Parabola
120236
If a line \(y=3 x+1\) cuts the parabola \(x^2-4 x-\) \(4 y+20=0\) at \(A\) and \(B\), then the tangent of the angle subtended by the line segment \(A B\) at origin is
1 \(\frac{8 \sqrt{3}}{205}\)
2 \(\frac{8 \sqrt{3}}{209}\)
3 \(\frac{8 \sqrt{3}}{215}\)
4 None of these
Explanation:
B Given,
Line : \(y=3 x+1\)
Parabola: \(x^2-4 x-4 y+20=0\)
Joint equation of \(\mathrm{OA}\) and \(\mathrm{OB}\) is found by making equation of parabola homogeneous using straight line :
\(x^2-4 x(y-3 x)-4 y-(y-3 x)+20(y-3 x)^2=0\)
\(x^2(1+12+180)-y^2(4-20)-x y(4-12+120)=\)
0
\(193 x^2+16 y^2-112 x y=0\)
Here,
\(\mathrm{a}=193, \mathrm{~b}=16,2 \mathrm{~h}=112\)
\(\therefore \quad \tan \theta=\frac{2 \sqrt{\mathrm{h}^2-\mathrm{ab}}}{\mathrm{a}+\mathrm{b}}\)
\(=\frac{2 \times \sqrt{(56)^2-193 \times 16}}{(193+16)}\)
\(\tan \theta=\frac{8 \sqrt{3}}{209}\)
AMU-2019
Parabola
120237
The equation of the common tangent to the parabola \(y^2=8 x\) and rectangular hyperbola \(x y\) \(=-1\) is
1 \(x-y+2=0\)
2 \(9 x-3 y+2=0\)
3 \(2 \mathrm{x}+\mathrm{y}+1=0\)
4 \(x+2 y-1=0\)
Explanation:
A The parabola \(\mathrm{y}^2=8 \mathrm{x}\)
Tangent to the curve \(\mathrm{y}=\mathrm{mx}+\frac{2}{\mathrm{~m}}\)
So, it must satisfy \(\mathrm{xy}=-1\)
\(x\left(m x+\frac{2}{m}\right)=-1\)
\(m x^2+\frac{2}{m} x+1=0\)
Now, quadratic equation then \(b^2-4 a c=0\)
\(\frac{4}{\mathrm{~m}^2}-4 \mathrm{~m}=0\)
\(\mathrm{~m}^3=1 \quad \mathrm{~m}=1\)
So, the common tangent \(\mathrm{y}=\mathrm{x}+2\)
\(x-y+2=0\)
AMU-2011
Parabola
120238
The line \(x+y=6\) is normal to the parabola \(y^2=8 x\) at the point
1 \((4,2)\)
2 \((2,4)\)
3 \((2,2)\)
4 \((3,3)\)
Explanation:
B \(y=-x+6\)
is normal to \(y^2=4 a x\) slope of line \((1)=-1\)
Now, equation of any normal
\(y^2=4 a x\) is \(y=m x-2 a m-a m^3\)
Comparing (i) and (ii), we get
\(\mathrm{mx}=-\mathrm{x} \Rightarrow \mathrm{m}=-1\)
\(6=-2 a(-1)-a(-1)^3\)
\(6=2 a+a=3 a \Rightarrow a=2\)
\(\therefore\) Point \(\left(\mathrm{am}^2,-2 \mathrm{am}\right)=\left(2(-1)^2,-2 \times 2 \times(-1)\right)=(2,4)\)
120235
The normal chord at a point \(t\) on the parabola \(y^2=4 a x\) subtends a right angle at the vertex. Then, \(t^2\) is equal to
1 2
2 \(\sqrt{2}\)
3 4
4 None of these
Explanation:
A : Given,
The equation of any normal to \(y^2=4 \mathrm{ax}\) at (at \(\left.{ }^2, 2 a t\right)\) is
\(y+t x=2 a t+a t^3\)
The equation of the line joining the vertex (origin) to the points of intersection of the parabola and equation (i) is,
\(y^2=4 a x\left(\frac{y+t x}{2 a t+a t^3}\right) \Rightarrow\left(2 t+t^3\right) y^2=4 x(y+t x)\)
If equation (i) makes a right angle at the vertex, then coefficient of \(x^2+\) coefficient of \(y^2=0\)
\(4 \mathrm{t}-2 \mathrm{t}-\mathrm{t}^3=0\)
\(\mathrm{t}^2=2\)
BCECE-2010
Parabola
120236
If a line \(y=3 x+1\) cuts the parabola \(x^2-4 x-\) \(4 y+20=0\) at \(A\) and \(B\), then the tangent of the angle subtended by the line segment \(A B\) at origin is
1 \(\frac{8 \sqrt{3}}{205}\)
2 \(\frac{8 \sqrt{3}}{209}\)
3 \(\frac{8 \sqrt{3}}{215}\)
4 None of these
Explanation:
B Given,
Line : \(y=3 x+1\)
Parabola: \(x^2-4 x-4 y+20=0\)
Joint equation of \(\mathrm{OA}\) and \(\mathrm{OB}\) is found by making equation of parabola homogeneous using straight line :
\(x^2-4 x(y-3 x)-4 y-(y-3 x)+20(y-3 x)^2=0\)
\(x^2(1+12+180)-y^2(4-20)-x y(4-12+120)=\)
0
\(193 x^2+16 y^2-112 x y=0\)
Here,
\(\mathrm{a}=193, \mathrm{~b}=16,2 \mathrm{~h}=112\)
\(\therefore \quad \tan \theta=\frac{2 \sqrt{\mathrm{h}^2-\mathrm{ab}}}{\mathrm{a}+\mathrm{b}}\)
\(=\frac{2 \times \sqrt{(56)^2-193 \times 16}}{(193+16)}\)
\(\tan \theta=\frac{8 \sqrt{3}}{209}\)
AMU-2019
Parabola
120237
The equation of the common tangent to the parabola \(y^2=8 x\) and rectangular hyperbola \(x y\) \(=-1\) is
1 \(x-y+2=0\)
2 \(9 x-3 y+2=0\)
3 \(2 \mathrm{x}+\mathrm{y}+1=0\)
4 \(x+2 y-1=0\)
Explanation:
A The parabola \(\mathrm{y}^2=8 \mathrm{x}\)
Tangent to the curve \(\mathrm{y}=\mathrm{mx}+\frac{2}{\mathrm{~m}}\)
So, it must satisfy \(\mathrm{xy}=-1\)
\(x\left(m x+\frac{2}{m}\right)=-1\)
\(m x^2+\frac{2}{m} x+1=0\)
Now, quadratic equation then \(b^2-4 a c=0\)
\(\frac{4}{\mathrm{~m}^2}-4 \mathrm{~m}=0\)
\(\mathrm{~m}^3=1 \quad \mathrm{~m}=1\)
So, the common tangent \(\mathrm{y}=\mathrm{x}+2\)
\(x-y+2=0\)
AMU-2011
Parabola
120238
The line \(x+y=6\) is normal to the parabola \(y^2=8 x\) at the point
1 \((4,2)\)
2 \((2,4)\)
3 \((2,2)\)
4 \((3,3)\)
Explanation:
B \(y=-x+6\)
is normal to \(y^2=4 a x\) slope of line \((1)=-1\)
Now, equation of any normal
\(y^2=4 a x\) is \(y=m x-2 a m-a m^3\)
Comparing (i) and (ii), we get
\(\mathrm{mx}=-\mathrm{x} \Rightarrow \mathrm{m}=-1\)
\(6=-2 a(-1)-a(-1)^3\)
\(6=2 a+a=3 a \Rightarrow a=2\)
\(\therefore\) Point \(\left(\mathrm{am}^2,-2 \mathrm{am}\right)=\left(2(-1)^2,-2 \times 2 \times(-1)\right)=(2,4)\)
NEET Test Series from KOTA - 10 Papers In MS WORD
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Parabola
120235
The normal chord at a point \(t\) on the parabola \(y^2=4 a x\) subtends a right angle at the vertex. Then, \(t^2\) is equal to
1 2
2 \(\sqrt{2}\)
3 4
4 None of these
Explanation:
A : Given,
The equation of any normal to \(y^2=4 \mathrm{ax}\) at (at \(\left.{ }^2, 2 a t\right)\) is
\(y+t x=2 a t+a t^3\)
The equation of the line joining the vertex (origin) to the points of intersection of the parabola and equation (i) is,
\(y^2=4 a x\left(\frac{y+t x}{2 a t+a t^3}\right) \Rightarrow\left(2 t+t^3\right) y^2=4 x(y+t x)\)
If equation (i) makes a right angle at the vertex, then coefficient of \(x^2+\) coefficient of \(y^2=0\)
\(4 \mathrm{t}-2 \mathrm{t}-\mathrm{t}^3=0\)
\(\mathrm{t}^2=2\)
BCECE-2010
Parabola
120236
If a line \(y=3 x+1\) cuts the parabola \(x^2-4 x-\) \(4 y+20=0\) at \(A\) and \(B\), then the tangent of the angle subtended by the line segment \(A B\) at origin is
1 \(\frac{8 \sqrt{3}}{205}\)
2 \(\frac{8 \sqrt{3}}{209}\)
3 \(\frac{8 \sqrt{3}}{215}\)
4 None of these
Explanation:
B Given,
Line : \(y=3 x+1\)
Parabola: \(x^2-4 x-4 y+20=0\)
Joint equation of \(\mathrm{OA}\) and \(\mathrm{OB}\) is found by making equation of parabola homogeneous using straight line :
\(x^2-4 x(y-3 x)-4 y-(y-3 x)+20(y-3 x)^2=0\)
\(x^2(1+12+180)-y^2(4-20)-x y(4-12+120)=\)
0
\(193 x^2+16 y^2-112 x y=0\)
Here,
\(\mathrm{a}=193, \mathrm{~b}=16,2 \mathrm{~h}=112\)
\(\therefore \quad \tan \theta=\frac{2 \sqrt{\mathrm{h}^2-\mathrm{ab}}}{\mathrm{a}+\mathrm{b}}\)
\(=\frac{2 \times \sqrt{(56)^2-193 \times 16}}{(193+16)}\)
\(\tan \theta=\frac{8 \sqrt{3}}{209}\)
AMU-2019
Parabola
120237
The equation of the common tangent to the parabola \(y^2=8 x\) and rectangular hyperbola \(x y\) \(=-1\) is
1 \(x-y+2=0\)
2 \(9 x-3 y+2=0\)
3 \(2 \mathrm{x}+\mathrm{y}+1=0\)
4 \(x+2 y-1=0\)
Explanation:
A The parabola \(\mathrm{y}^2=8 \mathrm{x}\)
Tangent to the curve \(\mathrm{y}=\mathrm{mx}+\frac{2}{\mathrm{~m}}\)
So, it must satisfy \(\mathrm{xy}=-1\)
\(x\left(m x+\frac{2}{m}\right)=-1\)
\(m x^2+\frac{2}{m} x+1=0\)
Now, quadratic equation then \(b^2-4 a c=0\)
\(\frac{4}{\mathrm{~m}^2}-4 \mathrm{~m}=0\)
\(\mathrm{~m}^3=1 \quad \mathrm{~m}=1\)
So, the common tangent \(\mathrm{y}=\mathrm{x}+2\)
\(x-y+2=0\)
AMU-2011
Parabola
120238
The line \(x+y=6\) is normal to the parabola \(y^2=8 x\) at the point
1 \((4,2)\)
2 \((2,4)\)
3 \((2,2)\)
4 \((3,3)\)
Explanation:
B \(y=-x+6\)
is normal to \(y^2=4 a x\) slope of line \((1)=-1\)
Now, equation of any normal
\(y^2=4 a x\) is \(y=m x-2 a m-a m^3\)
Comparing (i) and (ii), we get
\(\mathrm{mx}=-\mathrm{x} \Rightarrow \mathrm{m}=-1\)
\(6=-2 a(-1)-a(-1)^3\)
\(6=2 a+a=3 a \Rightarrow a=2\)
\(\therefore\) Point \(\left(\mathrm{am}^2,-2 \mathrm{am}\right)=\left(2(-1)^2,-2 \times 2 \times(-1)\right)=(2,4)\)
