120261
The Equation of the tangent to the parabola \(y\) \(=12 x\) at \((3,-6)\) is
1 \(x-y+9=0\)
2 \(x+y+3=0\)
3 \(x+y-3=0\)
4 \(x=3\)
Explanation:
B Given, parabola: \(\mathrm{y}^2=12 \mathrm{x}=4(3) \mathrm{x}\)
Here, \(a=3\),
Equation of tangent to this parabola is
\(\quad \mathrm{y}=\mathrm{mx}+\frac{\mathrm{a}}{\mathrm{m}}\)
\(\text { At point }(3,-6)\)
\(\quad-6=\mathrm{m}(3)+\frac{3}{\mathrm{~m}}\)
\(\quad-6 \mathrm{~m}=3 \mathrm{~m}^2+3\)
\(\Rightarrow \quad \mathrm{m}^2+2 \mathrm{~m}+1=0\)
\(\Rightarrow \quad \mathrm{m}+1)^2=0\)
\(\Rightarrow \quad \mathrm{m}=-1\)
Hence, equation of tangent is
\(\quad \mathrm{y}=(-1) \mathrm{x}+\frac{3}{(-1)}\)
\(\mathrm{x}+\mathrm{y}+3=0\)
At point \((3,-6)\)
Hence, equation of tangent is
\(\mathrm{y}=(-1) \mathrm{x}+\frac{3}{(-1)}\)
\(\mathrm{x}+\mathrm{y}+3=0\)
AP EAMCET-21.09.2020
Parabola
120262
If the line \(y=x\) is a tangent to the parabola \(y=\) \(a x^2+b x+c\) at the point \((1,1)\) and the curve passes through \((-1,0)\), then
C Given, parabola: \(y=a x^2+b x+c\)
And, \(\mathrm{y}=\mathrm{x}\) is tangent
Here, \(\mathrm{m}=1\)
Differentiate parabola w.r.t. \(x\)
\(\frac{\mathrm{dy}}{\mathrm{dx}}=2 \mathrm{ax}+\mathrm{b}\)
\(\mathrm{m}=2 \mathrm{ax}+\mathrm{b}\)
\(1=2 \mathrm{ax}+\mathrm{b}\)
At, \(\quad(1,1)\)
\(2 \mathrm{a}+\mathrm{b}=1\)
Also, parabola passes through \((1,1)\)
\(\therefore \quad 1=\mathrm{a}+\mathrm{b}+\mathrm{c}\)
Parabola passes through \((-1,0)\)
\(0=\mathrm{a}-\mathrm{b}+\mathrm{c}\)
From (i), (ii), and (iii)
\(\mathrm{a}=\mathrm{c}=\frac{1}{4} \text { and } \mathrm{b}=\frac{1}{2}\)
WB JEE-2020
Parabola
120263
From the point \((-1,-6)\), two tangents are drawn to \(y^2=4 x\). Then the angle between the two tangents is
1 \(\pi / 3\)
2 \(\pi / 4\)
3 \(\pi / 6\)
4 \(\pi / 2\)
Explanation:
D Given, from the point \((-1,-6)\) two tangents are drawn to \(\mathrm{y}^2=4 \mathrm{x}\)
Here, \(\quad \mathrm{a}=1\)
Equation of tangent
\(\mathrm{y}=\mathrm{mx}+\frac{1}{\mathrm{~m}} \quad\{\mathrm{a}=1\}\)
It passes through \((-1,-6)\)
\(-6=m(-1)+\frac{1}{(m)}\)
\(m^2-6 m-1=0\)
Here, \(\mathrm{m}_1+\mathrm{m}_2=6\) and \(\mathrm{m}_1 \cdot \mathrm{m}_2=-1\).
\(\left|\mathrm{m}_1-\mathrm{m}_2\right|=\sqrt{\left(\mathrm{m}_1+\mathrm{m}_2\right)^2}-4 \mathrm{~m}_1 \mathrm{~m}_2=2 \sqrt{10}\)
Angle between tangent,
\(\tan \theta =\left|\frac{\mathrm{m}_2-\mathrm{m}_1}{1+\mathrm{m}_1 \mathrm{~m}_2}\right|\)
\(\tan \theta =\frac{2 \sqrt{10}}{1+(-1)}=\infty\)
\(\therefore \quad \theta=\frac{\pi}{2}\)
WB JEE-2022
Parabola
120264
The equation of the normal to the parabola \(y^2=\) \(4 x\) which is perpendicular to \(x+3 y+1=0\) is
1 \(3 x-y=33\)
2 \(3 x-y+33=0\)
3 \(3 x+y=33\)
4 \(3 x+y+33=0\)
Explanation:
A Given, \(\mathrm{y}^2=4 \mathrm{x}\)
Line,
\(x+3 y+1=0\)
\(y=\frac{-1}{3} x-\frac{1}{3}\)
Slope, \(\mathrm{m}_1=\frac{-1}{3}\)
Normal slope, \(\mathrm{m}_1 \cdot \mathrm{m}_2=-1\)
\(\mathrm{m}_2=3\)
Now, equation of normal to parabola is
\(y=m_2 x-2 a m_2-a m_2^3\)
\(y=3 x-2(1)(3)-1 \times(3)^3\)
\(y=3 x-6-27\)
\(3 x-y=33\)
AP EAMCET-21.04.2019
Parabola
120265
If the angle between the tangents drawn through the point \((-2,-1)\) to the parabola \(y^2-\) \(4 x\) is \(\theta\), then \(\tan 2 \theta=\)
1 3
2 -3
3 \(-\frac{3}{4}\)
4 \(\frac{3}{4}\)
Explanation:
C Given, parabola, \(\mathrm{y}^2=4 \mathrm{x}\)
We know that, \(\mathrm{s}_1^2=\mathrm{s} \cdot \mathrm{s}_1 \quad\{\) at \((-2,-1)\}\)
\([\mathrm{y}(-1)-2(\mathrm{x}-2)]^2=(1+8)\left(\mathrm{y}^2-4 \mathrm{x}\right)\)
\((2 \mathrm{x}+\mathrm{y}-4)^2=9\left(\mathrm{y}^2-4 \mathrm{x}\right)\)
\(4 \mathrm{x}^2+4 \mathrm{xy}-8 \mathrm{y}^2+20 \mathrm{x}-8 \mathrm{y}+16=0\)
\(\tan \theta=\left|\frac{2 \sqrt{\mathrm{h}^2-\mathrm{ab}}}{\mathrm{a}+\mathrm{b}}\right|=\left|\frac{2 \sqrt{4+32}}{4-8}\right|=3\)
\(\tan 2 \theta=\frac{2 \tan \theta}{1-\tan ^2 \theta}=\frac{2 \times 3}{1-9}\)
\(\tan 2 \theta=\frac{-3}{4}\)
So, equation of common tangent is \(\mathrm{y}=\mathrm{x}+2\).
120261
The Equation of the tangent to the parabola \(y\) \(=12 x\) at \((3,-6)\) is
1 \(x-y+9=0\)
2 \(x+y+3=0\)
3 \(x+y-3=0\)
4 \(x=3\)
Explanation:
B Given, parabola: \(\mathrm{y}^2=12 \mathrm{x}=4(3) \mathrm{x}\)
Here, \(a=3\),
Equation of tangent to this parabola is
\(\quad \mathrm{y}=\mathrm{mx}+\frac{\mathrm{a}}{\mathrm{m}}\)
\(\text { At point }(3,-6)\)
\(\quad-6=\mathrm{m}(3)+\frac{3}{\mathrm{~m}}\)
\(\quad-6 \mathrm{~m}=3 \mathrm{~m}^2+3\)
\(\Rightarrow \quad \mathrm{m}^2+2 \mathrm{~m}+1=0\)
\(\Rightarrow \quad \mathrm{m}+1)^2=0\)
\(\Rightarrow \quad \mathrm{m}=-1\)
Hence, equation of tangent is
\(\quad \mathrm{y}=(-1) \mathrm{x}+\frac{3}{(-1)}\)
\(\mathrm{x}+\mathrm{y}+3=0\)
At point \((3,-6)\)
Hence, equation of tangent is
\(\mathrm{y}=(-1) \mathrm{x}+\frac{3}{(-1)}\)
\(\mathrm{x}+\mathrm{y}+3=0\)
AP EAMCET-21.09.2020
Parabola
120262
If the line \(y=x\) is a tangent to the parabola \(y=\) \(a x^2+b x+c\) at the point \((1,1)\) and the curve passes through \((-1,0)\), then
C Given, parabola: \(y=a x^2+b x+c\)
And, \(\mathrm{y}=\mathrm{x}\) is tangent
Here, \(\mathrm{m}=1\)
Differentiate parabola w.r.t. \(x\)
\(\frac{\mathrm{dy}}{\mathrm{dx}}=2 \mathrm{ax}+\mathrm{b}\)
\(\mathrm{m}=2 \mathrm{ax}+\mathrm{b}\)
\(1=2 \mathrm{ax}+\mathrm{b}\)
At, \(\quad(1,1)\)
\(2 \mathrm{a}+\mathrm{b}=1\)
Also, parabola passes through \((1,1)\)
\(\therefore \quad 1=\mathrm{a}+\mathrm{b}+\mathrm{c}\)
Parabola passes through \((-1,0)\)
\(0=\mathrm{a}-\mathrm{b}+\mathrm{c}\)
From (i), (ii), and (iii)
\(\mathrm{a}=\mathrm{c}=\frac{1}{4} \text { and } \mathrm{b}=\frac{1}{2}\)
WB JEE-2020
Parabola
120263
From the point \((-1,-6)\), two tangents are drawn to \(y^2=4 x\). Then the angle between the two tangents is
1 \(\pi / 3\)
2 \(\pi / 4\)
3 \(\pi / 6\)
4 \(\pi / 2\)
Explanation:
D Given, from the point \((-1,-6)\) two tangents are drawn to \(\mathrm{y}^2=4 \mathrm{x}\)
Here, \(\quad \mathrm{a}=1\)
Equation of tangent
\(\mathrm{y}=\mathrm{mx}+\frac{1}{\mathrm{~m}} \quad\{\mathrm{a}=1\}\)
It passes through \((-1,-6)\)
\(-6=m(-1)+\frac{1}{(m)}\)
\(m^2-6 m-1=0\)
Here, \(\mathrm{m}_1+\mathrm{m}_2=6\) and \(\mathrm{m}_1 \cdot \mathrm{m}_2=-1\).
\(\left|\mathrm{m}_1-\mathrm{m}_2\right|=\sqrt{\left(\mathrm{m}_1+\mathrm{m}_2\right)^2}-4 \mathrm{~m}_1 \mathrm{~m}_2=2 \sqrt{10}\)
Angle between tangent,
\(\tan \theta =\left|\frac{\mathrm{m}_2-\mathrm{m}_1}{1+\mathrm{m}_1 \mathrm{~m}_2}\right|\)
\(\tan \theta =\frac{2 \sqrt{10}}{1+(-1)}=\infty\)
\(\therefore \quad \theta=\frac{\pi}{2}\)
WB JEE-2022
Parabola
120264
The equation of the normal to the parabola \(y^2=\) \(4 x\) which is perpendicular to \(x+3 y+1=0\) is
1 \(3 x-y=33\)
2 \(3 x-y+33=0\)
3 \(3 x+y=33\)
4 \(3 x+y+33=0\)
Explanation:
A Given, \(\mathrm{y}^2=4 \mathrm{x}\)
Line,
\(x+3 y+1=0\)
\(y=\frac{-1}{3} x-\frac{1}{3}\)
Slope, \(\mathrm{m}_1=\frac{-1}{3}\)
Normal slope, \(\mathrm{m}_1 \cdot \mathrm{m}_2=-1\)
\(\mathrm{m}_2=3\)
Now, equation of normal to parabola is
\(y=m_2 x-2 a m_2-a m_2^3\)
\(y=3 x-2(1)(3)-1 \times(3)^3\)
\(y=3 x-6-27\)
\(3 x-y=33\)
AP EAMCET-21.04.2019
Parabola
120265
If the angle between the tangents drawn through the point \((-2,-1)\) to the parabola \(y^2-\) \(4 x\) is \(\theta\), then \(\tan 2 \theta=\)
1 3
2 -3
3 \(-\frac{3}{4}\)
4 \(\frac{3}{4}\)
Explanation:
C Given, parabola, \(\mathrm{y}^2=4 \mathrm{x}\)
We know that, \(\mathrm{s}_1^2=\mathrm{s} \cdot \mathrm{s}_1 \quad\{\) at \((-2,-1)\}\)
\([\mathrm{y}(-1)-2(\mathrm{x}-2)]^2=(1+8)\left(\mathrm{y}^2-4 \mathrm{x}\right)\)
\((2 \mathrm{x}+\mathrm{y}-4)^2=9\left(\mathrm{y}^2-4 \mathrm{x}\right)\)
\(4 \mathrm{x}^2+4 \mathrm{xy}-8 \mathrm{y}^2+20 \mathrm{x}-8 \mathrm{y}+16=0\)
\(\tan \theta=\left|\frac{2 \sqrt{\mathrm{h}^2-\mathrm{ab}}}{\mathrm{a}+\mathrm{b}}\right|=\left|\frac{2 \sqrt{4+32}}{4-8}\right|=3\)
\(\tan 2 \theta=\frac{2 \tan \theta}{1-\tan ^2 \theta}=\frac{2 \times 3}{1-9}\)
\(\tan 2 \theta=\frac{-3}{4}\)
So, equation of common tangent is \(\mathrm{y}=\mathrm{x}+2\).
120261
The Equation of the tangent to the parabola \(y\) \(=12 x\) at \((3,-6)\) is
1 \(x-y+9=0\)
2 \(x+y+3=0\)
3 \(x+y-3=0\)
4 \(x=3\)
Explanation:
B Given, parabola: \(\mathrm{y}^2=12 \mathrm{x}=4(3) \mathrm{x}\)
Here, \(a=3\),
Equation of tangent to this parabola is
\(\quad \mathrm{y}=\mathrm{mx}+\frac{\mathrm{a}}{\mathrm{m}}\)
\(\text { At point }(3,-6)\)
\(\quad-6=\mathrm{m}(3)+\frac{3}{\mathrm{~m}}\)
\(\quad-6 \mathrm{~m}=3 \mathrm{~m}^2+3\)
\(\Rightarrow \quad \mathrm{m}^2+2 \mathrm{~m}+1=0\)
\(\Rightarrow \quad \mathrm{m}+1)^2=0\)
\(\Rightarrow \quad \mathrm{m}=-1\)
Hence, equation of tangent is
\(\quad \mathrm{y}=(-1) \mathrm{x}+\frac{3}{(-1)}\)
\(\mathrm{x}+\mathrm{y}+3=0\)
At point \((3,-6)\)
Hence, equation of tangent is
\(\mathrm{y}=(-1) \mathrm{x}+\frac{3}{(-1)}\)
\(\mathrm{x}+\mathrm{y}+3=0\)
AP EAMCET-21.09.2020
Parabola
120262
If the line \(y=x\) is a tangent to the parabola \(y=\) \(a x^2+b x+c\) at the point \((1,1)\) and the curve passes through \((-1,0)\), then
C Given, parabola: \(y=a x^2+b x+c\)
And, \(\mathrm{y}=\mathrm{x}\) is tangent
Here, \(\mathrm{m}=1\)
Differentiate parabola w.r.t. \(x\)
\(\frac{\mathrm{dy}}{\mathrm{dx}}=2 \mathrm{ax}+\mathrm{b}\)
\(\mathrm{m}=2 \mathrm{ax}+\mathrm{b}\)
\(1=2 \mathrm{ax}+\mathrm{b}\)
At, \(\quad(1,1)\)
\(2 \mathrm{a}+\mathrm{b}=1\)
Also, parabola passes through \((1,1)\)
\(\therefore \quad 1=\mathrm{a}+\mathrm{b}+\mathrm{c}\)
Parabola passes through \((-1,0)\)
\(0=\mathrm{a}-\mathrm{b}+\mathrm{c}\)
From (i), (ii), and (iii)
\(\mathrm{a}=\mathrm{c}=\frac{1}{4} \text { and } \mathrm{b}=\frac{1}{2}\)
WB JEE-2020
Parabola
120263
From the point \((-1,-6)\), two tangents are drawn to \(y^2=4 x\). Then the angle between the two tangents is
1 \(\pi / 3\)
2 \(\pi / 4\)
3 \(\pi / 6\)
4 \(\pi / 2\)
Explanation:
D Given, from the point \((-1,-6)\) two tangents are drawn to \(\mathrm{y}^2=4 \mathrm{x}\)
Here, \(\quad \mathrm{a}=1\)
Equation of tangent
\(\mathrm{y}=\mathrm{mx}+\frac{1}{\mathrm{~m}} \quad\{\mathrm{a}=1\}\)
It passes through \((-1,-6)\)
\(-6=m(-1)+\frac{1}{(m)}\)
\(m^2-6 m-1=0\)
Here, \(\mathrm{m}_1+\mathrm{m}_2=6\) and \(\mathrm{m}_1 \cdot \mathrm{m}_2=-1\).
\(\left|\mathrm{m}_1-\mathrm{m}_2\right|=\sqrt{\left(\mathrm{m}_1+\mathrm{m}_2\right)^2}-4 \mathrm{~m}_1 \mathrm{~m}_2=2 \sqrt{10}\)
Angle between tangent,
\(\tan \theta =\left|\frac{\mathrm{m}_2-\mathrm{m}_1}{1+\mathrm{m}_1 \mathrm{~m}_2}\right|\)
\(\tan \theta =\frac{2 \sqrt{10}}{1+(-1)}=\infty\)
\(\therefore \quad \theta=\frac{\pi}{2}\)
WB JEE-2022
Parabola
120264
The equation of the normal to the parabola \(y^2=\) \(4 x\) which is perpendicular to \(x+3 y+1=0\) is
1 \(3 x-y=33\)
2 \(3 x-y+33=0\)
3 \(3 x+y=33\)
4 \(3 x+y+33=0\)
Explanation:
A Given, \(\mathrm{y}^2=4 \mathrm{x}\)
Line,
\(x+3 y+1=0\)
\(y=\frac{-1}{3} x-\frac{1}{3}\)
Slope, \(\mathrm{m}_1=\frac{-1}{3}\)
Normal slope, \(\mathrm{m}_1 \cdot \mathrm{m}_2=-1\)
\(\mathrm{m}_2=3\)
Now, equation of normal to parabola is
\(y=m_2 x-2 a m_2-a m_2^3\)
\(y=3 x-2(1)(3)-1 \times(3)^3\)
\(y=3 x-6-27\)
\(3 x-y=33\)
AP EAMCET-21.04.2019
Parabola
120265
If the angle between the tangents drawn through the point \((-2,-1)\) to the parabola \(y^2-\) \(4 x\) is \(\theta\), then \(\tan 2 \theta=\)
1 3
2 -3
3 \(-\frac{3}{4}\)
4 \(\frac{3}{4}\)
Explanation:
C Given, parabola, \(\mathrm{y}^2=4 \mathrm{x}\)
We know that, \(\mathrm{s}_1^2=\mathrm{s} \cdot \mathrm{s}_1 \quad\{\) at \((-2,-1)\}\)
\([\mathrm{y}(-1)-2(\mathrm{x}-2)]^2=(1+8)\left(\mathrm{y}^2-4 \mathrm{x}\right)\)
\((2 \mathrm{x}+\mathrm{y}-4)^2=9\left(\mathrm{y}^2-4 \mathrm{x}\right)\)
\(4 \mathrm{x}^2+4 \mathrm{xy}-8 \mathrm{y}^2+20 \mathrm{x}-8 \mathrm{y}+16=0\)
\(\tan \theta=\left|\frac{2 \sqrt{\mathrm{h}^2-\mathrm{ab}}}{\mathrm{a}+\mathrm{b}}\right|=\left|\frac{2 \sqrt{4+32}}{4-8}\right|=3\)
\(\tan 2 \theta=\frac{2 \tan \theta}{1-\tan ^2 \theta}=\frac{2 \times 3}{1-9}\)
\(\tan 2 \theta=\frac{-3}{4}\)
So, equation of common tangent is \(\mathrm{y}=\mathrm{x}+2\).
120261
The Equation of the tangent to the parabola \(y\) \(=12 x\) at \((3,-6)\) is
1 \(x-y+9=0\)
2 \(x+y+3=0\)
3 \(x+y-3=0\)
4 \(x=3\)
Explanation:
B Given, parabola: \(\mathrm{y}^2=12 \mathrm{x}=4(3) \mathrm{x}\)
Here, \(a=3\),
Equation of tangent to this parabola is
\(\quad \mathrm{y}=\mathrm{mx}+\frac{\mathrm{a}}{\mathrm{m}}\)
\(\text { At point }(3,-6)\)
\(\quad-6=\mathrm{m}(3)+\frac{3}{\mathrm{~m}}\)
\(\quad-6 \mathrm{~m}=3 \mathrm{~m}^2+3\)
\(\Rightarrow \quad \mathrm{m}^2+2 \mathrm{~m}+1=0\)
\(\Rightarrow \quad \mathrm{m}+1)^2=0\)
\(\Rightarrow \quad \mathrm{m}=-1\)
Hence, equation of tangent is
\(\quad \mathrm{y}=(-1) \mathrm{x}+\frac{3}{(-1)}\)
\(\mathrm{x}+\mathrm{y}+3=0\)
At point \((3,-6)\)
Hence, equation of tangent is
\(\mathrm{y}=(-1) \mathrm{x}+\frac{3}{(-1)}\)
\(\mathrm{x}+\mathrm{y}+3=0\)
AP EAMCET-21.09.2020
Parabola
120262
If the line \(y=x\) is a tangent to the parabola \(y=\) \(a x^2+b x+c\) at the point \((1,1)\) and the curve passes through \((-1,0)\), then
C Given, parabola: \(y=a x^2+b x+c\)
And, \(\mathrm{y}=\mathrm{x}\) is tangent
Here, \(\mathrm{m}=1\)
Differentiate parabola w.r.t. \(x\)
\(\frac{\mathrm{dy}}{\mathrm{dx}}=2 \mathrm{ax}+\mathrm{b}\)
\(\mathrm{m}=2 \mathrm{ax}+\mathrm{b}\)
\(1=2 \mathrm{ax}+\mathrm{b}\)
At, \(\quad(1,1)\)
\(2 \mathrm{a}+\mathrm{b}=1\)
Also, parabola passes through \((1,1)\)
\(\therefore \quad 1=\mathrm{a}+\mathrm{b}+\mathrm{c}\)
Parabola passes through \((-1,0)\)
\(0=\mathrm{a}-\mathrm{b}+\mathrm{c}\)
From (i), (ii), and (iii)
\(\mathrm{a}=\mathrm{c}=\frac{1}{4} \text { and } \mathrm{b}=\frac{1}{2}\)
WB JEE-2020
Parabola
120263
From the point \((-1,-6)\), two tangents are drawn to \(y^2=4 x\). Then the angle between the two tangents is
1 \(\pi / 3\)
2 \(\pi / 4\)
3 \(\pi / 6\)
4 \(\pi / 2\)
Explanation:
D Given, from the point \((-1,-6)\) two tangents are drawn to \(\mathrm{y}^2=4 \mathrm{x}\)
Here, \(\quad \mathrm{a}=1\)
Equation of tangent
\(\mathrm{y}=\mathrm{mx}+\frac{1}{\mathrm{~m}} \quad\{\mathrm{a}=1\}\)
It passes through \((-1,-6)\)
\(-6=m(-1)+\frac{1}{(m)}\)
\(m^2-6 m-1=0\)
Here, \(\mathrm{m}_1+\mathrm{m}_2=6\) and \(\mathrm{m}_1 \cdot \mathrm{m}_2=-1\).
\(\left|\mathrm{m}_1-\mathrm{m}_2\right|=\sqrt{\left(\mathrm{m}_1+\mathrm{m}_2\right)^2}-4 \mathrm{~m}_1 \mathrm{~m}_2=2 \sqrt{10}\)
Angle between tangent,
\(\tan \theta =\left|\frac{\mathrm{m}_2-\mathrm{m}_1}{1+\mathrm{m}_1 \mathrm{~m}_2}\right|\)
\(\tan \theta =\frac{2 \sqrt{10}}{1+(-1)}=\infty\)
\(\therefore \quad \theta=\frac{\pi}{2}\)
WB JEE-2022
Parabola
120264
The equation of the normal to the parabola \(y^2=\) \(4 x\) which is perpendicular to \(x+3 y+1=0\) is
1 \(3 x-y=33\)
2 \(3 x-y+33=0\)
3 \(3 x+y=33\)
4 \(3 x+y+33=0\)
Explanation:
A Given, \(\mathrm{y}^2=4 \mathrm{x}\)
Line,
\(x+3 y+1=0\)
\(y=\frac{-1}{3} x-\frac{1}{3}\)
Slope, \(\mathrm{m}_1=\frac{-1}{3}\)
Normal slope, \(\mathrm{m}_1 \cdot \mathrm{m}_2=-1\)
\(\mathrm{m}_2=3\)
Now, equation of normal to parabola is
\(y=m_2 x-2 a m_2-a m_2^3\)
\(y=3 x-2(1)(3)-1 \times(3)^3\)
\(y=3 x-6-27\)
\(3 x-y=33\)
AP EAMCET-21.04.2019
Parabola
120265
If the angle between the tangents drawn through the point \((-2,-1)\) to the parabola \(y^2-\) \(4 x\) is \(\theta\), then \(\tan 2 \theta=\)
1 3
2 -3
3 \(-\frac{3}{4}\)
4 \(\frac{3}{4}\)
Explanation:
C Given, parabola, \(\mathrm{y}^2=4 \mathrm{x}\)
We know that, \(\mathrm{s}_1^2=\mathrm{s} \cdot \mathrm{s}_1 \quad\{\) at \((-2,-1)\}\)
\([\mathrm{y}(-1)-2(\mathrm{x}-2)]^2=(1+8)\left(\mathrm{y}^2-4 \mathrm{x}\right)\)
\((2 \mathrm{x}+\mathrm{y}-4)^2=9\left(\mathrm{y}^2-4 \mathrm{x}\right)\)
\(4 \mathrm{x}^2+4 \mathrm{xy}-8 \mathrm{y}^2+20 \mathrm{x}-8 \mathrm{y}+16=0\)
\(\tan \theta=\left|\frac{2 \sqrt{\mathrm{h}^2-\mathrm{ab}}}{\mathrm{a}+\mathrm{b}}\right|=\left|\frac{2 \sqrt{4+32}}{4-8}\right|=3\)
\(\tan 2 \theta=\frac{2 \tan \theta}{1-\tan ^2 \theta}=\frac{2 \times 3}{1-9}\)
\(\tan 2 \theta=\frac{-3}{4}\)
So, equation of common tangent is \(\mathrm{y}=\mathrm{x}+2\).
120261
The Equation of the tangent to the parabola \(y\) \(=12 x\) at \((3,-6)\) is
1 \(x-y+9=0\)
2 \(x+y+3=0\)
3 \(x+y-3=0\)
4 \(x=3\)
Explanation:
B Given, parabola: \(\mathrm{y}^2=12 \mathrm{x}=4(3) \mathrm{x}\)
Here, \(a=3\),
Equation of tangent to this parabola is
\(\quad \mathrm{y}=\mathrm{mx}+\frac{\mathrm{a}}{\mathrm{m}}\)
\(\text { At point }(3,-6)\)
\(\quad-6=\mathrm{m}(3)+\frac{3}{\mathrm{~m}}\)
\(\quad-6 \mathrm{~m}=3 \mathrm{~m}^2+3\)
\(\Rightarrow \quad \mathrm{m}^2+2 \mathrm{~m}+1=0\)
\(\Rightarrow \quad \mathrm{m}+1)^2=0\)
\(\Rightarrow \quad \mathrm{m}=-1\)
Hence, equation of tangent is
\(\quad \mathrm{y}=(-1) \mathrm{x}+\frac{3}{(-1)}\)
\(\mathrm{x}+\mathrm{y}+3=0\)
At point \((3,-6)\)
Hence, equation of tangent is
\(\mathrm{y}=(-1) \mathrm{x}+\frac{3}{(-1)}\)
\(\mathrm{x}+\mathrm{y}+3=0\)
AP EAMCET-21.09.2020
Parabola
120262
If the line \(y=x\) is a tangent to the parabola \(y=\) \(a x^2+b x+c\) at the point \((1,1)\) and the curve passes through \((-1,0)\), then
C Given, parabola: \(y=a x^2+b x+c\)
And, \(\mathrm{y}=\mathrm{x}\) is tangent
Here, \(\mathrm{m}=1\)
Differentiate parabola w.r.t. \(x\)
\(\frac{\mathrm{dy}}{\mathrm{dx}}=2 \mathrm{ax}+\mathrm{b}\)
\(\mathrm{m}=2 \mathrm{ax}+\mathrm{b}\)
\(1=2 \mathrm{ax}+\mathrm{b}\)
At, \(\quad(1,1)\)
\(2 \mathrm{a}+\mathrm{b}=1\)
Also, parabola passes through \((1,1)\)
\(\therefore \quad 1=\mathrm{a}+\mathrm{b}+\mathrm{c}\)
Parabola passes through \((-1,0)\)
\(0=\mathrm{a}-\mathrm{b}+\mathrm{c}\)
From (i), (ii), and (iii)
\(\mathrm{a}=\mathrm{c}=\frac{1}{4} \text { and } \mathrm{b}=\frac{1}{2}\)
WB JEE-2020
Parabola
120263
From the point \((-1,-6)\), two tangents are drawn to \(y^2=4 x\). Then the angle between the two tangents is
1 \(\pi / 3\)
2 \(\pi / 4\)
3 \(\pi / 6\)
4 \(\pi / 2\)
Explanation:
D Given, from the point \((-1,-6)\) two tangents are drawn to \(\mathrm{y}^2=4 \mathrm{x}\)
Here, \(\quad \mathrm{a}=1\)
Equation of tangent
\(\mathrm{y}=\mathrm{mx}+\frac{1}{\mathrm{~m}} \quad\{\mathrm{a}=1\}\)
It passes through \((-1,-6)\)
\(-6=m(-1)+\frac{1}{(m)}\)
\(m^2-6 m-1=0\)
Here, \(\mathrm{m}_1+\mathrm{m}_2=6\) and \(\mathrm{m}_1 \cdot \mathrm{m}_2=-1\).
\(\left|\mathrm{m}_1-\mathrm{m}_2\right|=\sqrt{\left(\mathrm{m}_1+\mathrm{m}_2\right)^2}-4 \mathrm{~m}_1 \mathrm{~m}_2=2 \sqrt{10}\)
Angle between tangent,
\(\tan \theta =\left|\frac{\mathrm{m}_2-\mathrm{m}_1}{1+\mathrm{m}_1 \mathrm{~m}_2}\right|\)
\(\tan \theta =\frac{2 \sqrt{10}}{1+(-1)}=\infty\)
\(\therefore \quad \theta=\frac{\pi}{2}\)
WB JEE-2022
Parabola
120264
The equation of the normal to the parabola \(y^2=\) \(4 x\) which is perpendicular to \(x+3 y+1=0\) is
1 \(3 x-y=33\)
2 \(3 x-y+33=0\)
3 \(3 x+y=33\)
4 \(3 x+y+33=0\)
Explanation:
A Given, \(\mathrm{y}^2=4 \mathrm{x}\)
Line,
\(x+3 y+1=0\)
\(y=\frac{-1}{3} x-\frac{1}{3}\)
Slope, \(\mathrm{m}_1=\frac{-1}{3}\)
Normal slope, \(\mathrm{m}_1 \cdot \mathrm{m}_2=-1\)
\(\mathrm{m}_2=3\)
Now, equation of normal to parabola is
\(y=m_2 x-2 a m_2-a m_2^3\)
\(y=3 x-2(1)(3)-1 \times(3)^3\)
\(y=3 x-6-27\)
\(3 x-y=33\)
AP EAMCET-21.04.2019
Parabola
120265
If the angle between the tangents drawn through the point \((-2,-1)\) to the parabola \(y^2-\) \(4 x\) is \(\theta\), then \(\tan 2 \theta=\)
1 3
2 -3
3 \(-\frac{3}{4}\)
4 \(\frac{3}{4}\)
Explanation:
C Given, parabola, \(\mathrm{y}^2=4 \mathrm{x}\)
We know that, \(\mathrm{s}_1^2=\mathrm{s} \cdot \mathrm{s}_1 \quad\{\) at \((-2,-1)\}\)
\([\mathrm{y}(-1)-2(\mathrm{x}-2)]^2=(1+8)\left(\mathrm{y}^2-4 \mathrm{x}\right)\)
\((2 \mathrm{x}+\mathrm{y}-4)^2=9\left(\mathrm{y}^2-4 \mathrm{x}\right)\)
\(4 \mathrm{x}^2+4 \mathrm{xy}-8 \mathrm{y}^2+20 \mathrm{x}-8 \mathrm{y}+16=0\)
\(\tan \theta=\left|\frac{2 \sqrt{\mathrm{h}^2-\mathrm{ab}}}{\mathrm{a}+\mathrm{b}}\right|=\left|\frac{2 \sqrt{4+32}}{4-8}\right|=3\)
\(\tan 2 \theta=\frac{2 \tan \theta}{1-\tan ^2 \theta}=\frac{2 \times 3}{1-9}\)
\(\tan 2 \theta=\frac{-3}{4}\)
So, equation of common tangent is \(\mathrm{y}=\mathrm{x}+2\).
