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Question: Find the angle between two tangents drawn to the parabola \(y = {x^2}\) from the point (0 ,-2)....

Find the angle between two tangents drawn to the parabola y=x2y = {x^2} from the point (0 ,-2).

Explanation

Solution

Hint:- We should have knowledge that equation of pair of tangent is SS1=T2S{S_1} = {T^2} and also have knowledge that angle between two lines with slope m1{m_1} and m2{m_2} is α=tan1m1m21+m1m2\alpha = {\tan ^{ - 1}}\left| {\dfrac{{{m_1} - {m_2}}}{{1 + {m_1}{m_2}}}} \right|.

Complete step-by-step answer:
We have
Parabola: y=x2y = {x^2}
We have to draw tangent from (0 ,-2)

We know formulae of equation pair of tangents from a point is SS1=T2S{S_1} = {T^2}
Where SS is the equation of curve given and S1{S_1} is the equation obtained when we put a given point in the equation of curve from where we have to draw a pair of tangents. T is the equation of tangent to the curve from that given point.
Now putting the value of all terms in formula SS1=T2S{S_1} = {T^2} we get,
Here for writing equation of tangent we should know shortcut as like we write (y=y+y12,x2=xx1)\left( {y = \dfrac{{y + {y_1}}}{2},{x^2} = x{x_1}} \right)
We use these shortcuts for writing the equation of tangent to a curve where x1,y1{x_1},{y_1} are the points from which the tangent has to be drawn.
(T=x.0(y2)2=(y2)2)\left( {\therefore T = x.0 - \dfrac{{\left( {y - 2} \right)}}{2} = - \dfrac{{\left( {y - 2} \right)}}{2}} \right)
(x2y)(0(2))=[0(y2)2]2\Rightarrow \left( {{x^2} - y} \right)\left( {0 - \left( { - 2} \right)} \right) = {\left[ {0 - \dfrac{{\left( {y - 2} \right)}}{2}} \right]^2}
4(2x22y)=y2+44y\Rightarrow 4\left( {2{x^2} - 2y} \right) = {y^2} + 4 - 4y
8x24yy24=0 8x2(y2+4y+4)=0 8x2(y+2)2=0  \Rightarrow 8{x^2} - 4y - {y^2} - 4 = 0 \\\ \Rightarrow 8{x^2} - \left( {{y^2} + 4y + 4} \right) = 0 \\\ \Rightarrow 8{x^2} - {\left( {y + 2} \right)^2} = 0 \\\
Now use the formulae (a2b2=(a+b)(ab))\left( {{a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)} \right)
(22xy2)(22x+y+2)=0\Rightarrow \left( {2\sqrt 2 x - y - 2} \right)\left( {2\sqrt 2 x + y + 2} \right) = 0
So equations of tangents,
T1:22xy2=0{T_1}:2\sqrt 2 x - y - 2 = 0 and T2:22x+y+2=0{T_2}:2\sqrt 2 x + y + 2 = 0
Now for slope of a straight line we compare with ( y = mx + c )
Where m is the slope of the line.
Slope of T1=m1=22{T_1} = {m_1} = 2\sqrt 2
Slope of T2=m2=22{T_2} = {m_2} = - 2\sqrt 2
Now angle between these tangents, α=tan1[m1m21+m1m2]\alpha = {\tan ^{ - 1}}\left[ {\dfrac{{{m_1} - {m_2}}}{{1 + {m_1}{m_2}}}} \right]
α=tan122+2218\Rightarrow \alpha = {\tan ^{ - 1}}\left| {\dfrac{{2\sqrt 2 + 2\sqrt 2 }}{{1 - 8}}} \right|=tan1(427){\tan ^{ - 1}}\left( {\dfrac{{4\sqrt 2 }}{7}} \right)

Note: Whenever we get this type of question the key concept of solving is you have to write an equation of pair of tangents and from that equation we have to find slope of the tangents and then using formula find angle between tangents.