[tex]\qquad\qquad\huge\underline{{\sf Answer}}♨[/tex]
Let's solve ~
If it passes through x, then let's find x when y = 0
[tex]\qquad \tt \dashrightarrow \:0 = {x}^{2} - x - 6[/tex]
[tex]\qquad \tt \dashrightarrow \: {x}^{2} - 3x + 2x - 6 = 0[/tex]
[tex]\qquad \tt \dashrightarrow \:x(x - 3) + 2(x - 3) = 0[/tex]
[tex]\qquad \tt \dashrightarrow \:(x - 3) (x + 2) = 0[/tex]
So, required values of x are 3 and -2
Now, let's differentiate the equation to get slope slope for tangent ~
[tex]\qquad \tt \dashrightarrow \: m = \dfrac{d}{dx} ( {x}^{2} - x - 6)[/tex]
[tex]\qquad \tt \dashrightarrow \: m = 2 x - 1[/tex]
Now, plug in the values of x to find slopes of tangents
[tex]\qquad \tt \dashrightarrow \: m _1 = (2 \times 3) - 1[/tex]
[tex]\qquad \tt \dashrightarrow \: m_1 = 6 - 1[/tex]
[tex]\qquad \tt \dashrightarrow \: m _1 = 5[/tex]
and
[tex]\qquad \tt \dashrightarrow \:m_2 = (2 \times - 2) - 1[/tex]
[tex]\qquad \tt \dashrightarrow \:m_2 = - 4- 1[/tex]
[tex]\qquad \tt \dashrightarrow \:m_2 = - 5[/tex]
We know, tangent are normal are perpendicular. so let's find out slopes of normals m1' and m2'
[tex]\qquad \tt \dashrightarrow \:m _1\cdot m_1 ' = - 1[/tex]
[tex]\qquad \tt \dashrightarrow \:m _1' = \dfrac{- 1}{m_1}[/tex]
[tex]\qquad \tt \dashrightarrow \:m _1' = \dfrac{- 1}{ 5}[/tex]
and
[tex]\qquad \tt \dashrightarrow \:m _2\cdot m_2 ' = - 1[/tex]
[tex]\qquad \tt \dashrightarrow \:m _2' = \dfrac{- 1}{m_2}[/tex]
[tex]\qquad \tt \dashrightarrow \:m _2' = \dfrac{- 1}{ -5}[/tex]
[tex]\qquad \tt \dashrightarrow \:m _2' = \dfrac{1}{ 5}[/tex]
Now, write the equations of normals using point slope form :
Normal 1 : passing through (3 , 0), and slope = -1/5
[tex]\qquad \tt \dashrightarrow \:y - 0 = - \dfrac{1}{5} (x - 3)[/tex]
[tex]\qquad \tt \dashrightarrow \:5y = - (x - 3)[/tex]
[tex]\qquad \tt \dashrightarrow \:5y = - x + 3[/tex]
[tex]\qquad \tt \dashrightarrow \:x + 5y - 3 = 0[/tex]
and
Normal 2 : passing through (-2 , 0), and slope = 1/5
[tex]\qquad \tt \dashrightarrow \:y - 0 = \dfrac{1}{5} (x - ( - 2))[/tex]
[tex]\qquad \tt \dashrightarrow \:5y = x + 2[/tex]
[tex]\qquad \tt \dashrightarrow \ x - 5y + 2 = 0[/tex]
That's all for Aunty ~ hope it helps !
