Find the equation of the line which is tangent to the parabola y^2 =12x and forms an angle of 45° with the line y=3x−4.


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Question Number 183241 by depressiveshrek last updated on 23/Dec/22

$F i n d t h e e q u a t i o n o f t h e l i n e w h i c h$ $i s t a n g e n t t o t h e p a r a b o l a y^{2} = 12 x$ $a n d f o r m s a n a n g l e o f 45 ° w i t h$ $t h e l i n e y = 3 x - 4 .$
	Answered by cortano1 last updated on 24/Dec/22
	$Let the line is y=mx+c and 1=∣((m−3)/(1+3m))∣ ⇒∣m−3∣=∣3m+1∣ ⇒(4m−2)(2m+4)=0 ; m=−2; (1/2) the line which tangent to y^2 =12x ⇒(mx+c)^2 = 12x ⇒m^2 x^2 +(2mc−12)x+c^2 =0 ,Δ=0 ⇒(2mc−12)^2 −(2mc)^2 =0 ⇒(4mc−12)(−12)=0 ⇒c=(3/m) = −(3/2) ; 6 ∴ { ((y=−2x−(3/2))),((y=(1/2)x+6 )) :}$
	$L e t t h e l i n e i s y = m x + c$ $a n d 1 =∣ \frac{m - 3}{1 + 3 m} ∣ \Rightarrow∣ m - 3 ∣=∣ 3 m + 1 ∣$ $\Rightarrow (4 m - 2) (2 m + 4) = 0; m = - 2; \frac{1}{2}$ $t h e l i n e w h i c h t a n g e n t t o y^{2} = 12 x$ $\Rightarrow {(m x + c)}^{2} = 12 x$ $\Rightarrow m^{2} x^{2} + (2 m c - 12) x + c^{2} = 0, Δ = 0$ $\Rightarrow {(2 m c - 12)}^{2} - {(2 m c)}^{2} = 0$ $\Rightarrow (4 m c - 12) (- 12) = 0$ $\Rightarrow c = \frac{3}{m} = - \frac{3}{2}; 6$ $∴ {\begin{cases} y = - 2 x - \frac{3}{2} \\ y = \frac{1}{2} x + 6 \end{cases}$
	Commented by cortano1 last updated on 24/Dec/22

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