Find the equation of the tangents drawn from the point (1,3) to the parabola y^2 =−16x.


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Question Number 157487 by ZiYangLee last updated on 23/Oct/21

$Find the equation of the tangents drawn$ $from the point (1, 3) to the parabola$ $y^{2} = - 16 x .$
	Answered by mr W last updated on 23/Oct/21

	$s a y t h e t a n g e n t l i n e i s$ $x = 1 + k (y - 3)$ $y^{2} = - 16 [1 + k (y - 3)]$ $y^{2} + 16 k y + 16 (1 - 3 k) = 0$ $i t s h o u l d h a v e o n l y o n e r o o t .$ ${(16 k)}^{2} - 4 \times 16 (1 - 3 k) = 0$ $4 k^{2} + 3 k - 1 = 0$ $(4 k - 1) (k + 1) = 0$ $\Rightarrow k = - 1, \frac{1}{4}$ $x = 1 - (y - 3) \Rightarrow y = - x + 4$ $x = 1 + \frac{y - 3}{4} \Rightarrow y = 4 x - 1$
	Commented by mr W last updated on 23/Oct/21

	Answered by cortano last updated on 24/Oct/21
	$let (x,y) is contact point of tangent line . we have { ((m=y′=−(8/y))),((m=((y−3)/(x−1))=((y−3)/(−(y^2 /(16))−1)))) :} then −(8/y)=−((16(y−3))/(y^2 +16)) ⇒y^2 +16=2y(y−3) ⇒y^2 +16=2y^2 −6y ⇒y^2 −6y−16=0 ⇒(y−8)(y+2)=0 { ((y=8→x=−4)),((y=−2→x=−(1/4))) :} we get two of tangent line are { ((at (−4,8) ⇒m=−1; x+y=4)),((at (−(1/4),−2)⇒m=4 ; 4x−y=1)) :}$
	$l e t (x, y) i s c o n t a c t p o i n t o f t a n g e n t$ $l i n e . w e h a v e {\begin{cases} m = y^{'} = - \frac{8}{y} \\ m = \frac{y - 3}{x - 1} = \frac{y - 3}{- \frac{y^{2}}{16} - 1} \end{cases}$ $t h e n - \frac{8}{y} = - \frac{16 (y - 3)}{y^{2} + 16}$ $\Rightarrow y^{2} + 16 = 2 y (y - 3)$ $\Rightarrow y^{2} + 16 = 2 y^{2} - 6 y$ $\Rightarrow y^{2} - 6 y - 16 = 0$ $\Rightarrow (y - 8) (y + 2) = 0$ ${\begin{cases} y = 8 \to x = - 4 \\ y = - 2 \to x = - \frac{1}{4} \end{cases}$ $w e g e t t w o o f t a n g e n t l i n e a r e$ ${\begin{cases} a t (- 4, 8) \Rightarrow m = - 1; x + y = 4 \\ a t (- \frac{1}{4}, - 2) \Rightarrow m = 4; 4 x - y = 1 \end{cases}$
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