Find the point of intersection of the two tangents at the end points of the latusrectum of the parabola $(y+3)^2=8(x-2)$

Question

Equation of latusrectum

$$X-a=0$$$$x-2-2=0$$ $$x=4$$

Therefore $$(y+3)^2=8(4-2)$$ $$y=1,-7$$ Then point from which tangents are drawn are $(4,1)$ and $(4,-7)$.

Their point intersection will $(-4,-3)$, using the GM and AM property of point of intersection of tangent.

The answer is (0,-3). How is the right answer?

Answer 1

Note that the tangent of the parabola is,

$$y' = \frac4{y+3}$$

So, at the two ends of the latus rectum with $y=1,-7$, they are $y'= 1,-1$, respectively. Hence, the corresponding equations of the two tangent lines are,

$$y-1=x-4,\>\>\>\>\>\>y+7 = -1(x-4)$$

They intersect at the point $(0,-3)$.