The tangents to the curve y=(x-2)^2-1 at its points of intersection with the line x-y=3, i…

Question

The tangents to the curve $y=(x-2)^2-1$ at its points of intersection with the line $x-y=3$, intersect at the point:

Accepted Answer

Let the coordinates of C be (h, k)

So the chord of contact of C w.r.t y = (x-2)2 - 1

$$ \Rightarrow $$ y = x2 - 4x + 3 is T = 0

$$y+k \over 2$$ = xh + 3 - 2(x+h)

$$ \Rightarrow $$ 2(h-2)x - y = -(6-4h-k)

On comparing it with x - y = 3

2 (h -2) = 1 $$ \Rightarrow $$ h = $$5 \over 2$$

6 - 4h - k = -3 $$ \Rightarrow $$ k = -1

$$ \therefore $$ C = ($$5 \over 2$$, -1)

Solutions · 1