Applications of Derivatives Question - None

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

Solution:

Find points of intersection of $y = (x-2)^2 - 1$ and $x - y = 3$.

From line: $y = x - 3$

Substituting: $x - 3 = (x-2)^2 - 1$

$x - 3 = x^2 - 4x + 4 - 1$

$x - 3 = x^2 - 4x + 3$

$0 = x^2 - 5x + 6$

$0 = (x-2)(x-3)$

So $x = 2$ or $x = 3$

When $x = 2: y = -1$, point is $(2, -1)$

When $x = 3: y = 0$, point is $(3, 0)$

For curve $y = (x-2)^2 - 1$, we have $\frac{dy}{dx} = 2(x-2)$

At $(2, -1)$: slope = $0$

Tangent: $y = -1$

At $(3, 0)$: slope = $2$

Tangent: $y - 0 = 2(x - 3)$, i.e., $y = 2x - 6$

Intersection of $y = -1$ and $y = 2x - 6$:

$-1 = 2x - 6$

$2x = 5$

$x = \frac{5}{2}$

Point of intersection: $\left(\frac{5}{2}, -1\right)$

∴ Answer is (a).

Question: