Find the equation of the tangent to the curve x2 - xy - y2 + 4 = 0 at the point (2, 4).

Question

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To find the equation of the tangent to a curve at a given point, the point must lie on the curve. Let's first check if the point $k(2, 4)$ lies on the curve $x^2 - xy - y^2 + 4 = 0$ .

Substitute $x=2$ and $y=4$ into the equation:

(2)^2 - (2)(4) - (4)^2 + 4

4 - 8 - 16 + 4

-4 - 16 + 4

-20 + 4 = -16

Since $-16 \neq 0$ , the point $(2, 4)$ does not lie on the curve $x^2 - xy - y^2 + 4 = 0$ . Therefore, a tangent line at this specific point cannot be found in the standard way.

It is possible there is a typo in the question. If we assume $x=2$ , we can find the $y$ -coordinates that are on the curve:

2^2 - 2y - y^2 + 4 = 0

4 - 2y - y^2 + 4 = 0

-y^2 - 2y + 8 = 0

y^2 + 2y - 8 = 0

Factoring the quadratic equation:

(y+4)(y-2) = 0

This gives $y = -4$ or $y = 2$ . So, the points $(2, -4)$ and $(2, 2)$ are on the curve.

Let's assume the question intended for the point to be $(2, -4)$ and solve for the tangent at this point.

Differentiate the equation implicitly with respect to $x$ .
The equation is $x^2 - xy - y^2 + 4 = 0$ .
Differentiating each term:
$\frac{d}{dx}(x^2) - \frac{d}{dx}(xy) - \frac{d}{dx}(y^2) + \frac{d}{dx}(4) = \frac{d}{dx}(0)$ $2x - \left(1 \cdot y + x \cdot \frac{dy}{dx}\right) - 2y \frac{dy}{dx} + 0 = 0$ $2x - y - x\frac{dy}{dx} - 2y\frac{dy}{dx} = 0$
Rearrange the equation to solve for $\frac{dy}{dx}$ .
Group terms containing $\frac{dy}{dx}$ :
$2x - y = x\frac{dy}{dx} + 2y\frac{dy}{dx}$
Factor out $\frac{dy}{dx}$ :
$2x - y = (x + 2y)\frac{dy}{dx}$
Solve for $\frac{dy}{dx}$ :
$\frac{dy}{dx} = \frac{2x - y}{x + 2y}$
Evaluate $\frac{dy}{dx}$ at the point $(2, -4)$ to find the slope ( $m$ ) of the tangent.
Substitute $x=2$ and $y=-4$ :
$m = \frac{2(2) - (-4)}{2 + 2(-4)} = \frac{4 + 4}{2 - 8} = \frac{8}{-6} = -\frac{4}{3}$

Find the equation of the tangent to the curve x2 − xy − y2 + 4 = 0 at the point (2, 4).

Step-by-step solution

Differentiate the equation implicitly with respect to $x$ .

Rearrange the equation to solve for $\frac{dy}{dx}$ .

Evaluate $\frac{dy}{dx}$ at the point $(2, -4)$ to find the slope ( $m$ ) of the tangent.

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