Implicit Derivative of an Ellipse Calculator

Enter the two positive semi-axis lengths, the point coordinates, and an optional center. The result includes the ellipse equation, the formula for dy/dx, the tangent classification, and either a numeric slope and point-slope tangent line or a vertical tangent line. The point is checked before any slope is reported, preventing an unrelated coordinate from producing a plausible but invalid answer.

Describe the ellipse and the point

Use semi_axis_x for the horizontal semi-axis a and semi_axis_y for the vertical semi-axis b. Both are lengths, so they must be positive finite numbers. The optional center_x and center_y fields represent h and k; each defaults to zero, giving an ellipse centered at the origin. Finally, provide the x and y coordinates of the point where you want the tangent slope. The supported equation is (x − h)²/a² + (y − k)²/b² = 1, with axes parallel to the coordinate axes. Enter semi-axis lengths, not full width and height: an ellipse that is ten units wide and six units tall has a = 5 and b = 3. The calculator substitutes the supplied point into the normalized equation and requires the result to equal one within a small numerical tolerance. This check matters because the derivative formula describes slopes only on the curve. A coordinate near the ellipse may look reasonable in a graph but cannot define its tangent unless it actually belongs to the ellipse. Rotated ellipses and equations containing an xy term are outside this input model.

Follow the implicit differentiation

Start with (x − h)²/a² + (y − k)²/b² = 1 and differentiate both sides with respect to x. The first term becomes 2(x − h)/a². The second term contains y, which depends on x, so the chain rule produces 2(y − k)(dy/dx)/b². The derivative of the constant on the right is zero. After canceling the common factor of two, move the x term to the other side and isolate dy/dx. The result is dy/dx = −b²(x − h) divided by a²(y − k). The output preserves this structure in derivative_formula and then substitutes the supplied coordinates to compute slope. For a = 5, b = 3, center (0, 0), and point (4, 1.8), the slope is −0.8. The returned point-slope equation is a direct tangent-line representation, so you can expand it into slope-intercept or standard form if an exercise requires another format. Keeping the derivative formula beside the evaluated value makes it easier to audit signs, squared axes, center offsets, and chain-rule placement instead of treating the answer as an unexplained decimal.

Interpret finite, horizontal, and vertical tangents

Most valid points produce a finite numeric slope and a tangent line in point-slope form. When x equals the center coordinate h, the numerator of the derivative is zero; provided y differs from k, the tangent is horizontal and the calculator labels tangent_type as horizontal. At the leftmost or rightmost point, y equals k, so the derivative formula has a zero denominator. The tangent is vertical there, and an ordinary numeric dy/dx does not exist. In that case the output deliberately omits slope and returns a line such as x = 5. Omitting an undefined value is safer than encoding it as null, infinity, or an enormous approximation that downstream software might mistake for a real slope. The point check uses a dimensionless tolerance of 1e-9 to accommodate ordinary decimal coordinates, but it does not project approximate points onto the curve or repair inconsistent measurements. Extremely large values that overflow finite arithmetic are rejected. The algorithm uses no network, random values, saved state, or clock, so identical inputs always return identical structured results suitable for coursework checks and automated calculations.