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Chapter 14: Problem 315

Find the equation of the tangent line to the ellipse: \(4 x^{2}+9 y^{2}=40\), at the point \((1,2)\)

Short Answer

Expert verified
The equation of the tangent line to the ellipse \(4x^2 + 9y^2 = 40\) at the point (1,2) is \(y = -\frac{2}{9}x + \frac{20}{9}\).

Step by step solution

01

Calculate the gradient of the ellipse at the given point

To calculate the gradient, we will first implicitly differentiate the equation of the ellipse with respect to x and then substitute the point (1,2) into the result. The given ellipse equation is \(4x^2 + 9y^2 = 40\). Implicit differentiation with respect to x: \(8x + 18y \frac{dy}{dx} = 0\) Now, we plug in the point (1,2): \(8(1) + 18(2) \frac{dy}{dx} = 0\)
02

Solve for the gradient \(\frac{dy}{dx}\)

We now have a linear equation for \(\frac{dy}{dx}\), which we can solve for: \(8 + 36 \frac{dy}{dx} = 0\) \(\frac{dy}{dx} = -\frac{8}{36}\) \(\frac{dy}{dx} = -\frac{2}{9}\) The gradient at the point (1,2) is -2/9.
03

Use the point-slope form to find the equation of the tangent line

The point-slope form of a linear equation is \(y - y_1 = m(x - x_1)\), where m is the slope (gradient) and (x_1, y_1) is a point on the line. We know the slope is -2/9 and the point (1,2) is on the tangent line. Therefore: \(y - 2 = -\frac{2}{9}(x - 1)\) Now we will simplify the equation to the slope-intercept form \(y = mx + b\): \(y - 2 = -\frac{2}{9}x + \frac{2}{9}\) \(y = -\frac{2}{9}x + \frac{2}{9} + 2\) \(y = -\frac{2}{9}x + \frac{20}{9}\)
04

Final Answer

The equation of the tangent line to the ellipse at the point (1,2) is: \(y = -\frac{2}{9}x + \frac{20}{9}\)

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Most popular questions from this chapter

Find the angle of intersection of the circles: \(\mathrm{x}^{2}+\mathrm{y}^{2}-4 \mathrm{x}=1\) \(\mathrm{x}^{2}+\mathrm{y}^{2}-2 \mathrm{y}=9 .\)

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