Question: Question 15
Find the equation of the tangent line to the curve (a lemniscate) $$2(x^2 + y^2)^2 = 25 (x^2 - y^2)$$ at the point $$(-3,1)$$. The equation of this tangent line can be written in the form $$y = mx + b$$ where $$m$$ is:
and where $$b$$ is:
Question 16
Find the slope of the tangent line to the curve (a lemniscate)
$$2(x^2 + y^2)^2 = 25 (x^2 - y^2)$$
at the point $$(3,1)$$
slope =
1. **State the problem:** Find the equation of the tangent line to the lemniscate curve $$2(x^2 + y^2)^2 = 25(x^2 - y^2)$$ at the point $$(-3,1)$$ and find the slope of the tangent line at the point $$(3,1)$$.
2. **Recall the formula:** To find the slope of the tangent line, we implicitly differentiate the equation with respect to $$x$$.
3. **Differentiate implicitly:**
$$\text{Given } 2(x^2 + y^2)^2 = 25(x^2 - y^2)$$
Differentiate both sides:
$$\frac{d}{dx}\left[2(x^2 + y^2)^2\right] = \frac{d}{dx}\left[25(x^2 - y^2)\right]$$
Using chain rule:
$$2 \cdot 2(x^2 + y^2) \cdot \frac{d}{dx}(x^2 + y^2) = 25 \cdot \frac{d}{dx}(x^2 - y^2)$$
Calculate derivatives inside:
$$4(x^2 + y^2)(2x + 2y \frac{dy}{dx}) = 25(2x - 2y \frac{dy}{dx})$$
Simplify:
$$4(x^2 + y^2)(2x + 2y y') = 25(2x - 2y y')$$
Where $$y' = \frac{dy}{dx}$$.
4. **Expand and group terms:**
$$8x(x^2 + y^2) + 8y(x^2 + y^2) y' = 50x - 50y y'$$
Group $$y'$$ terms on one side:
$$8y(x^2 + y^2) y' + 50y y' = 50x - 8x(x^2 + y^2)$$
Factor $$y'$$:
$$y' \left[8y(x^2 + y^2) + 50y\right] = 50x - 8x(x^2 + y^2)$$
5. **Solve for $$y'$$:**
$$y' = \frac{50x - 8x(x^2 + y^2)}{8y(x^2 + y^2) + 50y}$$
6. **Evaluate slope at $$(-3,1)$$:**
Calculate $$x^2 + y^2 = (-3)^2 + 1^2 = 9 + 1 = 10$$
Substitute:
$$y' = \frac{50(-3) - 8(-3)(10)}{8(1)(10) + 50(1)} = \frac{-150 + 240}{80 + 50} = \frac{90}{130} = \frac{9}{13}$$
So, slope $$m = \frac{9}{13}$$.
7. **Find $$b$$ using point-slope form:**
Equation:
$$y = mx + b$$
Substitute $$x = -3$$, $$y = 1$$, and $$m = \frac{9}{13}$$:
$$1 = \frac{9}{13}(-3) + b$$
$$1 = -\frac{27}{13} + b$$
$$b = 1 + \frac{27}{13} = \frac{13}{13} + \frac{27}{13} = \frac{40}{13}$$
8. **Equation of tangent line at $$(-3,1)$$:**
$$y = \frac{9}{13}x + \frac{40}{13}$$
9. **Evaluate slope at $$(3,1)$$ for Question 16:**
Calculate $$x^2 + y^2 = 3^2 + 1^2 = 9 + 1 = 10$$
Substitute:
$$y' = \frac{50(3) - 8(3)(10)}{8(1)(10) + 50(1)} = \frac{150 - 240}{80 + 50} = \frac{-90}{130} = -\frac{9}{13}$$
**Final answers:**
- Question 15 slope $$m = \frac{9}{13}$$, intercept $$b = \frac{40}{13}$$.
- Question 16 slope $$= -\frac{9}{13}$$.