6.7 Q-2

Question Statement

Write the equation of the tangent to the given conic at the indicated point.

i. $3x^2 = -16y$ at the points where the ordinate $y = -3$
ii. $3x^2 - 7y^2 = 20$ at the point where $y = -1$

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Background and Explanation

In this problem, we are tasked with finding the equations of tangents to conics at specific points. The key steps involve:

1. Identifying the points on the conic that satisfy the given conditions.
2. Differentiating the conic equation to find the slope of the tangent at each point.
3. Using the point-slope form of the equation of a line to find the equation of the tangent.

The point-slope form of a line is:

$$y - y_1 = m(x - x_1)$$

where $m$ is the slope of the tangent, and $(x_1, y_1)$ is the point of tangency.

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Solution

i. $3x^2 = -16y$ at the points where $y = -3$

1. Find the points on the conic:
   Given the equation $3x^2 = -16y$, substitute $y = -3$:

$$3x^2 = -16(-3) \quad \Rightarrow \quad 3x^2 = 48 \quad \Rightarrow \quad x^2 = 16 \quad \Rightarrow \quad x = \pm 4$$

Therefore, the points on the conic are $(4, -3)$ and $(-4, -3)$.

2. Find the tangent at $(4, -3)$:
   To find the tangent, replace $x^2$ by $4x$ and $y$ by $\frac{1}{2}(y - 3)$ in the equation $3x^2 = -16y$:

$$3(4x) = -16\left(\frac{1}{2}(y - 3)\right)$$

Simplifying:

$$12x = -8(y - 3) \quad \Rightarrow \quad 3x = -2(y - 3) \quad \Rightarrow \quad 3x + 2y - 6 = 0$$

Therefore, the equation of the tangent at $(4, -3)$ is:

$$3x + 2y - 6 = 0$$

3. Find the tangent at $(-4, -3)$:
   Now, replace $x^2$ by $-4x$ and $y$ by $\frac{1}{2}(y - 3)$:

$$3(-4x) = -16\left(\frac{1}{2}(y - 3)\right)$$

Simplifying:

$$-12x = -8(y - 3) \quad \Rightarrow \quad 3x = 2(y - 3) \quad \Rightarrow \quad 3x - 2y + 6 = 0$$

Therefore, the equation of the tangent at $(-4, -3)$ is:

$$3x - 2y + 6 = 0$$

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ii. $3x^2 - 7y^2 = 20$ at the point where $y = -1$

1. Find the points on the conic:
   Substitute $y = -1$ into the equation $3x^2 - 7y^2 = 20$:

$$3x^2 - 7(1)^2 = 20 \quad \Rightarrow \quad 3x^2 - 7 = 20 \quad \Rightarrow \quad 3x^2 = 27 \quad \Rightarrow \quad x^2 = 9 \quad \Rightarrow \quad x = \pm 3$$

Therefore, the points on the conic are $(3, -1)$ and $(-3, -1)$.

2. Differentiate the equation:
   Differentiating $3x^2 - 7y^2 = 20$ with respect to $x$:

$$\frac{d}{dx}(3x^2 - 7y^2) = \frac{d}{dx}(20)$$

This gives:

$$6x - 14y \frac{dy}{dx} = 0 \quad \Rightarrow \quad 14y \frac{dy}{dx} = 6x \quad \Rightarrow \quad \frac{dy}{dx} = \frac{3x}{7y}$$

At the point $(3, -1)$, the slope is:

$$\frac{dy}{dx} = \frac{3(3)}{7(-1)} = \frac{9}{-7} = -\frac{9}{7}$$

At the point $(-3, -1)$, the slope is:

$$\frac{dy}{dx} = \frac{3(-3)}{7(-1)} = \frac{-9}{-7} = \frac{9}{7}$$

3. Find the tangent at $(3, -1)$:
   Using the point-slope form for the tangent line at $(3, -1)$:

$$y - (-1) = -\frac{9}{7}(x - 3)$$

Simplifying:

$$7y + 7 = -9x + 27 \quad \Rightarrow \quad 9x + 7y - 20 = 0$$

Therefore, the equation of the tangent at $(3, -1)$ is:

$$9x + 7y - 20 = 0$$

4. Find the tangent at $(-3, -1)$:
   Using the point-slope form for the tangent line at $(-3, -1)$:

$$y - (-1) = \frac{9}{7}(x - (-3))$$

Simplifying:

$$7y + 7 = 9x + 27 \quad \Rightarrow \quad 9x - 7y + 20 = 0$$

Therefore, the equation of the tangent at $(-3, -1)$ is:

$$9x - 7y + 20 = 0$$

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Key Formulas or Methods Used

1. Point-Slope Form of the Line:
   The equation of a line through a point $(x_1, y_1)$ with slope $m$ is:

$$y - y_1 = m(x - x_1)$$

2. Differentiation to Find the Slope of the Tangent:
   For a curve $f(x, y) = 0$, the slope of the tangent at a point is:

$$\frac{dy}{dx} = \frac{-\text{(partial derivative with respect to } x)}{\text{(partial derivative with respect to } y)}$$

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Summary of Steps

1. For part (i), substitute the given ordinate into the equation to find the corresponding x-values, then substitute into the equation for the tangent using the point-slope form.
2. For part (ii), differentiate the equation of the conic, substitute the coordinates of the points, and then use the point-slope form to find the tangent lines at the given points.
3. Simplify the equations for the tangents at each point to get the final answer.