6.7 Q-3

Question Statement

Find the equation of the tangent that passes through the given points to the given conics.

i. $x^2 + y^2 = 25$ through $(7, -1)$

ii. $y^2 = 12x$ through $(1, 4)$

iii. $x^2 - 2y^2 = 2$ through $(1, -2)$

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Background and Explanation

To solve this problem, we need to find the equation of the tangent line to each of the given conics at the specified points. This involves the following steps:

1. Differentiating the conic equation to find the slope of the tangent line at the point of interest.
2. Using the point-slope form of the equation of a line to write the equation of the tangent line.

The point-slope form of a line is given by:

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

where:

• $m$ is the slope of the tangent line at the point $(x_1, y_1)$,
• $(x_1, y_1)$ is the point on the conic.

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Solution

i. $x^2 + y^2 = 25$ through $(7, -1)$

1. Differentiate the equation of the circle:
   The equation of the circle is:

$$x^2 + y^2 = 25$$

Differentiating both sides with respect to $x$:

$$2x + 2y \frac{dy}{dx} = 0$$

Solving for $\frac{dy}{dx}$:

$$\frac{dy}{dx} = -\frac{x}{y}$$

2. Find the slope at the point $(7, -1)$:
   Substituting $x = 7$ and $y = -1$ into the derivative:

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

So, the slope of the tangent is $m = 7$.

3. Use the point-slope form to find the equation of the tangent:
   Using the point $(7, -1)$ and the slope $m = 7$, the equation of the tangent is:

$$y - (-1) = 7(x - 7)$$

Simplifying:

$$y + 1 = 7x - 49 \quad \Rightarrow \quad 7x - y - 50 = 0$$

Therefore, the equation of the tangent is:

$$7x - y - 50 = 0$$

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ii. $y^2 = 12x$ through $(1, 4)$

1. Differentiate the equation of the parabola:
   The equation of the parabola is:

$$y^2 = 12x$$

Differentiating both sides with respect to $x$:

$$2y \frac{dy}{dx} = 12$$

Solving for $\frac{dy}{dx}$:

$$\frac{dy}{dx} = \frac{6}{y}$$

2. Find the slope at the point $(1, 4)$:
   Substituting $y = 4$ into the derivative:

$$\frac{dy}{dx}(1, 4) = \frac{6}{4} = \frac{3}{2}$$

So, the slope of the tangent is $m = \frac{3}{2}$.

3. Use the point-slope form to find the equation of the tangent:
   Using the point $(1, 4)$ and the slope $m = \frac{3}{2}$, the equation of the tangent is:

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

Simplifying:

$$2y - 8 = 3x - 3 \quad \Rightarrow \quad 3x - 2y + 5 = 0$$

Therefore, the equation of the tangent is:

$$3x - 2y + 5 = 0$$

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iii. $x^2 - 2y^2 = 2$ through $(1, -2)$

1. Differentiate the equation of the hyperbola:
   The equation of the hyperbola is:

$$x^2 - 2y^2 = 2$$

Differentiating both sides with respect to $x$:

$$2x - 4y \frac{dy}{dx} = 0$$

Solving for $\frac{dy}{dx}$:

$$\frac{dy}{dx} = \frac{x}{2y}$$

2. Find the slope at the point $(1, -2)$:
   Substituting $x = 1$ and $y = -2$ into the derivative:

$$\frac{dy}{dx}(1, -2) = \frac{1}{2(-2)} = -\frac{1}{4}$$

So, the slope of the tangent is $m = -\frac{1}{4}$.

3. Use the point-slope form to find the equation of the tangent:
   Using the point $(1, -2)$ and the slope $m = -\frac{1}{4}$, the equation of the tangent is:

$$y - (-2) = -\frac{1}{4}(x - 1)$$

Simplifying:

$$4y + 8 = -x + 1 \quad \Rightarrow \quad x + 4y + 7 = 0$$

Therefore, the equation of the tangent is:

$$x + 4y + 7 = 0$$

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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:

$$\frac{dy}{dx} = \text{Slope of tangent line}$$

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Summary of Steps

1. Differentiate the equation of the conic to find the slope of the tangent at the given point.
2. Use the point-slope form of the equation of a line to write the equation of the tangent line.
3. Simplify the equation to get the final result.