6.7 Q-7

Question Statement

Find the equation of the common tangents to the following conics:

1. $x^2 = 80y$ and $x^2 + y^2 = 81$
2. $y^2 = 16x$ and $x^2 = 2y$

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Background and Explanation

To solve this problem, we need to find the common tangents to two different conics. The key steps involve:

1. Finding the equation of the common tangent: For each pair of conics, the tangent line must satisfy the equations of both conics. This requires solving for the slope and the intercept of the tangent.
2. Using discriminants: For a line to be tangent to both conics, the discriminant of the corresponding quadratic equation must be zero. This ensures that the line touches each conic at exactly one point.
3. Using the point-slope form of the line: The equation of the common tangent can then be derived using the point-slope form of the equation of a line.

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Solution

i. $x^2 = 80y$ and $x^2 + y^2 = 81$

1. Find the equation of the common tangent:
   Let the equation of the common tangent be:

$$y = mx + c \tag{3}$$

Substituting this into the equation of the first conic $x^2 = 80y$:

$$x^2 = 80(mx + c) \quad \Rightarrow \quad x^2 - 80mx - 80c = 0$$

This is a quadratic equation in $x$, and for the line to be tangent to the conic, the discriminant of this equation must be zero. The discriminant of a quadratic equation $ax^2 + bx + c = 0$ is given by $\Delta = b^2 - 4ac$.

2. Set the discriminant equal to zero:
   The discriminant of $x^2 - 80mx - 80c = 0$ is:

$$(-80m)^2 - 4(1)(-80c) = 0 \quad \Rightarrow \quad 6400m^2 + 320c = 0$$

Solving for $c$:

$$c = -20m^2$$

Therefore, the equation of the common tangent is:

$$y = mx - 20m^2 \tag{4}$$

3. Substitute into the second conic equation:
   Now, substitute $y = mx - 20m^2$ into the second conic equation $x^2 + y^2 = 81$:

$$x^2 + (mx - 20m^2)^2 = 81$$

Expanding:

$$x^2 + m^2x^2 - 40m^3x + 400m^4 = 81$$

Rearranging:

$$(1 + m^2)x^2 - 40m^3x + (400m^4 - 81) = 0$$

The discriminant of this quadratic in $x$ must also be zero for the line to be tangent. Therefore, the discriminant is:

$$(-40m^3)^2 - 4(1+m^2)(400m^4 - 81) = 0$$

Simplifying:

$$1600m^6 - 4(400m^6 + 400m^4 - 81m^2 - 81) = 0$$ $$1600m^6 - 1600m^6 - 1600m^4 + 324m^2 + 324 = 0$$ $$4(400m^4 - 81m^2 - 81) = 0$$

Solving for $m$:

$$400m^4 - 81m^2 - 81 = 0$$

Solving this quadratic equation for $m^2$ (let $t = m^2$):

$$400t^2 - 81t - 81 = 0$$

Using the quadratic formula:

$$t = \frac{-(-81) \pm \sqrt{(-81)^2 - 4(400)(-81)}}{2(400)}$$ $$t = \frac{81 \pm \sqrt{6561 + 129600}}{800} = \frac{81 \pm \sqrt{136161}}{800} = \frac{81 \pm 369}{800}$$ $$t = \frac{450}{800} \quad \text{or} \quad t = -\frac{288}{800}$$

Thus, $t = \frac{9}{16}$. Therefore:

$$m^2 = \frac{9}{16}, \quad m = \pm \frac{3}{4}$$

4. Find the equations of the tangents:
   Now, using $m = \pm \frac{3}{4}$ and $c = -20m^2$, we substitute to find $c$:

$$c = -20 \times \frac{9}{16} = -\frac{180}{16} = -\frac{45}{4}$$

Thus, the equation of the tangent is:

$$y = \pm \frac{3}{4}x - \frac{45}{4}$$

Multiplying through by 4 to eliminate the fractions:

$$4y = 3x - 45$$

Therefore, the two equations of the common tangents are:

$$3x - 4y - 45 = 0 \quad \text{and} \quad 3x - 4y + 45 = 0$$

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ii. $y^2 = 16x$ and $x^2 = 2y$

1. Find the equation of the common tangent:
   Let the equation of the common tangent be:

$$y = mx + c \tag{3}$$

This line will be tangent to $y^2 = 16x$, so substitute $y = mx + c$ into $y^2 = 16x$:

$$(mx + c)^2 = 16x$$

Expanding:

$$m^2x^2 + 2mcx + c^2 = 16x$$

Rearranging:

$$m^2x^2 + (2mc - 16)x + c^2 = 0$$

For this to be a tangent, the discriminant must be zero:

$$(2mc - 16)^2 - 4m^2c^2 = 0$$

Solving for $c$:

$$c = \frac{4}{m}$$

2. Solve the system of equations:
   Substitute $c = \frac{4}{m}$ into the second conic $x^2 = 2y$:

$$x^2 = 2(mx + \frac{4}{m})$$

Simplifying:

$$x^2 = 2mx + \frac{8}{m}$$

Rearranging:

$$x^2 - 2mx - \frac{8}{m} = 0$$

The discriminant of this equation must be zero for the line to be tangent:

$$(-2m)^2 - 4(1)(-\frac{8}{m}) = 0$$

Solving for $m$, we get $m = -2$.

3. Find the equation of the tangent:
   Substituting $m = -2$ into $c = \frac{4}{m}$, we get $c = -2$. Thus, the equation of the common tangent is:

$$y = -2x - 2$$

This can be written as:

$$2x + y + 2 = 0$$

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Key Formulas or Methods Used

1. Discriminant Condition for Tangency: For a line to be tangent to a conic, the discriminant of the corresponding quadratic equation must be zero.

$$\Delta = b^2 - 4ac = 0$$

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

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

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Summary of Steps

1. Set up the equation of the common tangent in the form $y = mx + c$.
2. Substitute this equation into the equations of both conics.
3. Apply the discriminant condition to find values for $m$ and $c$.
4. Solve the system of equations and simplify the resulting tangent equations.