Question Statement

Find the equation of the tangent to the hyperbola $2 x^{2} - 4 y^{2} + 6 x - 8 y - 7 = 0$ which is perpendicular to the line $x + 2 y + 5 = 0$ . Also, find the point of tangency.

Background and Explanation

To find the tangent to a conic section perpendicular to a given line, we first determine the required slope using the property $m_{1} \cdot m_{2} = - 1$ . We then substitute the general line equation $y = m x + c$ into the hyperbola's equation and apply the condition for tangency, which states that the discriminant of the resulting quadratic equation must be zero.

Solution

Step 1: Find the slope of the tangent

The given line is $x + 2 y + 5 = 0$ . Rewriting it in slope-intercept form ( $y = m x + b$ ): $2 y = - x - 5 ⟹ y = - \frac{1}{2} x - \frac{5}{2}$ The slope of this line is $m_{1} = - \frac{1}{2}$ . Since the tangent is perpendicular to this line, its slope $m$ is: $m = - \frac{1}{m _{1}} = 2$ Thus, the equation of the tangent line is: $y = 2 x + c$

Step 2: Substitute the tangent equation into the hyperbola

The equation of the hyperbola is $2 x^{2} - 4 y^{2} + 6 x - 8 y - 7 = 0$ . Substituting $y = 2 x + c$ : $2 x^{2} - 4 (2 x + c)^{2} + 6 x - 8 (2 x + c) - 7 = 0$ Expanding the terms: $2 x^{2} - 4 (4 x^{2} + 4 c x + c^{2}) + 6 x - 16 x - 8 c - 7 = 0$ $2 x^{2} - 16 x^{2} - 16 c x - 4 c^{2} - 10 x - 8 c - 7 = 0$ Grouping the terms into a quadratic in $x$ : $- 14 x^{2} + (- 16 c - 10) x + (- 4 c^{2} - 8 c - 7) = 0$

Step 3: Apply the condition for tangency

A line is tangent to a curve if the intersection results in exactly one point, meaning the discriminant ( $D = b^{2} - 4 a c$ ) of the quadratic equation (ii) must be zero: $[- (16 c + 10)]^{2} - 4 (- 14) (- 4 c^{2} - 8 c - 7) = 0$ $[2 (8 c + 5)]^{2} + 56 (- 4 c^{2} - 8 c - 7) = 0$ $4 (64 c^{2} + 80 c + 25) + 56 (- 4 c^{2} - 8 c - 7) = 0$ Dividing the entire equation by 4: $64 c^{2} + 80 c + 25 + 14 (- 4 c^{2} - 8 c - 7) = 0$ $64 c^{2} + 80 c + 25 - 56 c^{2} - 112 c - 98 = 0$ $8 c^{2} - 32 c - 73 = 0$

Solving for $c$ using the quadratic formula: $c = \frac{- ( - 32 ) \pm ( - 32 ) ^{2} - 4 ( 8 ) ( - 73 )}{2 ( 8 )}$ $c = \frac{32 \pm 1024 + 2336}{16}$ $c = \frac{32 \pm 3360}{16} = \frac{32 \pm 4 210}{16}$ $c = \frac{8 \pm 210}{4}$

Step 4: Equations of the tangent lines

Substituting the values of $c$ back into equation (i): $y = 2 x + \frac{8 \pm 210}{4}$ Multiplying by 4 to clear the fraction: $4 y = 8 x + 8 \pm 210$

Step 5: Find the point of tangency

For a quadratic $a x^{2} + b x + c = 0$ with $D = 0$ , the root is $x = \frac{- b}{2 a}$ . From equation (ii): $x = \frac{- ( - 16 c - 10 )}{2 ( - 14 )} = \frac{16 c + 10}{- 28} = - \frac{8 c + 5}{14}$ Substituting $x$ into $y = 2 x + c$ : $y = 2 (- \frac{8 c + 5}{14}) + c = - \frac{8 c + 5}{7} + c = \frac{- 8 c - 5 + 7 c}{7} = \frac{- c - 5}{7}$ Substituting $c = \frac{8 \pm 210}{4}$ into the expressions for $x$ and $y$ :

For $x$ : $x = - \frac{8 ( \frac{8 \pm 210}{4} ) + 5}{14} = - \frac{2 ( 8 \pm 210 ) + 5}{14} = - \frac{16 \pm 2 210 + 5}{14} = \frac{- 21 \mp 2 210}{14}$

For $y$ : $y = - \frac{( \frac{8 \pm 210}{4} ) + 5}{7} = - \frac{\frac{8 \pm 210 + 20}{4}}{7} = \frac{- 28 \mp 210}{28}$

The points of tangency are: $(\frac{- 21 \mp 2 210}{14}, \frac{- 28 \mp 210}{28})$

Key Formulas or Methods Used

Perpendicular Slopes: $m_{1} \cdot m_{2} = - 1$
Condition for Tangency: Discriminant $D = b^{2} - 4 a c = 0$
Quadratic Formula: $x = \frac{- b \pm b ^{2} - 4 a c}{2 a}$
Point of Tangency for $D = 0$ : $x = \frac{- b}{2 a}$

Summary of Steps

Determine the slope of the given line and find the perpendicular slope ( $m = 2$ ).
Set up the tangent equation $y = 2 x + c$ .
Substitute the tangent equation into the hyperbola equation to form a quadratic in $x$ .
Set the discriminant of the quadratic to zero to solve for the constant $c$ .
Write the final tangent equations using the calculated $c$ values.
Calculate the $x$ -coordinate using $x = - b /2 a$ and find the corresponding $y$ -coordinate.

Background and Explanation

To find the tangent to a conic section perpendicular to a given line, we first determine the required slope using the property

m_{1} \cdot m_{2} = - 1

. We then substitute the general line equation

y = m x + c

into the hyperbola's equation and apply the condition for tangency, which states that the discriminant of the resulting quadratic equation must be zero.

Step 3: Apply the condition for tangency

A line is tangent to a curve if the intersection results in exactly one point, meaning the discriminant (

D = b^{2} - 4 a c

) of the quadratic equation (ii) must be zero:

[- (16 c + 10)]^{2} - 4 (- 14) (- 4 c^{2} - 8 c - 7) = 0

[2 (8 c + 5)]^{2} + 56 (- 4 c^{2} - 8 c - 7) = 0

4 (64 c^{2} + 80 c + 25) + 56 (- 4 c^{2} - 8 c - 7) = 0

Dividing the entire equation by 4:

64 c^{2} + 80 c + 25 + 14 (- 4 c^{2} - 8 c - 7) = 0

64 c^{2} + 80 c + 25 - 56 c^{2} - 112 c - 98 = 0

8 c^{2} - 32 c - 73 = 0

Solving for

c

using the quadratic formula:

c = \frac{- ( - 32 ) \pm ( - 32 ) ^{2} - 4 ( 8 ) ( - 73 )}{2 ( 8 )}

c = \frac{32 \pm 1024 + 2336}{16}

c = \frac{32 \pm 3360}{16} = \frac{32 \pm 4 210}{16}

c = \frac{8 \pm 210}{4}

Step 5: Find the point of tangency

For a quadratic

a x^{2} + b x + c = 0

with

D = 0

, the root is

x = \frac{- b}{2 a}

. From equation (ii):

x = \frac{- ( - 16 c - 10 )}{2 ( - 14 )} = \frac{16 c + 10}{- 28} = - \frac{8 c + 5}{14}

Substituting

x

into

y = 2 x + c

y = 2 (- \frac{8 c + 5}{14}) + c = - \frac{8 c + 5}{7} + c = \frac{- 8 c - 5 + 7 c}{7} = \frac{- c - 5}{7}

Substituting

c = \frac{8 \pm 210}{4}

into the expressions for

x

and

y

For

x

x = - \frac{8 ( \frac{8 \pm 210}{4} ) + 5}{14} = - \frac{2 ( 8 \pm 210 ) + 5}{14} = - \frac{16 \pm 2 210 + 5}{14} = \frac{- 21 \mp 2 210}{14}

For

y

y = - \frac{( \frac{8 \pm 210}{4} ) + 5}{7} = - \frac{\frac{8 \pm 210 + 20}{4}}{7} = \frac{- 28 \mp 210}{28}

The points of tangency are:

(\frac{- 21 \mp 2 210}{14}, \frac{- 28 \mp 210}{28})

Summary of Steps

Determine the slope of the given line and find the perpendicular slope (

m = 2

Set up the tangent equation

y = 2 x + c

Substitute the tangent equation into the hyperbola equation to form a quadratic in

x

Set the discriminant of the quadratic to zero to solve for the constant

c

Write the final tangent equations using the calculated

c

values.

Calculate the

x

-coordinate using

x = - b /2 a

and find the corresponding

y

-coordinate.

7.9 Q-4

Question Statement

Background and Explanation

Solution

Step 1: Find the slope of the tangent

Step 2: Substitute the tangent equation into the hyperbola

Step 3: Apply the condition for tangency

Step 4: Equations of the tangent lines

Step 5: Find the point of tangency

Key Formulas or Methods Used

Summary of Steps

7.9 Q-4

Question Statement

Background and Explanation

Solution

Step 1: Find the slope of the tangent

Step 2: Substitute the tangent equation into the hyperbola

Step 3: Apply the condition for tangency

Step 4: Equations of the tangent lines

Step 5: Find the point of tangency

Key Formulas or Methods Used

Summary of Steps