Question Statement

Find the equation of the tangent to the ellipse $3 x^{2} + 4 y^{2} = 12$ which is perpendicular to the line $y + 2 x = 4$ .

Background and Explanation

To find the equation of a tangent line, we first determine its slope using the property that perpendicular lines have slopes that are negative reciprocals of each other. We then use the condition of tangency, which states that when a line and a curve intersect at exactly one point, the discriminant of the resulting quadratic equation must be zero.

Solution

The given line is $y + 2 x = 4$ , which can be written in slope-intercept form as: $y = - 2 x + 4$

The slope of this line is $m_{1} = - 2$ .

Since the tangent line is perpendicular to the given line, the slope ( $m$ ) of the tangent line is: $m = - \frac{1}{m _{1}} = - \frac{1}{- 2} = \frac{1}{2}$

Let the equation of the tangent line be: $y = \frac{1}{2} x + c \dots (i)$

To find the value of $c$ , we substitute this expression for $y$ into the equation of the ellipse $3 x^{2} + 4 y^{2} = 12$ :

$3 x^{2} + 4 (\frac{1}{2} x + c)^{2} 3 x^{2} + 4 (\frac{1}{4} x^{2} + c x + c^{2}) 3 x^{2} + x^{2} + 4 c x + 4 c^{2} 4 x^{2} + 4 c x + 4 c^{2} - 12 = 12 = 12 = 12 = 0$

We can simplify this by dividing the entire equation by 4: $x^{2} + c x + (c^{2} - 3) = 0$

Since the line is a tangent to the ellipse, the line must touch the ellipse at exactly one point. This means the discriminant ( $D = b^{2} - 4 a c$ ) of the quadratic equation above must be zero:

$c^{2} - 4 (1) (c^{2} - 3) c^{2} - 4 c^{2} + 12 - 3 c^{2} + 12 3 c^{2} c^{2} c = 0 = 0 = 0 = 12 = 4 = \pm 2$

Now, substitute the values of $c$ back into the tangent line equation $(i)$ : $y = \frac{1}{2} x \pm 2$

To clear the fraction, multiply the entire equation by 2: $2 y = x \pm 4$

Thus, the required equations of the tangent lines are: $x - 2 y + 4 = 0 and x - 2 y - 4 = 0$

Key Formulas or Methods Used

Slope of perpendicular lines: $m_{1} \cdot m_{2} = - 1$
Slope-intercept form of a line: $y = m x + c$
Condition for tangency: The discriminant of the intersection quadratic must be zero ( $b^{2} - 4 a c = 0$ ).
Standard form of an ellipse: $a x^{2} + b y^{2} = d$

Summary of Steps

Find the slope of the given line by rearranging it into $y = m x + b$ form.
Calculate the slope of the tangent line using the perpendicular slope formula.
Assume the tangent line has the form $y = m x + c$ and substitute it into the ellipse equation.
Simplify the resulting equation into a standard quadratic form ( $a x^{2} + b x + c = 0$ ).
Set the discriminant of the quadratic to zero to solve for the unknown constant $c$ .
Substitute the values of $c$ back into the line equation to get the final tangent equations.

Background and Explanation

Solution

The given line is

y + 2 x = 4

, which can be written in slope-intercept form as:

y = - 2 x + 4

The slope of this line is

m_{1} = - 2

Since the tangent line is perpendicular to the given line, the slope (

m

) of the tangent line is:

m = - \frac{1}{m _{1}} = - \frac{1}{- 2} = \frac{1}{2}

Let the equation of the tangent line be:

y = \frac{1}{2} x + c \dots (i)

To find the value of

c

, we substitute this expression for

y

into the equation of the ellipse

3 x^{2} + 4 y^{2} = 12

3 x^{2} + 4 (\frac{1}{2} x + c)^{2} 3 x^{2} + 4 (\frac{1}{4} x^{2} + c x + c^{2}) 3 x^{2} + x^{2} + 4 c x + 4 c^{2} 4 x^{2} + 4 c x + 4 c^{2} - 12 = 12 = 12 = 12 = 0

We can simplify this by dividing the entire equation by 4:

x^{2} + c x + (c^{2} - 3) = 0

Since the line is a tangent to the ellipse, the line must touch the ellipse at exactly one point. This means the discriminant (

D = b^{2} - 4 a c

) of the quadratic equation above must be zero:

c^{2} - 4 (1) (c^{2} - 3) c^{2} - 4 c^{2} + 12 - 3 c^{2} + 12 3 c^{2} c^{2} c = 0 = 0 = 0 = 12 = 4 = \pm 2

Now, substitute the values of

c

back into the tangent line equation

(i)

y = \frac{1}{2} x \pm 2

To clear the fraction, multiply the entire equation by 2:

2 y = x \pm 4

Thus, the required equations of the tangent lines are:

x - 2 y + 4 = 0 and x - 2 y - 4 = 0

Summary of Steps

Find the slope of the given line by rearranging it into

y = m x + b

form.

Calculate the slope of the tangent line using the perpendicular slope formula.

Assume the tangent line has the form

y = m x + c

and substitute it into the ellipse equation.

Simplify the resulting equation into a standard quadratic form (

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

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

c

Substitute the values of

c

back into the line equation to get the final tangent equations.

7.7 Q-4

Question Statement

Background and Explanation

Solution

Key Formulas or Methods Used

Summary of Steps

7.7 Q-4

Question Statement

Background and Explanation

Solution

Key Formulas or Methods Used

Summary of Steps