Question Statement

Find the equation of the tangent to the ellipse $x^{2} + 2 y^{2} - 3 x + 5 y - 3 = 0$ with slope 1.

Background and Explanation

To find the tangent to a curve with a given slope, we use the slope-intercept form of a line, $y = m x + c$ . A line is tangent to a conic section if it intersects the curve at exactly one point, which mathematically means the discriminant ( $D = b^{2} - 4 a c$ ) of the resulting quadratic equation must be zero.

Solution

The given equation of the ellipse is: $x^{2} + 2 y^{2} - 3 x + 5 y - 3 = 0$

We are looking for a tangent line with a slope of $1$ . Let the equation of the tangent be: $y = m x + c$

Given that the slope $m = 1$ : $y = x + c$

To find the value of $c$ , we substitute the expression for $y$ from equation (ii) into equation (i):

$x^{2} + 2 (x + c)^{2} - 3 x + 5 (x + c) - 3 = 0$

Expanding the terms: $\Rightarrow x^{2} + 2 (x^{2} + 2 c x + c^{2}) - 3 x + 5 x + 5 c - 3 = 0$ $\Rightarrow x^{2} + 2 x^{2} + 4 c x + 2 c^{2} - 3 x + 5 x + 5 c - 3 = 0$

Grouping the terms to form a quadratic equation in $x$ ( $a x^{2} + b x + c = 0$ ): $\Rightarrow 3 x^{2} + (4 c + 2) x + (2 c^{2} + 5 c - 3) = 0$

For the line to be a tangent, it must touch the ellipse at only one point. This occurs when the discriminant of the quadratic equation is equal to zero ( $D = 0$ ): $(4 c + 2)^{2} - 4 (3) (2 c^{2} + 5 c - 3) = 0$

Now, we solve for $c$ : $4 (2 c + 1)^{2} - 12 (2 c^{2} + 5 c - 3) = 0$

Divide the entire equation by 4 to simplify: $(2 c + 1)^{2} - 3 (2 c^{2} + 5 c - 3) = 0$

Expand the squares and distribute the $- 3$ : $4 c^{2} + 4 c + 1 - 6 c^{2} - 15 c + 9 = 0$ $- 2 c^{2} - 11 c + 10 = 0$

Multiply by $- 1$ to make the leading coefficient positive: $2 c^{2} + 11 c - 10 = 0$

Using the quadratic formula $c = \frac{- b \pm b ^{2} - 4 a c}{2 a}$ : $c = \frac{- 11 \pm ( 11 ) ^{2} - 4 ( 2 ) ( - 10 )}{2 ( 2 )}$ $c = \frac{- 11 \pm 121 + 80}{4}$ $c = \frac{- 11 \pm 201}{4}$

Finally, substitute the values of $c$ back into the equation of the line $y = x + c$ : $y = x + \frac{- 11 \pm 201}{4}$

These are the equations of the two possible tangent lines.

Key Formulas or Methods Used

Slope-Intercept Form: $y = m x + c$
Substitution Method: Substituting the line equation into the curve equation to find intersection points.
Condition for Tangency: For a quadratic equation $a x^{2} + b x + c = 0$ , the condition for a single point of contact (tangency) is the discriminant $D = b^{2} - 4 a c = 0$ .
Quadratic Formula: $x = \frac{- b \pm b ^{2} - 4 a c}{2 a}$

Summary of Steps

Set up the equation of the tangent line using the given slope: $y = 1 x + c$ .
Substitute $y = x + c$ into the ellipse equation.
Simplify the resulting expression into a standard quadratic equation in terms of $x$ .
Apply the tangency condition by setting the discriminant ( $b^{2} - 4 a c$ ) of the quadratic to zero.
Solve the resulting quadratic equation for the unknown constant $c$ .
Write the final equations of the tangents using the calculated values of $c$ .