Question Statement

Find the equation of the tangent to the hyperbola $x^2-2y^2+4x-6y+11=0$ which is parallel to the line $4x-8y+7=0$.

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Background and Explanation

To find a tangent line parallel to a given line, we use the fact that they share the same slope. When a line is tangent to a conic section, substituting the line's equation into the conic's equation results in a quadratic equation with exactly one solution, meaning its discriminant ($D=b^2-4ac$) must equal zero.

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Solution

Step 1: Determine the slope of the tangent line

The tangent line is parallel to the line $4x-8y+7=0$. We can find the slope by rewriting this line in slope-intercept form ($y=mx+b$): $8y=4x+7$ $y=\frac{4}{8}x+\frac{7}{8}⟹y=\frac{1}{2}x+\frac{7}{8}$ The slope $m$ is $\frac{1}{2}$. Since the tangent line is parallel, it will have the same slope. Let the equation of the tangent line be: $y=\frac{1}{2}x+c\dots \text{(i)}$

Step 2: Substitute the tangent equation into the hyperbola

We substitute $y=\frac{1}{2}x+c$ into the equation of the hyperbola $x^2-2y^2+4x-6y+11=0$: $x^2-2{\left(\frac{1}{2}x+c\right)}^2+4x-6\left(\frac{1}{2}x+c\right)+11=0$

Expanding the terms: $x^2-2\left(\frac{1}{4}{x}^2+cx+c^2\right)+4x-3x-6c+11=0$ $x^2-\frac{1}{2}{x}^2-2cx-2c^2+x-6c+11=0$

Combining like terms to form a quadratic equation in terms of $x$: $\frac{1}{2}{x}^2+(1-2c)x+(-2c^2-6c+11)=0$

Step 3: Apply the condition for tangency

For the line to be a tangent, the discriminant of the quadratic equation must be zero ($b^2-4ac=0$): $(1-2c{)}^2-4\left(\frac{1}{2}\right)(-2c^2-6c+11)=0$

Simplify the expression: $(4c^2-4c+1)-2(-2c^2-6c+11)=0$ $4c^2-4c+1+4c^2+12c-22=0$ $8c^2+8c-21=0$

Step 4: Solve for the constant $c$

Using the quadratic formula $c=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$: $c=\frac{-8\pm \sqrt{8^2-4(8)(-21)}}{2(8)}$ $c=\frac{-8\pm \sqrt{64+672}}{16}$ $c=\frac{-8\pm \sqrt{736}}{16}$

Simplifying the radical $\sqrt{736}=\sqrt{16\times 46}=4\sqrt{46}$: $c=\frac{-8\pm 4\sqrt{46}}{16}$ $c=\frac{-2\pm \sqrt{46}}{4}$

Step 5: Final Equation

Substitute the values of $c$ back into equation (i): $y=\frac{1}{2}x+\frac{-2\pm \sqrt{46}}{4}$

To write this in general form, multiply the entire equation by 4: $4y=2x+(-2\pm \sqrt{46})$ $2x-4y-2\pm \sqrt{46}=0$

The required equations of the tangent lines are $2x-4y-2+\sqrt{46}=0$ and $2x-4y-2-\sqrt{46}=0$.

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Key Formulas or Methods Used

• Slope-Intercept Form: $y=mx+c$
• Condition for Tangency: For a line and a conic, the discriminant $D=b^2-4ac$ of the resulting intersection equation must be $0$.
• Quadratic Formula: $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$
• Parallel Lines: Two lines are parallel if their slopes are equal ($m_1=m_2$).

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Summary of Steps

1. Find the slope ($m$) of the given line.
2. Set up the equation of the tangent line as $y=mx+c$.
3. Substitute this $y$ expression into the hyperbola's equation to get a quadratic in $x$.
4. Set the discriminant of that quadratic to zero to ensure there is only one point of contact.
5. Solve the resulting quadratic equation for $c$.
6. Substitute the values of $c$ back into the tangent line equation.