Question Statement

Prove that the line $3x+y-5=0$ is tangent to the parabola $y^2-2y+6x-6=0$. Also, find the point of contact.

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Background and Explanation

To prove a line is tangent to a curve, we substitute the linear equation into the quadratic equation to form a single quadratic equation in one variable. If the discriminant ($D=b^2-4ac$) of this resulting equation is zero, the line touches the curve at exactly one point, confirming it is a tangent.

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Solution

Given the equation of the line: $3x+y-5=0$

And the equation of the parabola: $y^2-2y+6x-6=0$

Step 1: Express one variable in terms of the other

From equation (1), we can isolate $x$: $3x=5-y$ $x=\frac{5-y}{3}$

Step 2: Substitute into the parabola equation

Substitute the expression for $x$ from equation (3) into equation (2): $y^2-2y+6\left(\frac{5-y}{3}\right)-6=0$

Simplify the term with the fraction: $y^2-2y+2(5-y)-6=0$ $y^2-2y+10-2y-6=0$ Combine like terms to form a quadratic equation: $y^2-4y+4=0$

Step 3: Prove tangency using the discriminant

A line is tangent to a curve if the intersection results in a single point. We check the discriminant ($D=b^2-4ac$) of the quadratic $y^2-4y+4=0$, where $a=1,b=-4,c=4$: $\text{Disc}=(-4{)}^2-4(1)(4)$ $\text{Disc}=16-16$ $\text{Disc}=0$

Since the discriminant is $0$, the line is indeed tangent to the parabola.

Step 4: Find the point of contact

To find the $y$-coordinate of the point of contact, solve the quadratic equation: $y^2-4y+4=0$ $(y-2{)}^2=0$ $y-2=0$ $y=2$

Now, substitute $y=2$ back into equation (1) to find the $x$-coordinate: $3x+(2)-5=0$ $3x-3=0$ $3x=3$ $x=1$

The point of contact is $(1,2)$.

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Key Formulas or Methods Used

• Substitution Method: Solving a system of equations by substituting one into the other.
• Discriminant Formula: $D=b^2-4ac$.
• Condition for Tangency: A line is tangent to a quadratic curve if the discriminant of their combined equation is zero ($D=0$).
• Perfect Square Identity: $a^2-2ab+b^2=(a-b{)}^2$.

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Summary of Steps

1. Isolate $x$ from the linear equation.
2. Substitute the expression for $x$ into the parabola's equation.
3. Simplify the resulting equation into a standard quadratic form $a{y}^2+by+c=0$.
4. Calculate the discriminant; $D=0$ proves the line is a tangent.
5. Solve the quadratic equation to find the $y$-coordinate of the point of contact.
6. Substitute the $y$-value back into the linear equation to find the $x$-coordinate.