Question

What is the tangent of prabola y2=4a(x-3) with slope 2

Answer 1

4x - 2y - 12 + a = 0

Answer 2

To find the tangent of the parabola $y^2 = 4a(x-3)$ with slope 2, we can use the standard form of the tangent equation.

1. Transform the given parabola to the standard form: The standard equation of a parabola is $Y^2 = 4AX$. Given parabola: $y^2 = 4a(x-3)$. Let $Y = y$ and $X = x-3$. Substituting these into the given equation, we get $Y^2 = 4aX$. Comparing this with the standard form $Y^2 = 4AX$, we identify the parameter $A$ for this specific parabola as $A = a$.

2. Recall the equation of a tangent to a parabola: For a parabola $Y^2 = 4AX$, the equation of a tangent with slope $m$ is given by: $Y = mX + \frac{A}{m}$

3. Substitute the values: We are given the slope $m=2$. We found $A=a$. And we have $Y=y$ and $X=x-3$. Substitute these into the tangent equation: $y = 2(x-3) + \frac{a}{2}$

4. Simplify the equation: $y = 2x - 6 + \frac{a}{2}$ To remove the fraction, multiply the entire equation by 2: $2y = 4x - 12 + a$ Rearrange the terms to the standard linear equation form ($Ax + By + C = 0$): $4x - 2y - 12 + a = 0$

This is the equation of the tangent to the given parabola with a slope of 2.