Find Equations Of Both Lines Through The Point (2, −3) That Are Tangent To The Parabola Y = X2 + X.
Find the equation of both lines that pass through the point (2, -3) and are tangent to the parabola y = x^2 + x.
The slope of the line tangent to the graph of the function `y(x)` at point `(x_0, y_0)` is the value of the derivative of `y(x)` at this point:
`m = y'(x_0)`.
For the given function `y(x) = x^2 + x`, the derivative `y'(x) = 2x + 1`, so the slope of the tangent line is as follows:
`m = 2x_0 + 1`.
If the tangent line passes through the point (2, -3), its equation in point-slope form is as follows:
`y - (-3) = m(x - 2)`.
Since the point `(x_0, y_0)` belongs to both the tangent line and the parabola, then, plugging in `y_0 = x_0^2 + x_0` and `m = 2x_0 + 1`, we get the following:
`x_0^2 + x_0 + 3 = (2x_0 + 1)(x_0 - 2)`
`x_0^2 + x_0 + 3 = 2x_0^2 + x_0 - 4x_0 - 2`
`x_0^2 - 4x_0 - 5 = 0`
`(x_0 - 5)(x_0 + 1) = 0`
From here, `x_0 = 5 `and `x_0 = -1`. So, the slope of the first tangent line is as follows:
`m = 2*5 + 1 = 11`and the slope of the second tangent line is
`m = 2(-1) + 1 = - 1`.
In point-slope form, the equations of the two tangent lines are as follows:
`y + 3 = 11(x - 2)`and `y + 3 = -(x -2)`.
In the slope-intercept form, these are as follows:
`y = 11x - 25` and `y = -x - 1`.
The equations of the lines passing through (2, -3) and tangent to the given parabola are as follows:
y = 11x - 25 and y = - x - 1.
We have to find the equations of both lines through the point (2,-3) that are tangent to the parabola `y=x^2+x`
Now we find two points on the two tangent lines to y using: `(x,x^2+x)` and (2,-3). This is to show that (x,f(x)) is a point on the original equation f(x).
Now we have to find the slope of these two points and equate it to the derivative or slope of the original equation.
So the equation of the two tangent lines to the parabola `y=x^2+x` passing through the point (2,-3) are:
The equation of the line through (2, -3) and tangential to the parabola y = x^2 + x has to be determined.
The slope of a tangent to a curve y = f(x) at the point x = c is f'(c)
y' = 2x + 1
Let the line that passes through (2, -3) be a tangent to the parabola at (x, x^2 + x)
(x^2 + x + 3)/(x - 2) = 2x + 1
=> x^2 + x + 3 = 2x^2 - 4x + x - 2
=> x^2 - 4x - 5 = 0
=> x^2 - 5x + x - 5 = 0
=> x(x - 5) + 1(x - 5) = 0
=> (x + 1)(x - 5) = 0
x = -1 and x = 5
The tangents are lines passing through (2, -3) and (-1, 0) and through (2, -3) and (5, 30)
The lines are (y + 3)/(x - 2) = (3/-3) = -1
=> y + 3 = 2 - x
=> x + y + 1 = 0
and (y + 3)/(x - 2) = (33/3)
=> y + 3 = 11x - 22
=> 11x - y - 25 = 0
The equation of the two tangent lines to the curve y = x^2 + x passing through (2, -3) are x + y + 1 = 0 and 11x - y - 25 = 0