Question

Find the equation of both lines that pass through the point (2, -3) and are tangent to the parabola y = x^2 + x.

Accepted Answer

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

Answer

We have to find the equations of both lines through the point (2,-3) that are tangent to the parabola y=x^2+x

First we will  find the derivative of y to find the slope m.
i.e. m=y'=2x+1
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.
i.e. 2x+1=\frac{x^2+x+3}{x-2}  
implies, x^2-4x-5=0

i.e. x=5 , -1

So now at x=5 the slope is m= 2(5)+1=11.So, the equation of the first tangent line through point (2,-3) is y+3=11(x-2)
i.e. \mathbf{y=11x-25}

at x=-1, the slope is m=2(-1)+1=-1 .So, the equation of the second tangent line through point (2,-3) is: y+3= -1(x-2).i.e.  \mathbf{y=-x-1}

So the equation of the two tangent lines to the parabola y=x^2+x passing through the point (2,-3) are:

y=11x-25 and y=-x-1

Answer

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 = 11and 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.

Math

Find Equations Of Both Lines Through The Point (2, −3) That Are Tangent To The Parabola Y = X2 + X.

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