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.

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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`

The slope of the line tangent...

(The entire section contains 3 answers and 572 words.)

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