# How do you find the equations of both lines through point (2,-3) that are tangent to the parabola #y=x^2+x#?

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The equations of the tangents that pass through

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The gradient of the tangent to a curve at any particular point is given by the derivative of the curve at that point. So for our curve (the parabola) we have

# y=x^2+x #

Differentiating wrt

# dy/dx=2x+1 #

Let

# m = 2alpha + 1 \ \ \ # (using the derivative)

And as P lies on the curve we also have:

# beta = alpha^2+alpha \ \ \ # (using the curve equation)

And so the tangent at

#y - (alpha^2+alpha) = (2alpha+1)(x-alpha)#

if this tangent also passes through

# \ \-3 - (alpha^2+alpha) = (2alpha+1)(2-alpha)#

# :. -3 - alpha^2-alpha = 3alpha-2alpha^2+2#

# :. alpha^2 -4alpha-5=0#

# :. (alpha-5)(alpha+1)=0#

# :. alpha =-1,5#

If

#y - 0 = (-1)(x+1)#

# :. \ \ y=-x-1 #

If

# \ \ \ \ \ y - 30 = (11)(x-5)#

# :. y - 30 = 11x-55#

# :. \ \ \ \ \ \ \ \ \y = 11x-25#

Hence the equations of the tangents that pass through

We can confirm this graphically:

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The two tgts. are

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Let

Using the **Slope-point Form** , we get, its eqn.,

Solving this eqn. with

the Parabola with the tgt. line.

But, a tgt. essentially intersects the Parabola in only one pt.,

For

**Respected Steve** has obtained!

**N.B.:**

**Enjoy Maths.!**

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