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

The equations of the tangents that pass through

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

#P# passes through#(alpha,alpha^2+alpha)# and has gradient#2alpha + 1# , so using the point/slope form#y−y_1=m(x−x_1)# the equation of the tangent at#P# is;

#y - (alpha^2+alpha) = (2alpha+1)(x-alpha)# if this tangent also passes through

#(2,-3)# then;

# \ \-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

#alpha =-1 => beta = 0 # , and the tangent equation becomes:

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

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

#alpha =5 => beta = 30# , and the tangent equation becomes:

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

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

# :. \ \ \ \ \ \ \ \ \y = 11x-25# Hence the equations of the tangents that pass through

#(2,-3)# are

# y=-x-1 # and#y = 11x-25# We can confirm this graphically:

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The two tgts. are

the Parabola with the tgt. line.

Enjoy Maths.!

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To find the equations of the lines tangent to the parabola y=x^2+x and passing through the point (2,-3), we can use the concept of derivatives.

First, we find the derivative of the parabola by differentiating y=x^2+x with respect to x. The derivative is given by dy/dx = 2x + 1.

Next, we substitute the x-coordinate of the given point (2,-3) into the derivative to find the slope of the tangent line at that point. Substituting x=2 into dy/dx = 2x + 1, we get the slope m = 2(2) + 1 = 5.

Since the tangent line passes through the point (2,-3), we can use the point-slope form of a linear equation to find the equation of the tangent line. Using the point-slope form y - y1 = m(x - x1), where (x1, y1) is the given point and m is the slope, we substitute the values to get y - (-3) = 5(x - 2).

Simplifying the equation, we have y + 3 = 5x - 10.

Therefore, the equation of the first line tangent to the parabola y=x^2+x and passing through the point (2,-3) is y = 5x - 13.

To find the equation of the second line, we need to consider the other tangent line. Since the parabola is symmetric, the second tangent line will have the same slope but a different y-intercept.

Using the point-slope form again, we substitute the values to get y - (-3) = 5(x - 2).

Simplifying the equation, we have y + 3 = 5x - 10.

Therefore, the equation of the second line tangent to the parabola y=x^2+x and passing through the point (2,-3) is y = 5x - 13.

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