Find the Equations of all lines tangent to Y = x^2 -6......?

Find the Equations of all lines tangent to Y = x^2 - 6 that pass through the point (4,6)





Some help on this one while figuring it out. The answer is not


y = 8x - 26.


We can't just plug in 4 as x to find the slope. As the line tangential to the curve at 4 is not passing throught (4,6).


My thoughts are we need to find where the tangential line would cross the curve. then find the slope and then the equation. i don't know how this would work.





I would like for you to show work. Cuz I can probably guess and check to find the answer. I just need to know how to get there with actual work.

Find the Equations of all lines tangent to Y = x^2 -6......?
Hi,





These tangent lines are y = 12x - 42 and y = 4x - 10.





If f(x) = x² - 6, then f"(x) = 2x. This means that 2x is the slope of a tangent line when "x" is the coordinate of the point on the graph (x,x² - 6).





Since slope = (y2 - y1)/(x2 - x1):





..........x² - 6 - 6


2x = --------------


............x - 4





This simplifies to:





2x(x - 4) = x² - 12


2x² - 8x = x² - 12


x² - 8x + 12 = 0


(x - 6)(x - 2) = 0


x = 6 or x = 2 are the two x values on the graph of y = x² - 6 that have tangents going through (4,6).





If x = 6, then the slope of 2x is 2(6) = 12. The equation in point-slope form would be y - 6 = 12(x - 4).





y - 6 = 12(x - 4)


y - 6 = 12x - 48


y = 12x - 42 %26lt;== FIRST TANGENT





If x = 2, then the slope of 2x is 2(2) = 4. The equation in point-slope form would be y - 6 = 4(x - 4).





y - 6 = 4(x - 4)


y - 6 = 4x - 16


y = 4x - 10 %26lt;== SECOND TANGENT





I hope that helps!! :-)
Reply:Y = x^2 - 6 that pass through the point (4,6)


dy/dx = 2x


at x = 4, dy/dx = 8


m = 8, m* = -1/8 since the shortest distance formula gives us this idea.


y - 6 = -1/8 (x - 4)


y - 6 = -1/8x + 0.5


y = -1/8x + 6.5





y = -1/8x +6.5
Reply:In order to get the equation for the tangent line at (4, 6), you need the slope of the curve at (4, 6). So take the derivative of the equation to get,





y' = 2x





From this we see the slope is 2.





So the equation for the line is going to be y = 2x +b. Plug in the values (4,6) to find out what b is. It is -2. So the equation you are looking for is





y = 2x - 2.