Question

Write the equations of two tangent lines y=4/x through P(3,1) (P is not on f(x) graph

Answer 1

@hartnn

Answer 2

well its easy let solve it . let find the slope of tangents . how do you find the slope of tangent to a curve y=4/x?

Answer 3

I know I have to find the derivative to find the slop of the function

Answer 4

but that's kind of it haha

Answer 5

okay lets find the slope first .. for at any point x on curve y= 4/x
slope of tangent at that point will be dy/dx

so take derivative with respect x of given equation can you find that?

Answer 6

-4x^-2

Answer 7

right so now slope of tangent at point x will -4/x^2
and we have a point (3,1)
so now can you find equation of tangent by slope point form?

Answer 8

is it y-1= -4/x^-2 (x-3) ? I'm not sure

Answer 9

if I plug x values into the original equation and then put these x and y values into the tangent line, I just get something like 5=5

Answer 10

i don't really get how to find the equations of the 2 tangent lines for the graph from here

Answer 11

okay just wait for a sec

Answer 12

okay thanks :)

Answer 13

okay i made a big mistake sorry for that really..

we have to find a tangent to given a curve which will pass through the point (3,1)

so we find slope of tangent which is equal to -4/x^2


so we have equation of tangent as

y-1= -4/x^2 (x-3)

we will write this as y-1=m(x-3) where m=-4/x^2

now as we want to find a point which on the graph and through we draw tangent which will pass through point (3,1)

so we will put y=4/x in equation (1)
(remember why we did that because its an important step)

so we get (4/x)-1=-4/x^2 (x-3)

by simplifying this we get a quadratic equation

x^2-8x+12=0

by solving that we get x=2 and x=6

now put this values of x in slope of tangent we get slope as -1 and -1/9

so now putting values of slope in equation of tangent we get
for m=-1

y-1=-1(x-3)

y-1=-x+3

x+y-4=0


and for m=-1/9

we have y-1=-1/9(x-3)

we get

9y-9=-x+3
that is

x+9y-12=0

this is your answer,,!! :)

Answer 14

okay i get it now :) thank you so much this is so helpful!

Answer 15

you are most welcome ..! :)