Question

find the eqns of tangents drawn to the curve y^2-2x^3-4y+8=0 from point (1,2)

Answer 1

@experimentX , @TuringTest , @Mani_Jha , @asnaseer

Answer 2

Similar way, Aravind. Find dy/dx at (1,2)-that is the slope of the tangent. Then use the two-point form to get the equation.

Answer 3

is 1,2 on the curve?

Answer 4

its a point

Answer 5

if not, then the question is asking for tangents of the curve that go thru 1,2

Answer 6

is 1,2 on the cave down or cave up side of the parabola?

Answer 7

u better check with wolf :P

Answer 8

err, inside or outside the cave that is

Answer 9

i aint got to chk it with anything :/

Answer 10

at x=1; y=3 ... the point 1,2 is under that; and the parabola has a +first coeef; so its on the outside and can intercept at most 2 tangents

Answer 11

opps, you aint got your exponents lined up do you

Answer 12

this is a cubic .....

Answer 13

y^2-2x^3-4y+8=0

2yy' -6x^2-4y' = 0

y' = 6x^2/(2y-4) for the slope of any point in the curve

Answer 14

thx

Answer 15

if we shift this and look at the graph from the origin

(y+2)^2-2(x+1)^3-4(y+2)+8=0

it looks like we can get a better view of it

Answer 16

\[y=2\pm\sqrt{2x^3-4}\]
\[y' = \frac{3x^2}{\sqrt{2x^3-4}}\]

slope of a line from 1,2 to x,y is:
\[\frac{y-2}{x-1}= \frac{3x^2}{\sqrt{2x^3-4}}\]

sub in for y to get it all in xs

\[\frac{\sqrt{2x^3-4}}{x-1}= \frac{3x^2}{\sqrt{2x^3-4}}\]
solve for x

Answer 17

\[2x^3-4 = 3x^3-3x^2\]

\[-x^3+3x^2-4 =0\]

Answer 18

x=-1, 2. but only 2 is in the domain to begin with; so when x=2 we have points of tangency with the line