# Forum

Forum breadcrumbs - You are here:

# Point of contact to given curve.

Find the sum of y- coordinate of point of contact to the curve

$y^2-4y-2x^3+8=0$. The tangents are drawn from the point $(1,2).$

The given equation can be written as

$(y-2)^2+4=2x^3$ Now let Y=y-2 hence equation will be

$Y^2+4=2x^3$ in this if you replace Y with -Y equation remain same, hence curve is symmetric about X axis i.e Y=0.

Hence if $Y_r=y_1-2$ will be one of y coordinate of point of contact then other would be $-Y_r=y_2-2$ hence sum $y_1+y_2=4\;---(1)$ for all $r=1,2,3...$

Now find all possible value of r

Any parametric coordinate on

$2x^3=Y^2+4$ will be $(t,\pm \sqrt{2t^3-4})$ where $t\ge 2^{\frac{1}{3}}$

Using symmetric about X axis finding t when Y is positive.

Hence $\cfrac{dy}{dx}=\cfrac{3t^2}{\sqrt{2t^3-4}}$

Hence tangent at t is

$Y-\sqrt{2t^3-4}=\cfrac{3t^2}{\sqrt{2t^3-4}}(x-t)$ the tangent passes through the point (1,0)

Hence we get $t^3-3t^2+4=0$

Whose roots are -1,2,2

Hence t=2 will give required tangent.

Hence from symmetry there will be only two point of contact above and below the X axis.

Hence from (1) sum of y- coordinate of point of contact will be 4.

Summary
Article Name
Discussion Forum
Description
While studying different concepts of mathematics we come across many questions which we want to find solutions. This forum is also to help all students and all experts or anyone who know the answer can give answer to any question. Throught this we get different solutions to same questions and different line of thoughts.
Author