# Point of contact to given curve.

Quote from Alok on August 19, 2020, 10:08 amFind 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).$

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).$

Quote from Hvmaths on August 19, 2020, 10:43 amThe 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.

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.