Answer to Question #104278 in Analytic Geometry for Gayatri Yadav

Question

A right circular cylinder passes through the point (1,−1,4) and has the axis along
the line x−1/2 =y−3/5 =z+1/3. Is this information sufficient to determine the equation of
the cylinder? If it is, determine the equation of the cylinder. Otherwise, state another condition so that the equation can be determined uniquely, and also find the equation.

Answer 1

Let $P(x_1, y_1, z_1)$ be any point on the cylinder. Draw PM perpendicular to the axis of the cylinder. Then $PM=r.$

Let $A(a, b, c)$ be the point which lies on the axis

$${x-a \over l}={y-b \over m}={z-c \over n}$$

Now

$$AP^2=(x_1-a)^2+(y_1-b)^2+(z_1-c)^2$$

$$MA = projection\ of\ AP\ on\ the\ axis$$ $$MA={l(x-a)+m(y-b)+n(z-c) \over \sqrt{l^2+m^2+n^2}}$$

Now, from the right angled â–³$AMP$ , we get

$$AP^2-MA^2=MP^2$$ $$(x_1-a)^2+(y_1-b)^2+(z_1-c)^2 -$$ $$-\big({l(x_1-a)+m(y_1-b)+n(z_1-c) \over \sqrt{l^2+m^2+n^2}}\big)^2=r^2$$

Then the required equation of the cylinder is

$$(x-a)^2+(y-b)^2+(z-c)^2 -$$ $$-\big({l(x-a)+m(y-b)+n(z-c) \over \sqrt{l^2+m^2+n^2}}\big)^2=r^2$$

Given $P(1, -1, 4),$ the line

$${x-1 \over 2}={y-3 \over 5}={z+1 \over 3}$$

$x_1=1, y_1=-1, z_1=4,$

$a=1,b=3,c=-1,$

$l=2,m=5,n=3$

$$r^2=(1-1)^2+(-1-3)^2+(4-(-1))^2 -$$ $$-\big({2(1-1)+5(-1-3)+3(4-(-1)) \over \sqrt{2^2+5^2+3^2}}\big)^2$$

$$r^2={1533\over 38}$$

The equation of the right circular cylinder

$$(x-1)^2+(y-3)^2+(z+1)^2 -$$ $$-{(2(x-1)+5(y-3)+3(z+1))^2 \over 38}={1533\over 38}$$