Answer to Question #104277 in Analytic Geometry for Gayatri Yadav

Question

a) If the equation of a parabola with the focus at (3,−4) and the directrix x+y = 2 is x^2 +y^2 −2xy−8x+20y+c = 0, then what is the value of c?
b) Find the equation of the plane passing through the line of intersection of the planes
2x+3y+z = 4 and x+y+z = 2, and which is perpendicular to the plane 2x+3y−z = 3.
c) What is the new equation of the conic x^2 +y^2 +4x−2y+3 = 0, when
i) the origin is shifted at (2,−1), followed by a rotation of axes through 45°?
ii) the axes are rotated through 45°, followed by the shifting of the origin at(2,−1)?
Are the equations in i) and ii) above the same?Why?
d) Does there pass a plane through the lines x+4/3 =y/2 =z−1/3
and x/2 =y−1/1 =z+1/1?Justify.
e) Find the equation of the right circular cone whose vertex is (1,0,1), the axis is
x−1 = y−2 = z−3, and the semi-vertical angle is 30°. Also, find the section of the cone by the coordinate planes.

Answer 1

a) A parabola is the set of points P whose distances from a fixed point F in the plane are equal to their distances from a fixed line l in the plane. The fixed point F is called focus and the fixed line l the directrix of the parabola. 

Given that $F(3, -4),$ directrix: $x+y=2$

Using the definition of parabola, we have 

$$\sqrt{(x-3)^2+(y-(-4))^2 }=\lvert{x+y-2 \over \sqrt{1^2+1^2}}\rvert$$ $$2x^2-12x+18+2y^2+16y+32=x^2+2xy-4x-4y+y^2+4$$ $$x^2+y^2-2xy-8x+20y+46=0$$

$c=46$

b) Any plane passing through the intersection of the planes $2x+3y+z-4=0$ and $x+y+z-2=0$ is of the form

$$2x+3y+z-4+\lambda(x+y+z-2)=0$$ $$(2+\lambda)x+(3+\lambda)y+(1+\lambda)z+(-4-2\lambda)=0$$

Since it is perpendicular to the plane $2x+3y-z-3=0,$ we get

$$(2+\lambda)\cdot2+(3+\lambda)\cdot3-(1+\lambda)\cdot1=0$$

$$4+2\lambda+9+3\lambda-1-\lambda=0$$ $$4\lambda=-12$$ $$\lambda=-3$$

Substitute

$$2x+3y+z-4-3(x+y+z-2)=0$$ $$2x+3y+z-4-3x-3y-3z+6=0$$ $$-x-2z-2=0$$

The equation of the plane passing through the line of intersection of the planes

2x+3y+z = 4 and x+y+z = 2, and which is perpendicular to the plane 2x+3y−z = 3 is

$$x+2z+2=0$$

c)

i)The shifted origin has the coordinates (2, -1). Let the coordinates of any point (x,y) on the given line changes to (x',y') on shifting the origin to (2,-1).

$$x=x'+2, y=y'-1$$ $$(x'+2)^2+(y'-1)^2+4(x'+2)-2(y'-1)+3=0$$ $$x'^2 +4x'+4+y'^2-2y'+1+4x'+8-2y'+2+3=0$$ $$x'^2+y'^2+8x'-4y'+18=0$$

Find a new representation after rotating through of axes through 45°

$$x'=x''\cos45\degree-y''\sin45\degree$$ $$y'=x''\sin45\degree+y''\cos45\degree$$

$$x'={x''-y'' \over \sqrt{2}}$$ $$y'={x''+y'' \over \sqrt{2}}$$

$$({x''-y'' \over \sqrt{2}})^2+({x''+y'' \over \sqrt{2}})^2+8({x''-y'' \over \sqrt{2}})-4({x''+y'' \over \sqrt{2}})+18=0$$

$${x''^2 \over 2}-x''y''+{y''^2 \over 2}+{x''^2 \over 2}+x''y''+{y''^2 \over 2}+$$ $$+4\sqrt{2}x''-4\sqrt{2}y''-2\sqrt{2}x''-2\sqrt{2}y''+18=0$$

$$x''^2+y''^2+2\sqrt{2}x''-6\sqrt{2}y''+18=0$$

ii) Find a new representation after rotating through of axes through 45°

$$x=x'\cos45\degree-y'\sin45\degree$$ $$y=x'\sin45\degree+y'\cos45\degree$$

$$x={x'-y' \over \sqrt{2}}$$ $$x={x'+y' \over \sqrt{2}}$$

$$({x'-y' \over \sqrt{2}})^2+({x'+y' \over \sqrt{2}})^2+4({x'-y' \over \sqrt{2}})-2({x'+y' \over \sqrt{2}})+3=0$$

$${x'^2 \over 2}-x'y'+{y'^2 \over 2}+{x'^2 \over 2}+x'y'+{y'^2 \over 2}+$$ $$+2\sqrt{2}x'-2\sqrt{2}y'-\sqrt{2}x'-\sqrt{2}y'+3=0$$

$$x'^2+y'^2+\sqrt{2}x'-3\sqrt{2}y'+3=0$$

The shifted origin has the coordinates (2, -1).

$$x'=x''+2, y'=y''-1$$

$$(x''+2)^2+(y''-1)^2+\sqrt{2}(x''+2)-3\sqrt{2}(y''-1)+3=0$$

$$x''^2+4x''+4+y''^2-2y''+1+$$ $$+\sqrt{2}x''+2\sqrt{2}-3\sqrt{2}y''+3\sqrt{2}+3=0$$

$$x''^2+y''^2+(4+\sqrt{2})x''-(2+3\sqrt{2})y''+5\sqrt{2}+8=0$$

The equations in i) and ii) above are not the same.The order of transformations is important.

d) Write the equations in parametric form.

$$L_1:x=2t, y=t+1, z=t-1$$ $$L_2:x=3s-4,y=2s,z=3s+1$$

The lines are not parallel since the vectors $v_1=(2,1,1)$ and $v_2=(3,2,3)$ are not parallel. Next we try to find intersection point by equating $x,y,$ and $z.$

$$2t=3s-4$$ $$t+1=2s$$ $$t-1=3s+1$$

$$2t-t-1-t+1=3s-4-2s$$ $$0=s-4$$ $$s=4$$

$$2t=12-4$$ $$t=8-1$$ $$t=12+2$$

We have contradiction. So there is no solution for $s$ and $t.$ Since the two lines are neither parallel nor intersecting, they are skew lines. No plane passes through the lines x+4/3 =y/2 =z−1/3

and x/2 =y−1/1 =z+1/1.

e) Find the equation of the right circular cone whose vertex is (1,2,3), the axis is x−1 = y−2 = z−3, and the semi-vertical angle is 30°. Also, find the section of the cone by the coordinate planes

Let $P(x,y,z)$ be any point on the cone with vertex at $V(1,2,3)$ and the axis

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

The direction ratios of $PV$ are $x-1,y-2,z-3$

$$\cos30\degree={x-1+y-2+z-3 \over \sqrt{1^2+1^2+1^2}\sqrt{(x-1)^2+(y-2)^2+(z-3)^2}}$$

$${\sqrt{3} \over 2}={x+y+z-6 \over \sqrt{3}\sqrt{(x-1)^2+(y-2)^2+(z-3)^2}}$$

$$3 \sqrt{(x-1)^2+(y-2)^2+(z-3)^2}=2(x+y+z-6)$$

$$9x^2-18x+9+9y^2-36y+36+9z^2-54z+81=$$ $$=4x^2+8xy+4y^2+4z^2-48z+144+8xz-48x+8yz-48y$$

The required equation of the right circular cone

$$5x^2+5y^2+5z^2-8xy-8xz-8yz+30x+12y-6z-18=0$$

$Oxy: z=0$

$$5x^2+5y^2-8xy+30x+12y-18=0,\ ellipse$$

$Oxz: y=0$

$$5x^2+5z^2-8xz+30x-6z-18=0,\ ellipse$$

$Oyz:x=0$

$$5y^2+5z^2-8yz+12y-6z-18=0,\ ellipse$$