Answer to Question #104277 in Analytic Geometry for Gayatri Yadav

Question #104277
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.
1
Expert's answer
2020-03-02T17:14:17-0500

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

Using the definition of parabola, we have 


(x3)2+(y(4))2=x+y212+12\sqrt{(x-3)^2+(y-(-4))^2 }=\lvert{x+y-2 \over \sqrt{1^2+1^2}}\rvert2x212x+18+2y2+16y+32=x2+2xy4x4y+y2+42x^2-12x+18+2y^2+16y+32=x^2+2xy-4x-4y+y^2+4x2+y22xy8x+20y+46=0x^2+y^2-2xy-8x+20y+46=0

c=46c=46


b) Any plane passing through the intersection of the planes 2x+3y+z4=02x+3y+z-4=0 and x+y+z2=0x+y+z-2=0 is of the form


2x+3y+z4+λ(x+y+z2)=02x+3y+z-4+\lambda(x+y+z-2)=0(2+λ)x+(3+λ)y+(1+λ)z+(42λ)=0(2+\lambda)x+(3+\lambda)y+(1+\lambda)z+(-4-2\lambda)=0

Since it is perpendicular to the plane 2x+3yz3=0,2x+3y-z-3=0, we get


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

4+2λ+9+3λ1λ=04+2\lambda+9+3\lambda-1-\lambda=04λ=124\lambda=-12λ=3\lambda=-3

Substitute


2x+3y+z43(x+y+z2)=02x+3y+z-4-3(x+y+z-2)=02x+3y+z43x3y3z+6=02x+3y+z-4-3x-3y-3z+6=0x2z2=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=0x+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=y1x=x'+2, y=y'-1(x+2)2+(y1)2+4(x+2)2(y1)+3=0(x'+2)^2+(y'-1)^2+4(x'+2)-2(y'-1)+3=0x2+4x+4+y22y+1+4x+82y+2+3=0x'^2 +4x'+4+y'^2-2y'+1+4x'+8-2y'+2+3=0x2+y2+8x4y+18=0x'^2+y'^2+8x'-4y'+18=0


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


x=xcos45°ysin45°x'=x''\cos45\degree-y''\sin45\degreey=xsin45°+ycos45°y'=x''\sin45\degree+y''\cos45\degree

x=xy2x'={x''-y'' \over \sqrt{2}}y=x+y2y'={x''+y'' \over \sqrt{2}}

(xy2)2+(x+y2)2+8(xy2)4(x+y2)+18=0({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

x22xy+y22+x22+xy+y22+{x''^2 \over 2}-x''y''+{y''^2 \over 2}+{x''^2 \over 2}+x''y''+{y''^2 \over 2}++42x42y22x22y+18=0+4\sqrt{2}x''-4\sqrt{2}y''-2\sqrt{2}x''-2\sqrt{2}y''+18=0

x2+y2+22x62y+18=0x''^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=xcos45°ysin45°x=x'\cos45\degree-y'\sin45\degreey=xsin45°+ycos45°y=x'\sin45\degree+y'\cos45\degree

x=xy2x={x'-y' \over \sqrt{2}}x=x+y2x={x'+y' \over \sqrt{2}}

(xy2)2+(x+y2)2+4(xy2)2(x+y2)+3=0({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

x22xy+y22+x22+xy+y22+{x'^2 \over 2}-x'y'+{y'^2 \over 2}+{x'^2 \over 2}+x'y'+{y'^2 \over 2}++22x22y2x2y+3=0+2\sqrt{2}x'-2\sqrt{2}y'-\sqrt{2}x'-\sqrt{2}y'+3=0


x2+y2+2x32y+3=0x'^2+y'^2+\sqrt{2}x'-3\sqrt{2}y'+3=0

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


x=x+2,y=y1x'=x''+2, y'=y''-1

(x+2)2+(y1)2+2(x+2)32(y1)+3=0(x''+2)^2+(y''-1)^2+\sqrt{2}(x''+2)-3\sqrt{2}(y''-1)+3=0

x2+4x+4+y22y+1+x''^2+4x''+4+y''^2-2y''+1++2x+2232y+32+3=0+\sqrt{2}x''+2\sqrt{2}-3\sqrt{2}y''+3\sqrt{2}+3=0

x2+y2+(4+2)x(2+32)y+52+8=0x''^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.


L1:x=2t,y=t+1,z=t1L_1:x=2t, y=t+1, z=t-1L2:x=3s4,y=2s,z=3s+1L_2:x=3s-4,y=2s,z=3s+1

The lines are not parallel since the vectors v1=(2,1,1)v_1=(2,1,1) and v2=(3,2,3)v_2=(3,2,3) are not parallel. Next we try to find intersection point by equating x,y,x,y, and z.z.


2t=3s42t=3s-4t+1=2st+1=2st1=3s+1t-1=3s+1


2tt1t+1=3s42s2t-t-1-t+1=3s-4-2s0=s40=s-4s=4s=4

2t=1242t=12-4t=81t=8-1t=12+2t=12+2

We have contradiction. So there is no solution for ss and t.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)P(x,y,z) be any point on the cone with vertex at V(1,2,3)V(1,2,3) and the axis


x11=y21=z31{x-1 \over 1}={y-2 \over 1}={z-3 \over 1}

The direction ratios of PVPV are x1,y2,z3x-1,y-2,z-3


cos30°=x1+y2+z312+12+12(x1)2+(y2)2+(z3)2\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}}

32=x+y+z63(x1)2+(y2)2+(z3)2{\sqrt{3} \over 2}={x+y+z-6 \over \sqrt{3}\sqrt{(x-1)^2+(y-2)^2+(z-3)^2}}

3(x1)2+(y2)2+(z3)2=2(x+y+z6)3 \sqrt{(x-1)^2+(y-2)^2+(z-3)^2}=2(x+y+z-6)

9x218x+9+9y236y+36+9z254z+81=9x^2-18x+9+9y^2-36y+36+9z^2-54z+81==4x2+8xy+4y2+4z248z+144+8xz48x+8yz48y=4x^2+8xy+4y^2+4z^2-48z+144+8xz-48x+8yz-48y

The required equation of the right circular cone


5x2+5y2+5z28xy8xz8yz+30x+12y6z18=05x^2+5y^2+5z^2-8xy-8xz-8yz+30x+12y-6z-18=0

Oxy:z=0Oxy: z=0


5x2+5y28xy+30x+12y18=0, ellipse5x^2+5y^2-8xy+30x+12y-18=0,\ ellipse

Oxz:y=0Oxz: y=0


5x2+5z28xz+30x6z18=0, ellipse5x^2+5z^2-8xz+30x-6z-18=0,\ ellipse

Oyz:x=0Oyz:x=0


5y2+5z28yz+12y6z18=0, ellipse5y^2+5z^2-8yz+12y-6z-18=0,\ ellipse


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment