Find the equation of the parabola,if
(i) the focus is at (-6,-6) and the vertex is at (-2,2)
(ii) the focus is at (0,-3) and the vertex is at (0,0)
(iii) the focus is at (0,-3) and the vertex is at (-1,-3)
(iv) the focus is at (a,0) and the vertex is at (a',0)
(v) the focus is at (0,0) and vertex is at the intersection of the lines x+y=1 and x-y=3.
Find the equation of the parabola,if
(i) the focus is at (-6,-6) and the vertex is at (-2,2)
(ii) the focus is at (0,-3) and the vertex is at (0,0)
(iii) the focus is at (0,-3) and the vertex is at (-1,-3)
(iv) the focus is at (a,0) and the vertex is at (a',0)
(v) the focus is at (0,0) and vertex is at the intersection of the lines x+y=1 and x-y=3.
(i) Given focus (-6,-6) Vertex (-2,2)
Slope of line connecting vertex and focus is 2+6−2+6=2
Slope of directrix will be −12,because both lines are perpendicular
Vertex is the midpoint of focus and point on directrix which passes through axis
−2=−6+x2;2=−6+y2(x,y)=(2,10)
Equation of directrix is given by
y−10=−12(x−2)2y−10=−x+2x+2y=22
Now let P(x,y) be any point on the parabola whose focus is S(-6,-6) and the directrix is 2y+x-22=0
Draw PM perpendicular to 2y+x-22=0
Then we have SP=PM
⇒SP2=PM2
Equation of Parabola is (x+6)2+(y+6)2=(x+2y−22)25
5[x2+y2+36+36+12x+12y]=[x2+4y2+484+4xy−88y−44x]4x2+y2−124−4xy+104x+148y=0(2x−y)2+4(26x+37y−31)=0
(ii) In parabola,vertex is the mid-point of the focus and the point of the intersection of the axis and directrix so,let (x1,y1) be the coordinate of the point of intersection of the axis and directrix.
Then (0,0) is the mid-point of the line segment joining (0,-3) and (x1,y1).
∴ x1+02=0 and y1−32=0⇒x1=0 andy1=3
Thus,the directrix meets the axis at (0,3)
∴ The equation of the directrix is y=3
Claerly,the required parabola is of the form x2=−4ay,where a=3
∴ equation of parabola is x2=−4×3×y⇒ x2=−12y
(iii) In parabola,vertex is the mid-point of the focus and the point of the intersection of the axis and directrix so,let (x1,y1) be the coordinate of the point of intersection of the axis and directrix.
Then (-1,-3) is the mid-point of the line segment joining (0,-3) and (x1,y1).
∴ x1+02=−1 and y1−32=−3⇒x1=−2 andy1=−3
Thus,the directrix meets the axis at (-2,-3)
Let A be the vertex and S be the focus of the required parabola.
Then,
m1=slope of AS=−3−(−3)0−(−1)=0∴ slope of.the directrix=−10=∞
∴ Directrix is perpendicular to the axis
Thus,the directrix passes through (-2,-3) and has slope ∞.so its equation is y-(-3)= ∞{x-(-2)}
y+3∞=x+2⇒ x+2=0
Let p(x,y) be a point on the parabola.
Then,PS=Distance of P from the directrix.
√(x−0)2+(y+3)2=∣∣
∣∣x+2√(1)2∣∣
∣∣⇒x2+(y+3)2=(x+2)2⇒x2+y2+9+6y=x2+4+4xy2−4x+6y+9−4=0⇒y2−4x+6y+5=0Hence,the required equation of parabola isy2−4x+6y+5=0
(iv) In parabola,vertex is the mid-point of the focus and the point of the intersection of the axis and directrix so,let (x1,y1) be the coordinate of the point of intersection of the axis and directrix.
Then (a',0) is the mid-point of the line segment joining (a,0) and (x1,y1).
∴ x1+a2=a′ and y1+02=0⇒x1=2a′−a andy1=0
Thus,the directrix meets the axis at (2a'-a)
Let p(x,y) be a point on the parabola.
Then, SP=PM.
⇒SP2=PM2
Equation of Parabola is (x−a)2+(y−0)2=[x−2a′+a√12]2⇒x2+a2−2ax+y2=(x−2a′+a)2⇒ x2+a2−2ax+y2=x2+(−2a)2+(a)2+2×x×(−2a)+2×(−2a)×a+2×a×x⇒ x2+a2−2ax+y2=x2+4(a′)2+a2−4xa′−4a′a+2ax⇒y2=x2−x2+a2−a2+2ax+4(a′)2−4xa′−4a′a+2ax⇒y2=4ax=4xa+4(a′)2−4a′a⇒y2=x2−x2+a2−a2+2a+4(a′)2−4xa′−4a′a+2ax⇒y2=4ax−4xa′+4(a′)2−4a′a⇒y2=4a′x−4a′a−4xa′+4(a′)2=4a(x−a′)−4a′(x−a′)=(4a−4a′)(x−a′)=4(a−a′)(x−a′)∴ y2=4(a−a′)(x−a′)⇒y2=−4(a′−a)(x−a′)Hence,required equation of parabola isy2=−4(a′−a)(x−a′)
(v) x+y=1 and x-y=3
Intersecting point of above lines is (x,y)=(2,-1) ...vertex
Focus (0,0)
Vertex is the midpoint of focus and point on directrix which passes through 2=0+x2;−1=0+y2(x,y)=(4,−2)
Slope of line passing through focus and vertex is −12
Slope of directrix is 2,as both are perependicular lines
y+2=2(x−4)2x−y=10 directrixSP2=PM25(x2+y2)=(2x−y−10)2x2+4y2−100+4xy−20y+40x=0(x+2y)2+20(2x−y−5)=0
(i) Given focus (-6,-6) Vertex (-2,2)
Slope of line connecting vertex and focus is 2+6−2+6=2
Slope of directrix will be −12,because both lines are perpendicular
Vertex is the midpoint of focus and point on directrix which passes through axis
−2=−6+x2;2=−6+y2(x,y)=(2,10)
Equation of directrix is given by
y−10=−12(x−2)2y−10=−x+2x+2y=22
Now let P(x,y) be any point on the parabola whose focus is S(-6,-6) and the directrix is 2y+x-22=0
Draw PM perpendicular to 2y+x-22=0
Then we have SP=PM
⇒SP2=PM2
Equation of Parabola is (x+6)2+(y+6)2=(x+2y−22)25
5[x2+y2+36+36+12x+12y]=[x2+4y2+484+4xy−88y−44x]4x2+y2−124−4xy+104x+148y=0(2x−y)2+4(26x+37y−31)=0
(ii) In parabola,vertex is the mid-point of the focus and the point of the intersection of the axis and directrix so,let (x1,y1) be the coordinate of the point of intersection of the axis and directrix.
Then (0,0) is the mid-point of the line segment joining (0,-3) and (x1,y1).
∴ x1+02=0 and y1−32=0⇒x1=0 andy1=3
Thus,the directrix meets the axis at (0,3)
∴ The equation of the directrix is y=3
Claerly,the required parabola is of the form x2=−4ay,where a=3
∴ equation of parabola is x2=−4×3×y⇒ x2=−12y
(iii) In parabola,vertex is the mid-point of the focus and the point of the intersection of the axis and directrix so,let (x1,y1) be the coordinate of the point of intersection of the axis and directrix.
Then (-1,-3) is the mid-point of the line segment joining (0,-3) and (x1,y1).
∴ x1+02=−1 and y1−32=−3⇒x1=−2 andy1=−3
Thus,the directrix meets the axis at (-2,-3)
Let A be the vertex and S be the focus of the required parabola.
Then,
m1=slope of AS=−3−(−3)0−(−1)=0∴ slope of.the directrix=−10=∞
∴ Directrix is perpendicular to the axis
Thus,the directrix passes through (-2,-3) and has slope ∞.so its equation is y-(-3)= ∞{x-(-2)}
y+3∞=x+2⇒ x+2=0
Let p(x,y) be a point on the parabola.
Then,PS=Distance of P from the directrix.
√(x−0)2+(y+3)2=∣∣ ∣∣x+2√(1)2∣∣ ∣∣⇒x2+(y+3)2=(x+2)2⇒x2+y2+9+6y=x2+4+4xy2−4x+6y+9−4=0⇒y2−4x+6y+5=0Hence,the required equation of parabola isy2−4x+6y+5=0
(iv) In parabola,vertex is the mid-point of the focus and the point of the intersection of the axis and directrix so,let (x1,y1) be the coordinate of the point of intersection of the axis and directrix.
Then (a',0) is the mid-point of the line segment joining (a,0) and (x1,y1).
∴ x1+a2=a′ and y1+02=0⇒x1=2a′−a andy1=0
Thus,the directrix meets the axis at (2a'-a)
Let p(x,y) be a point on the parabola.
Then, SP=PM.
⇒SP2=PM2
Equation of Parabola is (x−a)2+(y−0)2=[x−2a′+a√12]2⇒x2+a2−2ax+y2=(x−2a′+a)2⇒ x2+a2−2ax+y2=x2+(−2a)2+(a)2+2×x×(−2a)+2×(−2a)×a+2×a×x⇒ x2+a2−2ax+y2=x2+4(a′)2+a2−4xa′−4a′a+2ax⇒y2=x2−x2+a2−a2+2ax+4(a′)2−4xa′−4a′a+2ax⇒y2=4ax=4xa+4(a′)2−4a′a⇒y2=x2−x2+a2−a2+2a+4(a′)2−4xa′−4a′a+2ax⇒y2=4ax−4xa′+4(a′)2−4a′a⇒y2=4a′x−4a′a−4xa′+4(a′)2=4a(x−a′)−4a′(x−a′)=(4a−4a′)(x−a′)=4(a−a′)(x−a′)∴ y2=4(a−a′)(x−a′)⇒y2=−4(a′−a)(x−a′)Hence,required equation of parabola isy2=−4(a′−a)(x−a′)
(v) x+y=1 and x-y=3
Intersecting point of above lines is (x,y)=(2,-1) ...vertex
Focus (0,0)
Vertex is the midpoint of focus and point on directrix which passes through 2=0+x2;−1=0+y2(x,y)=(4,−2)
Slope of line passing through focus and vertex is −12
Slope of directrix is 2,as both are perependicular lines
y+2=2(x−4)2x−y=10 directrixSP2=PM25(x2+y2)=(2x−y−10)2x2+4y2−100+4xy−20y+40x=0(x+2y)2+20(2x−y−5)=0
