Coordinate Geometry- Parabola MCQs for A Level Exam

Given the vertex of the parabola is (0,0) and focus is at (0,-2).
This gives the axis of the parabola is the positive y− axis.
Then the equation of the parabola will be x^2 = 4ay where a = -2.
So the equation of the parabola is x2 = -8y.

f(0,1),d(x+2=0)
Distance of any point on parabola and focus is equal to distance of point and directrix.
fP=(h−0)²+(k−1)²= (h²+k²+1−2k)^1/2
Distance of point (h,k) and line x+2=0
Using point line distance formula.
dP=h+2
[h²+k²+1−2k]^1/2=h+2
h²+k²+1−2k = h²+4+4h
k²−2k+1−4−4h=0
replacing h→x,k→y  y²−2y+1−4−4x=0
(y−1)²=4(x+1)     …(1)
Let Y=y−1,X=x+1 then (1) becomes 
Y^2=4aX²
Here a=1 any point on this parabola will be of the form (at²,2at)=(t²,2at)
⇒X=t² ⇒x+1=t²
⇒x=t²−1
⇒Y²=2t
⇒y−1 = 2t ⇒ y = 2t+1
∴ Any point on the parabola (y−1)²=4(x+1) is 
= (t²−1,2t+1)

Slope of OP x slope of OQ = -1 
⇒ 4t₁.4t₂ = -1
Eq of tangent at 

Eq of tangent at 

Let (x₁ , y₁) is the point of intersection

ky=[4(x+h)]/2
=> 2ky=2(x+h)
2ky=4x+4h  =>4x−2ky+4h=0
4x−7y+10=0
4h=10  => h=5/2
2k=7 => k=7/2
point of intersection of tan⁡gent at p and q is (5/2,7/2)

y²+3=2(2x+y) represents parabola.
y²+3=4x+2y
y²−2y+3=4x
y²−2y+1+3=4x+1
(y−1)²=4x−2
(y−1)²=4(x−1/2)
So, the vertex of parabola=(1/2,1) and axis is parallel to x axis.
a=1
Focus=(1/2+1,1)
=(3/2,1)

Given the equation of parabola is
y² = 4ax
Let P(at², 2at) be any point on the parabola.
Equation of tangent at point P is ty = x + at²  where slope of the tangent is 1/t.  
Equation of line perpendicular to the tangent passes through (a,0) is given as
∴ y−0 = −t(x−a) 
or y = t(a−x)                        .....(i)
Equation of OP is given by
y−0 = 2/t(x−0) = 0 
⇒ y = 2x/t                            .....(ii)
Eliminating 't' from equations (i) and (ii), we get
y² = 2x(a−x)
or  2x² + y² − 2ax = 0

Focus of parabola y² = 8x is (2,0). Equation of circle with centre (2,0) is (x−2)² + y² = r²
Let AB is common chord and Q is mid point i.e. (1,0)
AQ² = y² = 8x
= 8×1 = 8
∴ r² = AQ² + QS²
= 8 + 1 = 9
So required circle is (x−2)² + y² = 9

As the quadrants lies in first and second quadrant
y = -a    Focus(0,a)
x^{2 = 4ay}
 

x+y=3 m=−1
(−2,1)m = 1
y−1 = 1(x+2)
y−1 = x+2
x−y+3 = 0
x−y+3 = 0
x+y−3 = 0
2x = 0
x = 0; y = 3
Vertex=
Midpoint of focus and directrix = (0−2)/2,(3+1)/2
​=(−1,2)

The chord of a parabola through the focus and perpendicular to the axis is called the latus rectum.

If the parabola is Y²= 4ax
 
take the focal chord which is easy for calculation e.x. LR (latus rectum)
then coordinates of extremities would be (a,2a) and (a,-2a)
 
equation of tangent of parabola at (a,2a) :
T=0 : 2ay = 2a(x+a)
y = x+a................................. (1)
equation of tangent of parabola at (a,-2a) :
T=0 -2ay = 2a (x+a)
y = -x-a .........................(2)
point of intersection of both tangents is (X₁, Y₁)
after solving eq1 and eq2 X₁ = -a and Y1 = 0
so ( -a, 0)
eqn of normal of parabola at (a, 2a)
y = -x +3a ...............................(3)
 
eqn of normal of parabola at (a, -2a)
y = x -3a..............................(4)
so point of intersection of normal's : (X₂, Y₂)
after solving eq3 and eq4 X₂= 3a and Y₂ = 0
so we conclude... for y₂= 4ax
Y₁= Y₂

x² − 2 = −2cost
⇒ x² = 2 − 2cost
⇒x² = 2(1−cost)
⇒x² = 2(1−(1−2sin² t/2))
⇒x² = 4sin² t/2
We have y = 4cos² t/2
⇒cos² t/2= y/4
We know the identity, sin² t/2 + cos² t/2 = 1
⇒ x²/4 + y/4 = 1
⇒ x² = 4−y represents a parabolic profile.

Parametric coordinates of the parabola y² = 4ax are (at², 2at).

Given parabola is, y²+4y−6x−2=0
⇒ y²+4y+4=6x+6=6(x+1)
⇒ (y+2)² = 6(x+1)
shifting origin to (−1,−2)
Y² = 4aX  where a = 3/2
We know locus of point of intersection of perpendicular tangent is directrix of the parabola itself
Hence required locus is X=−a ⇒ x+1=−3/2
⇒ 2x+5=0

Let the point be (h, k)

Now equation of tangent to the parabola y² = 4ax whose slope is m is

as it passes through (h, k)]

Comparing the equation with the standard form ax²+2hxy+by²+2gx+2fy+c=0
a=2,h=−3/2,b=5,g=3,f=−3/2,c=5
Δ=abc+2fgh−af²−bg²−ch²
=(2)(5)(5)+2(−3/2)(3)(−3/2)−(2)(−3/2)²−(5)(3)²−(5)(−3/2)²
=50+27/2−9/2−45−225/4
=−169/4 is not equal to 0
Descriminant =h²−ab
=(−3/2)²−(2)(5)
= 9/4−10
= −31/4<0
So, the curve represents either a circle or an ellipse
a is not equal to b and  
Δ/a+b = −(169/4)/2+5
=−169/28<0
So, the curve represents a ellipse.

Method to Solve :

x+y-1=0. or. y=1-x         (1)

y^2. = kx                        .(2)

On putting y=1-x from eq.(1)

(1-x)^2=k.x

1–2x+x^2 =kx

x^2 - (k+2)x+1 = 0

The given line is tangent to the parabola which touches the parabola one and only one point. means both the roots are same.

D or. b^ - 4 a.c. =0

{-(k+2)}^2. - 4.1.1. =0

(k+2)^2. = 4

(k+2) = +/-(2)

k= -2 +/-2

k= -2+2 or. -2–2.

k =0. , - 4. Answer.