Parabola - 3 - Free MCQ Practice Test with solutions, JEE Maths

The parabolas are y² = 4sin² α(x + sin²α) and y² = 4cos² α(x + cos²α), hence the locus is x + cos²α + sin²α = 0 ⇒ x + 1 = 0
Assume the point   as origin and line joining it to the centre as x-axis, the equation to the circle becomes x² + y² - 2x = 0, center is A₁ (1, 0) and the second circle has the equation x² + y² - 2√2 = 0 center A₀ (0, √2)
Similarly A₃(-2, 0), A₄ (0, -2√2) etc.  

Solving x + y = k + 1 and y = x(1 – x) ⇒ x² – 2x + k + 1 = 0  and Δ = 0 

The x-axis touches at A(1, 0) and x = y touches at B(1, 1). Hence the equation to the curve through these points is given by y(y – x) + k(x – 1)² = 0. For this to represent a parabola, 4k = 1. The equation is x² – 4xy + 4y² – 2x + 1 = 0. Vertex  focus

Let the parabola is y² = 4x. The equations  (x - t²) + (y - 2t)² + 2λ (x - yt + t²) = 0 and (x - 1)² + y² + 2μ(mx - y - m) = 0 should represent the same circle touching mx – y – m = 0 at (1, 0) and the parabola at (t², 2t). Eliminating λ and μ, we get mt³ - 3t² + 3mt + 1 = 0 will give three values of t for any given m.  

It lies on y = 0.
∴ 

Eliminating t, we get (3x - 4y + 2)² = 16x + 12y - 27. Vertex is 
Hence  k = 29 and the area is 3π.

Taking the new axes as  we see that the parabola can be  with the required condition |X| <  200

Center of such circle in the case of parabolas  y² = 4ax, x² = 4ay is 

where d is the distance between lines whose equations are ax+by+C₁ ​ =0 & ax+by+C₂ ​= 0

= 5
d = 5
If the distance between vertex and latus rectum = distance of vertex from directrix = a then d = 2a = 5
⇒ Length of latus rectum = 4a = 10

Let P be (t², 2t) area of Δ PAB

it is maximum at t = 1/2. 

If P(t) is the point and Q(T) is another point in the question, we have   and 

can have only one real root, there will be only one such point P. 

(y - 3)² = -4 (x - 3) (given equation of parabola.
∴ equation of latus rectum is x = 2. 

(y – 3)² = –4 (x – 3) (given equation of parabola.
∴ equation of latus rectum is x = 2. 

x = t² +t +1andy = t² -t +1
⇒ x + y - 2 = 2t² and x - y = 2t
⇒ 2(x + y -2) = (x - y)² ⇒ x² + y² -2xy -2x -2y + 4 = 0
Comparing with the equation
ax² + 2hxy + by² + 2gx + 2fy + c = 0, 
∴ abc + 2fgh - af² - bg² - ch² ≠ and h² = ab

The equation of the parabola is 

The equation of any tangent to this parabola is  
If it passes through (3, 4), then 
⇒ 11m² -12m - 4 = 0
Let m₁ & m₂ be the roots of this equation. Then, 

Let θ be the angle between the tangents. Then,

if it is common tangent, then m³ = 8 m = 2. 

If the tangents at P and Q intersect at T, then axis of parabola is parallel to TR, where R is the mid point of P and Q.  So, slope of the axis is 1.
∴ slope of the directrix = – 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

P(x₀ , y₀) : pt on curve nearest to line.
Normal at P is perpendicular to the line Normal at P has slope 
∴ y₀ = 2 and x₀ = 11; P(11, -2)

 which is of the following  PM² = SP²
∴ Focus= (1,1) . Directrix  is x+y+2 =0
Axis is x-y=0 and z= (-1,-1)
Vertex = (0,0). Parameter a = √2
The distance of the point from vertex and lie on axis  from which  3  normals can be drawn must be greater 2a = 2√2. Hence point on axis at a distance 2√2 is (2,2), Hence h>2

Equation of tangent at (x₁, y₁) on the parabola y² = 4x is
2x - y₁ + 2x₁ = 0      
length of perpendicular from centre of the circle (x - 3)² + y² = 9 is


Hence required equation 

Slope of the normal is given to be -1. We know that, foot of the normal is (am², -2am). Here a = 2, m = -1. Hence the required point is (2, 4). 

Eq. of tangent 2y = x + 4
∴ Q ≡ (-4,0)
Eq. of normal is y - 4 = -2 (x - 4)
⇒ y + 2x = 12
Clearly QR is diameter of the required circle.
⇒ (x + 4) (x – 6) + y² = 0
⇒ x² + y² – 2x – 24 = 0
centre (1, 0) 

Let y = mx+ c be such chord with extremities A and B .
∴ The combined equation of the pair of lines OA and OB is 
∴ Coeff of x² + Coeff of y² = 0

∴ c = -4 m
∴ The chord equation is y = m (x- 4) .

Any point on the given parabola is (t², 2t). The equation of the tangent at (1,2) is x-y +1 = 0. The image (h,k) of the point (t²,2t) in x-y + 1 = 0 is given by 

∴ h = t² - t² + 2t - 1 = 2t - 1
Eliminating t from h = 2t – 1 and k = t²+1
we get, (h+1)² = 4(k-1)
The required equation of reflection is (x+1)² = 4(y-1)

λx + y - 5 = 0 is the focal chord
Since it passes through (0, 5)
Let h, k be the mid point of all such chord h = 5(t₁ + t₂), 
 
On eliminating t₁, t₂
x² = 10 (y - 5).