Test: Parabola- 2 - JEE MCQ

Given directrix of parabola ⇒ x+y=2
and force is origin vertex is A(0,0)
We know that perpendicular distance from vertex  of the parabola to directrix is equal to 'a'  where 4a is the Latus Rectum of the 
Parabola 

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.

Given (t² , 2t) be one end of focal chord then other end be (1/t² , −2t )

Length of focal chord = [(t² - 1/t²)² + (2t + t/2)²]^½

= ( t + 1/t)²

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

For intersection of both the curve we must have,

Therefore, the point of intersection is (2,2)(2,2)
Hence, option 'B' is correct.

The equation of parabola be y²=4ax
let the point P be (at²,2at)
PN is ordinate ⇒N(at²,0)
Equation of straight line bisecting NP is
y=at
substituting y in equation of parabola
a²t² = 4ax
⇒x = 4at²
So the coordinates of Q are (4at² ,at)
Equation of NQ is y−0 = (at−0)/(at^2/4 - at²)(x−at²)
y= −4/3t(x−at²)
Put x=0
y = −4/3t(0−at^2)
⇒y=4at/3
⇒AT = 4at/3​
NP = 2at
AT/NP = (4at/3)/2at
= ⅔
AT = 2/3NP

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)]

The Equation of tangent at vertex to parabola x²+4x+2y=0 is :
x²+4x+2y+4−4=0
(x+2)² = −2(y−2)
x²=−2y
Equation of tangent at vertex is y=0
y−2=0
y=2

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 (k,k+3) be the point on the line x−y+3=0
Equation of chord of contact is S₁=0
⇒yy1=4(x+x1)
⇒y(k+3)=4(x+k)
⇒4x−3y−k(y−4)=0
Therefore, straight line passes through fixed point (3,4)

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)

Since the semi latus rectum of a parabola is the harmonic mean between the segment of any focal chord of a parabola, therefore,SP,4,SQ are in H.P.
⇒4=2(SP.SQ)/(SP+SQ)
⇒4=2*6.SQ/(6+SQ)
⇒SQ=3

 Let P(x₁ , y₁) be point of contact of two parabola. tangents at P of the two parabolas are