Test: Parabola - 2 - JEE MCQ

If y = -x + 1 is tangent to y² = kx, then
c = a / m

c = (k / 4) / (-1)

1 = (k / 4) / (-1)
k = -4

 Distance from focus (0,0) to directrix x + y = 2 is |0 + 0 – 2|/√2 = 2/√2 = √2.
Thus 2a = √2,
so a = √2/2.
Latus rectum = 4a = 4 · (√2/2) = 2√2.

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