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Test: Parabola- 4 - JEE MCQ


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15 Questions MCQ Test Chapter-wise Tests for JEE Main & Advanced - Test: Parabola- 4

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Test: Parabola- 4 - Question 1

The straight line joining any point P on the parabola y2 = 4ax to the vertex and perpendicular from the focus to the tangent at P, intersect at R, then the equation of the locus of R is

Detailed Solution for Test: Parabola- 4 - Question 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

Test: Parabola- 4 - Question 2

PQ is a normal chord of the parabola y2 = 4ax at P, A being the vertex of the parabola. Through P a line is drawn parallel to AQ meeting the x-axis in R. Then the length of AR is

Detailed Solution for Test: Parabola- 4 - Question 2

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Test: Parabola- 4 - Question 3

If the tangents and normals at the extremities of a focal chord of a parabola intersect at (x1, y1) and (x2, y2) respectively, then

Detailed Solution for Test: Parabola- 4 - Question 3

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

Test: Parabola- 4 - Question 4

From point P two tangents are drawn from it to the parabola y2 = 4x such that the slope of one tangent is three times the slope of the other. The locus of P is

Test: Parabola- 4 - Question 5

The equation of a straight line passing through the point (3, 6) and cutting the curve y = √x orthogonally is

Detailed Solution for Test: Parabola- 4 - Question 5

The curve y=(x)1/2 is the part of curve y2=x
Equation of normal at P(t2/4, t/2) is
y+tx=t/2+t3/4..... (1)
The equation will pass through (3,6)
6+3t=t/2+t3/4
t3−10t−24=0
Solving, we get t=4
So equation of line which is orthogonal and passes through (3,6) is
y+4x=18
4x+y-18 = 0

Test: Parabola- 4 - Question 6

The tangent and normal at P (t), for all real positive t, to the parabola y2 = 4ax meet the axis of the parabola in T and G respectively, then the angle at which the tangent at P to the parabola is inclined to the tangent at P to the circle through the points P, T and G is

Detailed Solution for Test: Parabola- 4 - Question 6

Equation of tangent and normal at P(at2,2at) on y2=4ax are
ty=x+at2.....(1)
y+tx=2at+at3...(2)
So, T(−at2,0) & G(2a+at2,0) 
Equation of circle passing through P,T and G is
(x+at2)(x−(2a+at2))+(y−0)(y−0)=0
x2+y2−2ax−at2(2a+at2)=0
Equation of tangent on the above circle at P(at2,2at) is at2x+2aty−a(x+at2)−at2(2a+at2)=0
Slope of line which is tangent to circle at P

Test: Parabola- 4 - Question 7

AB, AC are tangents to a parabola y2 = 4ax. p1 p2 and p3 are the lengths of the perpendiculars from A, B and C respectively on any tangent to the curve, then p2, p1, p3 are in

Detailed Solution for Test: Parabola- 4 - Question 7

Test: Parabola- 4 - Question 8

Through the vertex O of the parabola, y2 = 4ax two chords OP and OQ are drawn and the circles on OP and OQ as diameter intersect in R. If q1, q2 and f are the angles made with the axis by the tangent at P and Q on the parabola and by OR then the value of cotq1 + cotq2 equals

*Multiple options can be correct
Test: Parabola- 4 - Question 9

Let A be the vertex and L the length of the latus rectum of parabola, y2 – 2y – 4x – 7 = 0. The equation of the parabola with point A as vertex, 2L as the length of the latus rectum and the axis at right angles to that of the given curve is

*Multiple options can be correct
Test: Parabola- 4 - Question 10

Tangent to the parabola y2 = 4ax at point P meets the tangents at vertex A at point B and the axis of parabola at T, Q is any point on this tangent and N as the foot of perpendicular from Q on SP, where S is focus, M is the foot of perpendicular from Q on the directrix then

*Multiple options can be correct
Test: Parabola- 4 - Question 11

The parametric coordinates of any point on the parabola y2 = 4ax can be

Detailed Solution for Test: Parabola- 4 - Question 11

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

*Multiple options can be correct
Test: Parabola- 4 - Question 12

The length of the chord of the parabola y2 = x which is bisected at the point (2, 1) is less than

Detailed Solution for Test: Parabola- 4 - Question 12

Equation of chord to the given parabola with given mid point (2,1) is given by, 
T = S1

*Multiple options can be correct
Test: Parabola- 4 - Question 13

The locus of the mid point of the focal radii of a variable point moving on the parabola, y2 = 4ax is a parabola whose

*Multiple options can be correct
Test: Parabola- 4 - Question 14

A variable circle is described to passes through the point (1, 0) and tangent to the curve y = tan(tan-1x). The locus of the centre of the circle is a parabola whose

*Multiple options can be correct
Test: Parabola- 4 - Question 15

Two parabolas have the same focus. If their directrices are the x-axis & the y-axis respectively, then the slope of their common chord is

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