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JEE Exam  >  JEE Tests  >  35 Years Chapter wise Previous Year Solved Papers for JEE  >  Test: Comprehension Based Questions: Conic Sections - JEE MCQ

Test: Comprehension Based Questions: Conic Sections - JEE MCQ


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14 Questions MCQ Test 35 Years Chapter wise Previous Year Solved Papers for JEE - Test: Comprehension Based Questions: Conic Sections

Test: Comprehension Based Questions: Conic Sections for JEE 2024 is part of 35 Years Chapter wise Previous Year Solved Papers for JEE preparation. The Test: Comprehension Based Questions: Conic Sections questions and answers have been prepared according to the JEE exam syllabus.The Test: Comprehension Based Questions: Conic Sections MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Comprehension Based Questions: Conic Sections below.
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Test: Comprehension Based Questions: Conic Sections - Question 1
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PASSAGE 1

Consider the circle x2 + y2 = 9 and the parabola y2 = 8x. They intersect at P and Q in the first and the fourth quadrants, respectively. Tangents to the curcle at P and Q intersect the x-axis at R and tangents to the parabola at P and Q intersect the x-axis at S. (2007 -4 marks)

Q. The ratio of the areas of the triangles PQS and PQR is

Detailed Solution for Test: Comprehension Based Questions: Conic Sections - Question 1

Test: Comprehension Based Questions: Conic Sections - Question 2
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PASSAGE 1

Consider the circle x2 + y2 = 9 and the parabola y2 = 8x. They intersect at P and Q in the first and the fourth quadrants, respectively. Tangents to the curcle at P and Q intersect the x-axis at R and tangents to the parabola at P and Q intersect the x-axis at S. (2007 -4 marks)

Q. The radius of the circumcircle of the triangle PRS is (2007 -4 marks)

Detailed Solution for Test: Comprehension Based Questions: Conic Sections - Question 2

For ΔPRS,

b = PR = 6 , c = SR = 10

∴ Radius of circumference

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Test: Comprehension Based Questions: Conic Sections - Question 3
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PASSAGE 1

Consider the circle x2 + y2 = 9 and the parabola y2 = 8x. They intersect at P and Q in the first and the fourth quadrants, respectively. Tangents to the curcle at P and Q intersect the x-axis at R and tangents to the parabola at P and Q intersect the x-axis at S. (2007 -4 marks)

Q. The radius of the incircle of the triangle PQR is (2007 -4 marks)

Detailed Solution for Test: Comprehension Based Questions: Conic Sections - Question 3

Radius of incircle

We have a = PR  = 

c = PQ = 

and 

  

Test: Comprehension Based Questions: Conic Sections - Question 4
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PASSAGE 2

The circle x2 + y2 – 8x = 0 and hyperbola  intersect atthe points A and B. (2010)

Q. Equation of a common tangent with positive slope to the circle as well as to the hyperbola is
Detailed Solution for Test: Comprehension Based Questions: Conic Sections - Question 4

Any tangent to  is 

It touches circle with center (4,0) and radius = 4

⇒ 16 sec2α – 24 secα + 9 =

⇒ 12 sec2α + 8 sec– 15= 0

but sec α  is not possible

∴ sec α = –3/2

∴  slope of tangent = .

(for +ve value of m)

∴ Equation of tangent is 

or 2 x – y + 4=0

Test: Comprehension Based Questions: Conic Sections - Question 5
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PASSAGE 2

The circle x2 + y2 – 8x = 0 and hyperbola  intersect atthe points A and B. (2010)

Q. Equation of the circle with AB as its diameter is
Detailed Solution for Test: Comprehension Based Questions: Conic Sections - Question 5

The intersection points of given circle x2 + y2 – 8x= 0...(1) .
and hyperbola 4 x2 –9y2 –36 = 0 ...(2)
can be obtained by solving these equations
Substituting value of y2 from eqn (1) in eqn (2), we get 4 x2 – 9(8x –x2) = 36 ⇒ 13x2 – 72x – 36= 0

⇒(not possible)

 are points of intersection.
So eqn of required circle is

⇒ x2 + 36 – 12 x +y2 – 12= 0

⇒ x2+ y2 – 12x + 24=0

Test: Comprehension Based Questions: Conic Sections - Question 6
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PASSAGE 3
Tangents are drawn from the point P(3, 4) to the ellipse  touching the ellipse at points A and B. (2010)

Q. The coordinates of A and B are
Detailed Solution for Test: Comprehension Based Questions: Conic Sections - Question 6

Tangent to

at  the point (3 cosθ, 2 sinθ) is

As it passes throught (3,4), we get cosθ + 2 sinθ=1
⇒ 4 sin2θ = 1 + cos2θ – 2 cosθ
⇒ 5cos2θ – 2cosθ – 3= 0

∴ Required points are A (3, 0) and B

Test: Comprehension Based Questions: Conic Sections - Question 7
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PASSAGE 3
Tangents are drawn from the point P(3, 4) to the ellipse  touching the ellipse at points A and B. (2010)

Q. The orthocenter of the triangle PAB is
Detailed Solution for Test: Comprehension Based Questions: Conic Sections - Question 7

Let H be the  orthocentre of ΔPAB, then as BH ⊥ AP, BH is a horizontal line  through B.

∴ y- cordinate of B = 8/5

Let H has coordinater (α, 8 5)

Then slope of PH

and slope of AB 

But PH ⊥ AB ⇒ 

⇒ 4 = –5α + 15 or α = 11/5

Hence

Test: Comprehension Based Questions: Conic Sections - Question 8
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PASSAGE 3
Tangents are drawn from the point P(3, 4) to the ellipse  touching the ellipse at points A and B. (2010)

Q. The equation of the locus of the point whose distances from the point P and the line AB are equal, is
Detailed Solution for Test: Comprehension Based Questions: Conic Sections - Question 8

Clearly the moving point traces a parabola with focus at P(3, 4) and directrix as

or x + 3y – 3=0

∴ Equation of parabola is

(x– 3)2 + (y–4)2

or 9x2 + y2 – 6xy – 54x – 62y + 241= 0

Test: Comprehension Based Questions: Conic Sections - Question 9
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 PASSAGE 4

Let PQ be a focal chord of the parabola y2 = 4ax. The tangents to the parabola at P and Q meet at a point lying on the line y = 2x + a, a > 0.

Q. Length of chord PQ is (JEE Adv. 2013)
Detailed Solution for Test: Comprehension Based Questions: Conic Sections - Question 9

 

Test: Comprehension Based Questions: Conic Sections - Question 10
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 PASSAGE 4

Let PQ be a focal chord of the parabola y2 = 4ax. The tangents to the parabola at P and Q meet at a point lying on the line y = 2x + a, a > 0.

Q. If chord PQ subtends an angle θ at the vertex of y2 = 4ax, then tan q =            (JEE Adv. 2013)
Detailed Solution for Test: Comprehension Based Questions: Conic Sections - Question 10

As PQ is the focal chord of  y2 = 4ax

∴ Coordinates of P and Q can be taken as

P(at2 , 2at) and 

Tangents at P and Q are

which intersect each other  at

As R lies on the y = 2x + a,  a > 0

Test: Comprehension Based Questions: Conic Sections - Question 11
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PASSAGE 5
Let a, r, s, t be nonzero real numbers. Let P (at2, 2at), Q, R (ar2, 2ar) and S (as2, 2as) be distinct points on the parabola y2 = 4ax. Suppose that PQ is the focal chord and lines QR and PK are parallel, where K is the point (2a, 0) (JEE Adv. 2014)

Q. The value of r is
Detailed Solution for Test: Comprehension Based Questions: Conic Sections - Question 11

∵ PQ is a focal chord, 

Also QR || PK ⇒ mQR = mPK

[otherwise Q will coincide with R]

⇒ 2 - t2 = 1- tr ⇒ 

Test: Comprehension Based Questions: Conic Sections - Question 12
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PASSAGE 5
Let a, r, s, t be nonzero real numbers. Let P (at2, 2at), Q, R (ar2, 2ar) and S (as2, 2as) be distinct points on the parabola y2 = 4ax. Suppose that PQ is the focal chord and lines QR and PK are parallel, where K is the point (2a, 0) (JEE Adv. 2014)

Q. If st = 1, then the tangent at P and the normal at S to the parabola meet at a point whose ordinate is
Detailed Solution for Test: Comprehension Based Questions: Conic Sections - Question 12

Tangent at P is  
ty = x+ at2 ....(i)
Normal at S sx + y = 2as + as3 ....(ii)

But given st = 1 ⇒ 

⇒ xt2 = yt3 = 2at2 + a

Putting value of x from equation (i) in above equation we get

⇒ t2 ( ty - at2) + yt3 = 2at2 + a

⇒ (t3 + t3) y - at4 = 2at2 + a

⇒ 2t3y = a (t4 + 2t2+1)

Test: Comprehension Based Questions: Conic Sections - Question 13
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PASSAGE 6
Let F1(x1, 0) and F2(x2, 0) for x1 < 0 and x2 > 0, be the foci of the ellipse   Suppose a parabola having vertex at the origin and focus at F2 intersects the ellipse at point M in the first quadrant and at point N in the fourth quadrant.

Q. The orthocentre of th e triangle F1MN is (JEE Adv. 2016)
Detailed Solution for Test: Comprehension Based Questions: Conic Sections - Question 13

For ellipse

∴ F1(– 1, 0) and F2 (1, 0) Parabola with vertex at (0, 0) and focus at F2(1, 0) is y2 = 4x.
Intersection points of ellipse and parabola are and 

For orthocentre of ΔF1MN, clearly one altitude is x-axis i.e. y = 0 and altitude from M to F1N is

Putting y = 0 in above equation, we get

∴ Orthocentre 

Test: Comprehension Based Questions: Conic Sections - Question 14
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PASSAGE 6
Let F1(x1, 0) and F2(x2, 0) for x1 < 0 and x2 > 0, be the foci of the ellipse   Suppose a parabola having vertex at the origin and focus at F2 intersects the ellipse at point M in the first quadrant and at point N in the fourth quadrant.

Q. If the tangen ts to the ellipse at M an d N meet at R and the normal to the parabola at M meets the x-axis at Q, then the ratio of area of the triangle MQR to area of the quadrilateral MF1NF2 is   (JEE Adv. 2016)
Detailed Solution for Test: Comprehension Based Questions: Conic Sections - Question 14

For ellipse

∴ F1(– 1, 0) and F2 (1, 0) Parabola with vertex at (0, 0) and focus at F2(1, 0) is y2 = 4x.
Intersection points of ellipse and parabola are and 

Tangents to ellipse at M and N are

and 

Their intersection point is R (6, 0)

Also normal to parabola at  is 

Its intersection with x-axis is 

Now ar (ΔMQR)  =

Also area (MF1NF2) = 2 × Ar (F1MF2)

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