Test: Comprehension Based Questions: Conic Sections
14 Questions MCQ Test Maths 35 Years JEE Main & Advanced Past year Papers | Test: Comprehension Based Questions: Conic Sections
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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
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)
For ΔPRS,
b = PR = 6
, c = SR = 10
∴ Radius of circumference
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)
Radius of incircle
We have a = PR = 
c = PQ = 
and 
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
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
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
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
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
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
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
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
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
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
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)
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)
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
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
∵ PQ is a focal chord, 
Also QR || PK ⇒ mQR = mPK
[
otherwise Q will coincide with R]
⇒ 2 - t2 = 1- tr ⇒ 
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
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)
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)
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 
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)
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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