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DIRECTIONS : Each question contains statements given in two columns, which have to be matched. The statements in ColumnI are labelled A, B, C and D, while the statements in ColumnII are labelled p, q, r, s and t. Any given statement in ColumnI can have correct matching with ONE OR MORE statement(s) in ColumnII. The appropriate bubbles corresponding to the answers to these questions have to be darkened as illustrated in the following example :
If the correct matches are Ap, s and t; Bq and r; Cp and q; and Ds then the correct darkening of bubbles will look like the given.
Q.1. Match the following : (3, 0) is the pt. from which three normals are drawn to the parabola y^{2} = 4x which meet the parabola in the points P, Q and R. Then (2006  6M)
Ans. (A)(p), (B)(q), (C)(s), (D)(r)
Sol. Let y = mx – 2m – m^{3} be the equation of normal to y^{2} = 4x.
As it passes through (3, 0), we get m = 0, 1, – 1
Then three points on parabola are given by (m^{2}, – 2m) for m = 0, 1, – 1
∴ P (0, 0), Q (1, 2), R (1, – 2)
∴ Area of ΔPQR = = 2 sq. units
Radius of circumcircle,
NOTE THIS STEP
(where, a, b, c are the sides of ΔPQR)
Centroid of ΔPQR
Circumcentre
Thus, (A)  (p); (B)  (q); (C) – (s); (D) – (r)
DIRECTIONS : Each question contains statements given in two columns, which have to be matched. The statements in ColumnI are labelled A, B, C and D, while the statements in ColumnII are labelled p, q, r, s and t. Any given statement in ColumnI can have correct matching with ONE OR MORE statement(s) in ColumnII. The appropriate bubbles corresponding to the answers to these questions have to be darkened as illustrated in the following example :
If the correct matches are Ap, s and t; Bq and r; Cp and q; and Ds then the correct darkening of bubbles will look like the given.
2. Match the statements in Column I with the properties in Column II and indicate your answer by darkening the appropriate bubbles in the 4 × 4 matrix given in the ORS. (2007 6 marks)
Ans. (A)p, q; (B)p, q ; (C)q, r ; (D)q, r
Sol. (A)  p, q
It is clear from the figure that two intersecting circles have a common tangent and a common normal joining the centres (B)  p, q
(C) – q, r Two circle when one is completely inside the other have a common normal C_{1}C_{2} but no common tangent.
(D) – q, r Two branches of hyperbola have no common tangent but have a common normal joining S_{1}S_{2}
Matrix Match (A) – p, q; (B) – p, q; (C) – q, r; (D) – q, r
DIRECTIONS : Each question contains statements given in two columns, which have to be matched. The statements in ColumnI are labelled A, B, C and D, while the statements in ColumnII are labelled p, q, r, s and t. Any given statement in ColumnI can have correct matching with ONE OR MORE statement(s) in ColumnII. The appropriate bubbles corresponding to the answers to these questions have to be darkened as illustrated in the following example :
If the correct matches are Ap, s and t; Bq and r; Cp and q; and Ds then the correct darkening of bubbles will look like the given.
3. Match the conics in Column I with the statements/expressions in Column II. (2009)
Ans. (A)p; (B)s, t ; (C)r ; (D)q, s
Sol. (p) As the line hx + ky= 1, touches the circle x^{2} +y^{2}= 4
∴ Length of perpendicular from centre (0 , 0) of circle to line = radius of the circle
∴ Locus of (h, k) is , which is a circle.
(q) We know that if  z z_{1}    z z_{2}= k
where  k  <  z_{1}z_{2} then z traces a hyperbola.
Here  z + 2   z  2  =±3
∴ Locus of z is a hyperbola.
(r) We have
and
On squaring and adding, we get
or
which is the equation of an ellipse.
(s) We know eccentricity for a parabola = 1
for an ellipse < 1
for a hyperbola > 1
∴ The conics whose eccentricity lies in 1 ≤ x <∞ are parabola and hyperbola.
(t) Let z = x+ iy then
Re [( x + 1) + iy]^{2} = x^{2}+y^{2}+1
⇒ (x + 1)^{2}  y^{2} = x^{2} +y^{2}+1 ⇒ y^{2} =x , which is a parabola.
DIRECTIONS (Q. 4) : Following question has matching lists. The codes for the list have choices (a), (b), (c) and (d) out of which ONLY ONE is correct.
Q. 4. A line L : y = mx + 3 meets y – axis at E(0, 3) and the arc of the parabola y^{2} = 16x, 0 ≤ y ≤ 6 at the point F(x_{0}, y_{0}). The tangent to the parabola at F(x_{0}, y_{0}) intersects the yaxis at G(0, y_{1}). The slope m of the line L is chosen such that the area of the triangle EFG has a local maximum. (JEE Adv. 2013) Match List I with List II and select the correct answer using the code given below the lists :
Ans. (a)
Sol. (a) Equation of tangent to y^{2} = 16x at F (x_{0}, y_{0}) yy_{0} = 8(x + x_{0})
Area of ΔEFG
⇒
Also y_{0} = mx_{0} + 3
∴ 4 = m + 3 or m = 1 maximum area of ΔEFG
∴ (P) → (4), (Q) → (1), (R) → (2), (S) → (3)
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