A quadratic curve (parabola) is the path traced in the xy plane by the function C(t)=∑r=02(2r)tr(1−t)2−rOP⇀r,(0≤t≤1) where [(nr )means nCr]C(t)=OP⇀=xi^+yj^,OP0⇀=i^+j^OP1⇀=i^+7j^OP2⇀=7i^+j^Then match the following List- I List – IIA)Slope of axis of parabola has absolute valueP)2B)Dot product of C(0) and C(1)Q)4C)Modulus of cross product of C(0) and C(1)R)9D)Length of projection of 5. C(12) on 2i^+j^S)6 T)8

C(t)=(x,y)=(1−t2)(1,1)+2t(1−t)(1,7)+t2(7,1)⇒x=6t2+1,y=−12t2+12t+1∴ 2x+y=12t+3⇒t=2x+y−312(1)⇒x−16=(2x+y−3)2144⇒(2x+y−3)2=24(x−1)⇒(2x+y)2=36x+6y−9⇒(2x+y+k)2=(36+4k)x+(6+2k)+k2L1⊥L2⇒ (2)(36+4k)+1(6+2k)=0⇒10k+78=0⇒k=−395(2x+y−395)2=245x−485y+(39)225(2x+y−395)2=245(x−2y+|152|120)

A quadratic curve (parabola) is the path traced in the xy plane by the function C(t)=∑r=02(2r)tr(1−t)2−rOP⇀r,(0≤t≤1) where [(nr )means nCr]C(t)=OP⇀=xi^+yj^,OP0⇀=i^+j^OP1⇀=i^+7j^OP2⇀=7i^+j^Then match the following List- I List – IIA)Slope of axis of parabola has absolute valueP)2B)Dot product of C(0) and C(1)Q)4C)Modulus of cross product of C(0) and C(1)R)9D)Length of projection of 5. C(12) on 2i^+j^S)6 T)8

C(t)=(x,y)=(1−t2)(1,1)+2t(1−t)(1,7)+t2(7,1)⇒x=6t2+1,y=−12t2+12t+1∴ 2x+y=12t+3⇒t=2x+y−312(1)⇒x−16=(2x+y−3)2144⇒(2x+y−3)2=24(x−1)⇒(2x+y)2=36x+6y−9⇒(2x+y+k)2=(36+4k)x+(6+2k)+k2L1⊥L2⇒ (2)(36+4k)+1(6+2k)=0⇒10k+78=0⇒k=−395(2x+y−395)2=245x−485y+(39)225(2x+y−395)2=245(x−2y+|152|120)

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A quadratic curve (parabola) is the path traced in the $x y$ plane by the function $C (t) = \sum_{r = 0}^{2} (\begin{matrix} 2 \\ r \end{matrix}) t^{r} {(1 - t)}^{2 - r} {\overset{⇀}{O P}}_{r}, (0 \leq t \leq 1)$ where $[(\begin{matrix} n \\ r \end{matrix}) m e a n s^{n} C_{r}]$
$\begin{array}{l} C (t) = \overset{⇀}{O P} = x \hat{i} + y \hat{j}, \\ \overset{⇀}{O P_{0}} = \hat{i} + \hat{j} \\ \overset{⇀}{O P_{1}} = \hat{i} + 7 \hat{j} \\ \overset{⇀}{O P_{2}} = 7 \hat{i} + \hat{j} \end{array}$
Then match the following

List- I

List – II
A) Slope of axis of parabola has absolute value P) 2
B) Dot product of $C (0)$ and $C (1)$ Q) 4
C) Modulus of cross product of $C (0)$ and $C (1)$ R) 9
D) Length of projection of $\sqrt{5} . C (\frac{1}{2})$ on $2 \hat{i} + \hat{j}$ S) 6
T) 8

	List- I		List – II
A)	Slope of axis of parabola has absolute value	P)	2
B)	Dot product of $C (0)$ and $C (1)$	Q)	4
C)	Modulus of cross product of $C (0)$ and $C (1)$	R)	9
D)	Length of projection of $\sqrt{5} . C (\frac{1}{2})$ on $2 \hat{i} + \hat{j}$	S)	6
		T)	8

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answer is C.

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Detailed Solution

$\begin{array}{l} C (t) = (x, y) = (1 - t^{2}) (1, 1) + 2 t (1 - t) (1, 7) + t^{2} (7, 1) \\ \Rightarrow x = 6 t^{2} + 1, y = - 12 t^{2} + 12 t + 1 \\ ∴ 2 x + y = 12 t + 3 \Rightarrow t = \frac{2 x + y - 3}{12} \\ (1) \Rightarrow \frac{x - 1}{6} = \frac{{(2 x + y - 3)}^{2}}{144} \Rightarrow {(2 x + y - 3)}^{2} = 24 (x - 1) \\ \Rightarrow {(2 x + y)}^{2} = 36 x + 6 y - 9 \\ \Rightarrow {(2 x + y + k)}^{2} = (36 + 4 k) x + (6 + 2 k) + k^{2} \\ L_{1} ⊥ L_{2} \Rightarrow (2) (36 + 4 k) + 1 (6 + 2 k) = 0 \\ \Rightarrow 10 k + 78 = 0 \Rightarrow k = - \frac{39}{5} \\ {(2 x + y - \frac{39}{5})}^{2} = \frac{24}{5} x - \frac{48}{5} y + \frac{{(39)}^{2}}{25} \\ {(2 x + y - \frac{39}{5})}^{2} = \frac{24}{5} (x - 2 y + \frac{| 152 |}{120}) \end{array}$