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Area between Two Curves

Given below a...

Question

Given below are Matching Type Questions, with two columns(each having some items) each.Each item of column $I$ has to be matched with the items of column $I I$ , by encircling the correct matches:Note:An item of Column $I$ can be matched with more than one items in Column $I I$ .All the items of Column $I I$ have to be matched.

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Solution

$(A) ∵ 25 (x^{2} + y^{2}) = {(4 x + 3 y - 12)}^{2}$
or $25 x^{2} + 25 y^{2} = 16 x^{2} + 9 y^{2} + 24 x y - 96 x - 72 y + 144$
$\Rightarrow 9 x^{2} + 16 y^{2} - 24 x y = - 96 x - 72 y + 144$
$\Rightarrow {(3 x - 4 y)}^{2} = - 24 (4 x + 3 y - 6)$
$\Rightarrow 25 \times {(\frac{3 x - 4 y}{5})}^{2} = - 24 (\frac{4 x + 3 y - 6}{5}) \times 5$
$\Rightarrow {(\frac{3 x - 4 y}{5})}^{2} = - \frac{24}{5} (\frac{4 x + 3 y - 6}{5})$
Let $\frac{3 x - 4 y}{5} = Y$ and $\frac{4 x + 3 y - 6}{5} = X$
$∴ Y^{2} = - \frac{24}{5} X$
On comparing with $Y^{2} = - 4 ρ X$
$∴ 4 ρ = \frac{24}{5} \Rightarrow ρ = \frac{6}{5}$
Equation of directrix is $X - ρ = 0$
$\Rightarrow \frac{4 x + 3 y - 6}{5} - \frac{6}{5} = 0$
or $4 x + 3 y = 12$
$(B)$
$∵ 25 (x^{2} + y^{2}) = {(4 x - 3 y + 12)}^{2}$
or $25 x^{2} + 25 y^{2} = 16 x^{2} + 9 y^{2} - 24 x y + 96 x - 72 y + 144$
$\Rightarrow 9 x^{2} + 16 y^{2} + 24 x y = 96 x - 72 y + 144$
$\Rightarrow {(3 x + 4 y)}^{2} = 24 (4 x + 3 y + 6)$
$\Rightarrow 25 \times {(\frac{3 x + 4 y}{5})}^{2} = 24 (\frac{4 x - 3 y + 6}{5}) \times 5$
$\Rightarrow {(\frac{3 x + 4 y}{5})}^{2} = \frac{24}{5} (\frac{4 x - 3 y + 6}{5})$
Let $\frac{3 x + 4 y}{5} = Y$ and $\frac{4 x - 3 y + 6}{5} = X$
$∴ Y^{2} = \frac{24}{5} X$
On comparing with $Y^{2} = 4 ρ X$
$∴ 4 ρ = \frac{24}{5} \Rightarrow ρ = \frac{6}{5}$
Equation of directrix is $X + ρ = 0$
$\Rightarrow \frac{4 x - 3 y + 6}{5} + \frac{6}{5} = 0$
or $4 x - 3 y + 12 = 0$
and axis of the parabola is $Y = 0$
$\Rightarrow \frac{3 x + 4 y}{5} = 0$
or $3 x + 4 y = 0$
$(C)$
$∵ 25 (x^{2} + y^{2}) = {(3 x - 4 y + 12)}^{2}$
or $25 x^{2} + 25 y^{2} = 9 x^{2} + 16 y^{2} - 24 x y + 72 x - 96 y + 144$
$\Rightarrow 16 x^{2} + 9 y^{2} + 24 x y = 72 x - 96 y + 144$
$\Rightarrow 25 \times {(\frac{4 x + 3 y}{5})}^{2} = 24 (\frac{3 x - 4 y + 6}{5}) \times 5$
$\Rightarrow {(\frac{4 x + 3 y}{5})}^{2} = \frac{24}{5} (\frac{3 x - 4 y + 6}{5})$
Let $\frac{4 x + 3 y}{5} = Y$ and $\frac{3 x - 4 y + 6}{5} = X$
$∴ Y^{2} = \frac{24}{5} X$
On comparing with $Y^{2} = 4 ρ X$
$∴ 4 ρ = \frac{24}{5} \Rightarrow ρ = \frac{6}{5}$
Equation of directrix is $X + ρ = 0$
$\Rightarrow \frac{3 x - 4 y + 6}{5} + \frac{6}{5} = 0$
or $3 x - 4 y + 12 = 0$
and axis of the parabola is
$Y = 0$
or $4 x + 3 y = 0$

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a x^{2} + 2 h x y + b y^{2} + 2 g x + 2 f y + c = 0

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⎛ ⎜ ⎝ \begin{matrix} a & h & g h & b & f g & f & c \end{matrix} ⎞ ⎟ ⎠

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