You visited us 15 times! Enjoying our articles? Unlock Full Access!

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.

Open in App

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 $4 x + 3 y = 0$

Suggest Corrections

Similar questions

I

I I

I

I I

I I

I

I I

:

I

I I

I I

I

I I

I

I I

:

I

I I

I I

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

△

△ = a b c + 2 f g h - a f^{2} - b g^{2} - c h^{2}

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

I

I I

Join BYJU'S Learning Program