Exercise 5.6 - Chapter 5 Two Dimensional Analytical Geometry–II 12th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

Ex 5.6
Choose the correct or the most suitable answer from the given four alternatives.
Question 1.
The equation of the circle passing through $(1,5)$ and $(4,1)$ and touching $y$-axis is $x^2+y^2-5 x-6 y+9+\lambda$ $(4 x+3 y-19)=0$ where $\lambda$ is equal to ..........
(a) $0,-\frac{40}{9}$
(b) 0
(c) $\frac{40}{9}$
(d) $-\frac{40}{9}$
Solution:
(a) $0,-\frac{40}{9}$
Hint:
$
\begin{aligned}
& x^2+y^2-5 x-6 y+9+\lambda(4 x+3 y-19)=0 \\
& \text { ie., } x^2+y^2+x(4 \lambda-5)+y(3 \lambda-6)+9-19 \lambda=0
\end{aligned}
$
Since it touches $y$ axis $\mathrm{x}=0$
$
\Rightarrow y^2+y(3 \lambda-6)+9-19 \lambda=0
$
It is a quadratic in $\mathrm{y}$,
Since the circle touches $y$ axis the roots must be equal $\Rightarrow b^2-4 a c=0$
ie., $(3 \lambda-6)^2-4(1)(9-19 \lambda)=0$
Solving we get $\lambda=0$ or $-\frac{40}{9}$
Question 2.
The eccentricity of the hyperbola whose latus rectum is 8 and conjugate axis is equal to half the distance between the foci is.
(a) $\frac{4}{3}$
(b) $\frac{4}{\sqrt{3}}$
(c) $\frac{2}{\sqrt{3}}$
(d) $\frac{3}{2}$
Solution:
(c) $\frac{2}{\sqrt{3}}$
Hint:

Given $\frac{2 b^2}{a}=8$ and $2 \mathrm{~b}=\mathrm{ae}$
$
\begin{array}{rlrl}
\Rightarrow \quad b & =\frac{a e}{2} \quad & \Rightarrow b^2=\frac{a^2 e^2}{4} \\
\text { but } b^2 & =a^2\left(e^2-1\right)=a^2 e^2-a^2 \\
a^2 e^2-a^2 & =\frac{a^2 e^2}{4} \quad & \Rightarrow a^2 e^2-\frac{a^2 e^2}{4}=a^2 \\
\frac{3 a^2 e^2}{4} & =a^2 \\
e & =\sqrt{\frac{4}{3}}=\frac{2}{\sqrt{3}} &
\end{array}
$

Question 3.
The circle $x^2+y^2=4 x+8 y+5$ intersects the line $3 x-4 y=m$ at two distinct points if
(a) $15<\mathrm{m}<65$
(b) $35<\mathrm{m}<85$
(c) $-85<\mathrm{m}<-35$
(d) $-35<$ m $<15$
Solution:
(d) $-35<\mathrm{m}<15$
Hint:
$x^2+y^2-4 x-8 y-5=0$
$3 x-4 y=m$ Solving (1) and (2) we get
from (2) $\Rightarrow 3 x=4 y+m$
$\mathrm{x}=\frac{4 y+m}{3}$
Substituting $\mathrm{x}$ value in (1) we get
$
\left(\frac{4 y+m}{3}\right)^2+y^2-4\left(\frac{4 y+m}{3}\right)-8 y-5=0
$
It is a quadratic in $y$ and given that the roots are distinct
$
\Rightarrow b^2-4 a c>0
$
On simplifying we get
$
\begin{aligned}
& \Rightarrow-9 \mathrm{~m}^2-18 \mathrm{w}+4725>0 \Rightarrow \mathrm{m}^2+20 \mathrm{~m}-525<0 \\
& \Rightarrow(\mathrm{m}+35)(\mathrm{m}-15) \leq 0 \Rightarrow \mathrm{m} \text { lies between }-35 \text { and } 15 \text { ie, },-35<\mathrm{m}<15
\end{aligned}
$
Question 4.
The length of the diameter of the circle which touches the $\mathrm{x}$-axis at the point $(1,0)$ and passes through the point $(2,3) \ldots \ldots \ldots$
(a) $\frac{6}{5}$
(b) $\frac{5}{3}$
(c) $\frac{10}{3}$
(d) $\frac{3}{5}$
Solution:
(c) $\frac{10}{3}$
Hint:

Since radius is $\perp \mathrm{r}$ to tangent it passes through the centre $(1, \mathrm{y})$
$\mathrm{AC}=\mathrm{CB}=$ radius
$\mathrm{AC}^2=\mathrm{CB}^2 \Rightarrow \mathrm{y}^2=1+(\mathrm{y}-3)^2$
$6 \mathrm{y}=10 \Rightarrow \mathrm{y}=\frac{5}{3}$
Diameter $2 \mathrm{y}=\frac{10}{3}$

Question 5.
The radius of the circle $3 x^2+b y^2+4 b x-6 b y+b^2=0$ is
(a) 1
(b) 3
(c) $\sqrt{10}$
(d) $\sqrt{11}$
Solution:
(c) $\sqrt{10}$
Hint:
Here $\mathrm{b}=3$
$
\begin{aligned}
& 3 x^2+3 y^2+12 x-18 y+9=0 \\
& (\div 3) z z>x^2+y^2+4 x-6 y+3=0
\end{aligned}
$
Comparing with general form
$
\begin{aligned}
& \mathrm{g}=2, \mathrm{f}=-3, \mathrm{c}=3 \\
& \therefore \text { radius }=\sqrt{g^2+f^2-c}=\sqrt{4+9-3}=\sqrt{10}
\end{aligned}
$
Question 6.
The centre of the circle inscribed in a square formed by the lines $x^2-8 x-12=0$ and $y^2-14 y+45=0$ is
(a) $(4,7)$
(b) $(7,4)$
(c) $(9,4)$
(d) $(4,9)$

Solution:
(a) $(4,7)$
Hint:
$
\begin{aligned}
& (x-6)(x-2)=0 \Rightarrow x=2 \text { or } 6 \\
& (y-9)(y-5)=0 \Rightarrow y=5 \text { or } 9
\end{aligned}
$
$
\begin{aligned}
& P=\text { mid point of } \mathrm{AB}=\left(\frac{2+6}{2}, \frac{5+5}{2}\right)=(4,5) \\
& \mathrm{Q}=\text { mid point of } \mathrm{CD}=\left(\frac{2+6}{2}, \frac{9+9}{2}\right)=(4,9)
\end{aligned}
$
Centre of the circle $=$ Mid point of $\mathrm{PQ}$
$
=\left(\frac{4+4}{2}, \frac{5+9}{2}\right)=(4,7)
$

Question 7.
The equation of the normal to the circle $x^2+y^2-2 x-2 y+1=0$ which is parallel to the line $2 x+4 y=3$ is .
(a) $x+2 y=3$
(b) $x+2 y+3=3$
(c) $2 x+4 y+3=0$
(d) $x-2 y+3=0$
Solution:
(a) $x+2 y=3$
Hint:
Centre of the circle $=(1,1)$ and radius $=r=1$
Normal is parallel to $2 x+4 y=3$
So equation of normal will be the form $2 \mathrm{x}+4 \mathrm{y}=\mathrm{k}$
It passes throug the centre $(1,1) \Rightarrow 2+4=6=\mathrm{k}$
So equation of the normal is $x+4 y=6$
$(\div$ by 2$) x+2 y=3$
Question 8.
If $\mathrm{P}(\mathrm{x}, \mathrm{y})$ be any point on $16 \mathrm{x}^2+25 \mathrm{y}^2=400$ with foci $F_1(3,0)$ and $F_2(-3,0)$ then $P F_1+\mathrm{PF}_2$ is
(a) 8
(b) 6
(c) 10

(d) 12
Solution:
(c) 10
Hint:
$
\mathrm{F}_1 \mathrm{P}+\mathrm{F}_2 \mathrm{P}=2 \mathrm{a}
$
Here the equation is $16 x^2+25 y^2=400$
$
\begin{aligned}
& (\div \mathrm{by} 400) \Rightarrow \frac{16 x^2}{400}+\frac{25 y^2}{400}=1 \Rightarrow \frac{x^2}{25}+\frac{y^2}{16}=1 \\
& \mathrm{a}^2=25 \Rightarrow \mathrm{a}=5 \\
& \therefore 2 \mathrm{a}=10
\end{aligned}
$
Question 9.
The radius of the circle passing through the point $(6,2)$ two of whose diameter are $x+y=6$ and $x+2 y=$ 4 is ..............
(a) 10
(b) $2 \sqrt{5}$
(c) 6
(d) 4
Solution:
(b) $2 \sqrt{5}$
Hint:
The point of intersection of the diameters is the centre.
Now solving $\mathrm{x}+\mathrm{y}=6$.
$
\mathrm{x}+2 \mathrm{y}=4 \ldots \ldots \ldots \ldots \text { (2) }
$
(1) $-(2) \Rightarrow-y=2 \Rightarrow y=-2$
Substituting $y=-6$ in (1)
$
\begin{aligned}
& x-2=6 \Rightarrow x=8 \\
& \text { Centre }=(8,-2) \\
& \text { The circle passes through }(6,2) \\
& \therefore \text { radius }=\sqrt{(8-6)^2+(-2-2)^2}=\sqrt{4+16} \\
& =\sqrt{20}=\sqrt{4 \times 5}=2 \sqrt{5}
\end{aligned}
$
Question 10.
The area of quadrilateral formed with foci of the hyperbolas $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ and $\frac{x^2}{a^2}-\frac{y^2}{b^2}=-1$ is
(a) $4\left(a^2+b^2\right)$
(b) $2\left(a^2+b^2\right)$
(c) $\mathrm{a}^2+\mathrm{b}^2$
(d) $\frac{1}{2}\left(\mathrm{a}^2+\mathrm{b}^2\right)$

Solution:
(b) $2\left(a^2+b^2\right)$
Hint:
The foci are $(\pm a e, 0)$ and $(0, \pm a e)$

The above diagram is a rhombus.
Its area $=\frac{1}{2} \mathrm{~d}_1 \mathrm{~d}_2$
$
=\frac{1}{2} \text { (2ae) }(2 \mathrm{ae})=2 \mathrm{a}^2 \mathrm{e}^2
$
But we know $b^2=a^2\left(e^2-1\right)$
$
\begin{aligned}
& \text { i.e., } b^2=a^2 \mathrm{e}^2-\mathrm{a}^2 \\
& \Rightarrow \mathrm{b}^2+\mathrm{a}^2=\mathrm{a}^2 \mathrm{e}^2
\end{aligned}
$
So area $=2\left(a^2+b^2\right)$
Question 11.
If the normals of the parabola $y^2=4 x$ drawn at the end points of its latus rectum are tangents to the circle $(\mathrm{x}-3)^2+(\mathrm{y}+2)^2=\mathrm{r}^2$, then the value of $\mathrm{r}^2$ is
(a) 2
(b) 3
(c) 1
(d) 4
Solution:
(a) 2
Hint:
The normals are $x \pm y=3$
Distance from $(3,-2)$ on both normal is $r$
$
\text { (i.e.) } r=\left|\frac{3-2-3}{\sqrt{2}}\right|=\sqrt{2} \Rightarrow r^2=2
$
Question 12.
If $x+y=k$ is a normal to the parabola $y^2=12 x$, then the value of $k$ is
(a) 3
(b) $-1$
(c) 1
(d) 9
Solution:
(d) 9
Hint:
Slope of normal $=-1$ and Slope of tangent $=-1$
The condition for $\mathrm{y}=\mathrm{mx}+\mathrm{c}$ to be a tangent to the parabola
$
\mathrm{y}^2=4 \mathrm{ax} \text { is } \mathrm{c}=\frac{a}{m}
$
Here the parabola is $y^2=12 x \Rightarrow a=3$ and $\mathrm{m}=1$

$\therefore \mathrm{c}=\frac{3}{1}=3$
$\therefore$ Equation of tangent is $\mathrm{y}=\mathrm{x}+3$
Solving the parabola and tangent we get the points $(3,6)$ and $(3,-6)$ $\therefore \mathrm{k}=-3$ or $\mathrm{k}=9$
Question 13.
The ellipse $\mathrm{E}_1: \frac{x^2}{9}+\frac{y^2}{4}=1$ is inscribed in a rectangle $\mathrm{R}$ whose sides are parallel to the coordinate axes. Another ellipse $E_2$ passing through the point $(0,4)$ circumscribes the rectangle $R$. The eccentricity of the ellipse is ..............
(a) $\frac{\sqrt{2}}{2}$
(b) $\frac{\sqrt{3}}{2}$
(c) $\frac{1}{2}$
(d) $\frac{3}{4}$
Solution:
(c) $\frac{1}{2}$
Hint:
$e=\frac{1}{2}$

Question 14.
Tangents are drawn to the hyperbola $\frac{x^2}{9}-\frac{y^2}{4}=1$ parallel to the straight line $2 x-y=1$. One of the points of contact of tangents on the hyperbola is .............
(a) $\left(\frac{9}{2 \sqrt{2}}, \frac{-1}{\sqrt{2}}\right)$
(b) $\left(\frac{-9}{2 \sqrt{2}}, \frac{1}{\sqrt{2}}\right)$
(c) $\left(\frac{9}{2 \sqrt{2}}, \frac{1}{\sqrt{2}}\right)$
$(d)(3 \sqrt{3},-2 \sqrt{2})$
Solution:
(c) $\left(\frac{9}{2 \sqrt{2}}, \frac{1}{\sqrt{2}}\right)$
Hint:
The tangents are parallel to $2 x-y=1$
So equation of tangents will be of the form $2 \mathrm{x}-\mathrm{y}=\mathrm{k} \Rightarrow \mathrm{y}=2 \mathrm{x}-\mathrm{k}$
Comparing this equation with $\mathrm{y}=\mathrm{mx}+\mathrm{c}$ we get $\mathrm{m}=2$ and $\mathrm{c}=-\mathrm{k}$
The given hyperbola is $\frac{x^2}{9}-\frac{y^2}{4}=1$
$\Rightarrow \mathrm{a}^2=9$ and $\mathrm{b}^2=4$

Now the condition for the line $\mathrm{y}=\mathrm{mx}+\mathrm{c}$ to be a tangent to
$
\begin{aligned}
\frac{x^2}{a^2}-\frac{y^2}{b^2}=1 \text { is } c^2 & =a^2 m^2-b^2 \\
\text { LHS }=c^2=(-k)^2 & =k^2 \\
\text { RHS }=a^2 m^2-b^2 & =9(2)^2-4=32 \\
\text { Now, LHS } & =\text { RHS } \\
k^2=32 \Rightarrow k & =\pm \sqrt{32}=\pm 4 \sqrt{2}
\end{aligned}
$
The point of contact is
$
\begin{aligned}
\left(\frac{-a^2 m}{c}, \frac{-b^2}{c}\right) & =\left(\frac{-a^2 m}{-k}, \frac{-b^2}{-k}\right)=\left(\frac{a^2 m}{k}, \frac{b^2}{k}\right)=\left(\frac{9(2)}{\pm 4 \sqrt{2}}, \frac{4}{\pm 4 \sqrt{2}}\right) \\
& =\left(\frac{9}{2 \sqrt{2}}, \frac{1}{\sqrt{2}}\right)
\end{aligned}
$

Question 15.
The equation of the circle passing through the foci of the ellipse $\frac{x^2}{16}-\frac{y^2}{9}=1$ having centre at $(0,3)$ is
(a) $x^2+y^2-6 y-7=0$
(b) $x^2+y^2-6 y+7=0$
(c) $x^2+y^2-6 y-5=0$
(d) $x^2+y^2-6 y+5=0$
Solution:
(a) $x^2+y^2-6 y-7=0$
The given ellipse is $\frac{x^2}{16}+\frac{y^2}{9}=1$
$
\begin{aligned}
a^2=16 \text { and } b^2 & =9 \\
e^2 & =\frac{a^2-b^2}{a^2}=\frac{16-9}{16}=\frac{7}{16} \Rightarrow e=\frac{\sqrt{7}}{4}
\end{aligned}
$
Now $a=4$ and $e=\frac{\sqrt{7}}{4} \Rightarrow a e=\sqrt{7}$
Now the foci are $(\pm a e, 0)=(\pm \sqrt{7}, 0)$
Centre $=(0,3)$
Passing through the foci $=(\pm \sqrt{7}, 0)$
The distance between centre and foci $=$ radius of the circle
ie., $r=\sqrt{(0 \pm \sqrt{7})^2+(3-0)^2}=\sqrt{7+9}=\sqrt{16}=4$
Now Centre $=(0,3) ; r=4$
Equation of the circle is $(\mathrm{x}-0) 2+(\mathrm{y}-3) 2=42$
ie., $x^2+y^2-6 y+9-16=0$
$x^2+y^2-6 y-7=0$

Question 16.
Let $\mathrm{C}$ be the circle with centre at $(1,1)$ and radius $=1$. If $\mathrm{T}$ is the circle centered at $(0, \mathrm{y})$ passing through the origin and touching the circle $\mathrm{C}$ externally, then the radius of $\mathrm{T}$ is equal
(a) $\frac{\sqrt{3}}{\sqrt{2}}$
(b) $\frac{\sqrt{3}}{2}$
(c) $\frac{1}{2}$
(d) $\frac{1}{4}$
Solution:
(d) $\frac{1}{4}$
Hint:
Equation of circle is $x^2+y^2+2 g x+2 f y+c=0$
i.e., $\left(1+y^2\right)^2=(1-y)^2=1$
Solving we get $y=\frac{1}{4}$
Question 17.
Consider an ellipse whose centre is of the origin and its major axis is along $x$-axis. If its eccentricity is $\frac{3}{5}$ and the distance between its foci is 6 , then the area of the quadrilateral inscribed in the ellipse with diagonals as major and minor axis of the ellipse is
(a) 8
(b) 32
(c) 80
(d) 40
Solution:
(d) 40
Hint:
$
\begin{aligned}
& e=\frac{3}{5} \\
& \begin{aligned}
a e & =3 \\
a^2 & =25
\end{aligned} \Rightarrow a\left(\frac{3}{5}\right)=3 \Rightarrow a=5 \\
&
\end{aligned}
$
$
\text { Now } \begin{aligned}
b^2 & =a^2\left(1-e^2\right) \Rightarrow b^2=25\left(1-\frac{9}{25}\right)=16 \\
a^2 & =25 \Rightarrow a=5 ; 2 a=10 \\
b^2 & =16 \Rightarrow b=4 ; 2 b=8
\end{aligned}
$
Now area of the quadrilateral is $\frac{1}{2}\left(d_1 \times d_2\right)=\frac{1}{2}(10)(8)=40$

Question 18.
Area of the greatest rectangle inscribed in the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is
(a) $2 \mathrm{ab}$
(b) ab
(c) $\sqrt{a b}$
(d) $\frac{a}{b}$
Solution:
(a) $2 \mathrm{ab}$
Hint:
Area of the greatest rectangle is $(\sqrt{2} a)(\sqrt{2} b)=2 a b$
Question 19.
An ellipse has $O B$ as semi minor axes, $F$ and F' its foci and the angle FBF' is a right angle. Then the eccentricity of the ellipse is ............
(a) $\frac{1}{\sqrt{2}}$
(b) $\frac{1}{2}$
(c) $\frac{1}{4}$
$(d) \frac{1}{\sqrt{3}}$
Solution:

(a) $\frac{1}{\sqrt{2}}$
Hint:
Here $\mathrm{F}_1 \mathrm{~B}+\mathrm{FB}=\mathrm{FF}^{\prime}$
ie., $\left(\sqrt{a^2 e^2+b^2}\right)^2+\left(\sqrt{a^2 e^2+b^2}\right)^2=(2 a e)^2$
$\Rightarrow \quad e=\frac{1}{2}$

Question 20.
The eccentricity of the ellipse $(x-3)^2+(y-4)^2=\frac{y^2}{9}$ is ............
(a) $\frac{\sqrt{3}}{2}$
(b) $\frac{1}{3}$
(c) $\frac{1}{3 \sqrt{2}}$
(d) $\frac{1}{\sqrt{3}}$
Solution:
(b) $\frac{1}{3}$
Hint:
The given equation is of the form $\mathrm{FP}^2=\mathrm{e}^2 \mathrm{PM}^2$
$
\begin{aligned}
& \text { i.e., }(\mathrm{x}-3)^2+(\mathrm{y}-4)^2=\frac{1}{9}\left(\mathrm{y}^2\right) \\
& \Rightarrow \mathrm{e}^2=\frac{1}{9} \Rightarrow \mathrm{e}=\frac{1}{3}
\end{aligned}
$
Question 21.
If the two tangents drawn from a point $P$ to the parabola $y^2=4 x$ are at right angles then the locus of $P$ is
(a) $2 x+1=0$
(b) $x=-1$
(c) $2 x-1=0$
(d) $x=1$
Solution:
(b) $x=-1$
Hint:
When the tangents at drawns from a point on the directrix are at right angles.
So equation of directrix to $y^2=4 x$ is $x=-1$

Question 22.
The circle passing through $(1,-2)$ and touching the axis of $x$ at $(3,0)$ passing through the point
(a) $(-5,2)$
(b) $(2,-5)$
(c) $(5,-2)$
(d) $(-2,5)$
Solution:

(c) $(5,-2)$
Hint:
Let the centre be $(3, \mathrm{~h})$ $\mathrm{r}=\mathrm{k}$
Equation of the circle is
$
(\mathrm{x}-3)^2+(\mathrm{y}-\mathrm{h})^2=\mathrm{r}^2=(\mathrm{k})^2
$
It passes through $(1,-2) \Rightarrow \mathrm{k}^2=8$
Substituting $(5,-2)$ the equation satisfies.

Question 23.
The locus of a point whose distance from $(-2,0)$ is $\frac{2}{3}$ times its distance from the line $x=\frac{-9}{2}$ is
(a) a parabola
(b) a hyperbola
(c) an ellipse
(d) a circle
Solution:
(c) an ellipse
Hint:
Here $\mathrm{e}=\frac{2}{3}<1$
So the conic is an ellipse
Question 24.
The values of $m$ for which the line $y=m x+2 \sqrt{5}$ touches the hyperbola $16 x^2-9 y^2=144$ are the roots of $x^2-(a+b) x-4=0$, then the value of $(a+b)$ is
(a) 2
(b) 4
(c) 0
(d) $-2$
Solution:
(c) 0
Hint:
$
\begin{aligned}
& 16 x^2-9 y^2=144 \Rightarrow \frac{16 x^2}{144}-\frac{9 y^2}{144}=1 \\
& \text { ie., } \frac{x^2}{9}-\frac{y^2}{16}=1
\end{aligned}
$
Here $a^2=9 ; b^2=16$
The condition for the line $y=m x+c$ to be a tangent to $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is

$
c^2=a^2 m^2-b^2
$
Here the given line is $\mathrm{y}=\mathrm{mx}+2 \sqrt{5}$
$\Rightarrow \mathrm{m}=\mathrm{m}$ and $\mathrm{c}=2 \sqrt{5}$
The condition is $\mathrm{c}^2=\mathrm{a}^2 \mathrm{~m}^2-\mathrm{b}^2$
ie., $(2 \sqrt{5})^2=9\left(\mathrm{~m}^2\right)-16$
$\Rightarrow 9 \mathrm{~m}^2=20+16=36 \Rightarrow \mathrm{m}^2=4$
i.e., $\mathrm{m}=\pm 2$
$\mathrm{a}=2, \mathrm{~b}=-2$ (say)
So $a+b=2-2=0$
Question 25.
If the coordinates at one end of a diameter of the circle $x^2+y^2-8 x-4 y+c=0$ are $(11,2)$, the coordinates of the other end are
(a) $(-5,2)$
(b) $(-3,2)$
(c) $(5,-2)$
(d) $(-2,5)$
Solution:
(b) $(-3,2)$
Hint:

Centre $=(4,2)$ one end of diameter $(11,2) \quad(11,2) \quad(4,2) \quad(x, y)$
Let the other end be $(x, y)$
Now centre $=$ mid point of the end points of diameter.
$\Rightarrow(4,2)=\left(\frac{11+x}{2}, \frac{2+y}{2}\right)$

$\begin{aligned}
& \Rightarrow \begin{array}{l|l}
\Rightarrow \frac{11+x}{2}=4 & \frac{2+y}{2}=2
\end{array} \\
& \begin{array}{l|l}
11+x=8 & 2+y=4
\end{array} \\
& \begin{array}{l|l}
x=-3 & y=4-2=2
\end{array} \\
&
\end{aligned}$