Given x = a cos³ \(\theta\) , y = a sin³ \(\theta\)
\(\Rightarrow \quad \frac { x }{ a } ={ cos }^{ 3 }\theta \quad and\quad \frac { y }{ a } ={ sin }^{ 3 }\quad \theta \)
Taking power \(\left( \frac { 2 }{ 3 } \right) \)for both the equations, we get
\({ \left( \frac { x }{ a } \right) }^{ \frac { 2 }{ 3 } }={ \left( { cos }^{ 3 }\theta \right) }^{ \frac { 2 }{ 3 } }and\)
\({ \left( \frac { y }{ a } \right) }^{ \frac { 2 }{ 3 } }={ { (sin }^{ 3 }\theta })^{ \frac { 2 }{ 3 } }\)

\({ \left( \frac { x }{ a } \right) }^{ \frac { 2 }{ 3 } }={ cos }^{ 2 }\theta \quad and\quad { \left( \frac { y }{ a } \right) }^{ \frac { 2 }{ 3 } }={ sin }^{ 2 }\theta \)
We know that cos² \(\theta\) + sin² \(\theta\) = 1
\(\therefore { \left( \frac { x }{ a } \right) }^{ \frac { 2 }{ 3 } }+{ \left( \frac { y }{ a } \right) }^{ \frac { 2 }{ 3 } }=1\)
\(\Rightarrow \quad \frac { { x }^{ \frac { 2 }{ 3 } } }{ { x }^{ \frac { 2 }{ 3 } } } +\frac { { x }^{ \frac { 2 }{ 3 } } }{ { x }^{ \frac { 2 }{ 3 } } } =1\)
\(\Rightarrow \quad \frac { { x }^{ \frac { 2 }{ 3 } }+{ y }^{ \frac { 2 }{ 3 } } }{ { a }^{ \frac { 2 }{ 3 } } } =1\)
\(\Rightarrow \quad { x }^{ \frac { 2 }{ 3 } }+{ y }^{ \frac { 2 }{ 3 } }={ a }^{ \frac { 2 }{ 3 } }\)
\(\therefore \) The required point is (0, 12)

Let P(1, 2) and B(5, 3) are the given points.
By the property of reflector \(\angle \) XAB = \(\angle \) OAP = \(\theta\)
Let m₁ be the slope of the x-axis, m₂ and m₃ be the slopes of the lines AP and AB.
To find XAB,
Clearly m₁ = 0 Since it represents slope of x-axis.
\({ m }_{ 2 }=\frac { 2-0 }{ 1-x } =\frac { 2 }{ 1-x } and\left[ m=\frac { { y }_{ 2 }-{ y }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } } \right] \)
\({ m }_{ 3 }=\frac { 3-0 }{ 5-x } =\frac { 3 }{ 5-x } \)

\(tan\ \theta =\left| \frac { { m }_{ 1 }-{ m }_{ 3 } }{ 1+{ m }_{ 1 }.{ m }_{ 3 } } \right| \)
\(tan\ \theta =\left| \frac { 0-\frac { 3 }{ 5-x } }{ 1+0\left( \frac { 3 }{ 5 } -x \right) } \right| =\frac { 3 }{ 5-x } \)
To find \(\angle \)OAP,
\(tan(-\theta )=\left| \frac { { m }_{ 1 }-{ m }_{ 2 } }{ 1+{ m }_{ 1 }{ m }_{ 2 } } \right| =\left| \frac { 0-\frac { 2 }{ 1-x } }{ 1+0\left( \frac { 2 }{ 1-x } \right) } \right| =\frac { 2 }{ 1-x } \)  [Since OAP is in the clockwise direction]
\(\Rightarrow \ -tan\theta =\frac { 2 }{ 1-x } \ \quad [\because tan\theta (-\theta )=tan\quad \theta ]\)
\(\Rightarrow \quad tan\quad \theta =\frac { -2 }{ 1-x } \)
From (1) and (2),
\(\frac { 3 }{ 5-x } =\frac { -2 }{ 1-x } \Rightarrow 3-3x=-10+2x\)
\(\Rightarrow\)  3 + 10 = 2x + 3x 
\(\Rightarrow\)  13 = 5x 
\(\Rightarrow \ x=\frac { 13 }{ 5 } \)
The required co-ordinates of A is \( \left( \frac { 13 }{ 5 } ,0 \right) \)

(i) the distance between the place and the target.

Let d be the distance between the place and the target.
Since the speed IS constant and speed = \(\frac{distance}{time}\) we get

⇒ 6(d-1400) = 5(d-800)
⇒ 6(d-1400) = 5(d-800)
⇒ 6d-8400 = 5d-4000
⇒ d = 4400 m
(ii) the distance covered by it in 15 seconds.

using two point form

⇒ y-15 = \(\frac{x-300}{200}\)
When time = 15 sec
15-15 = \(\frac{x-300}{200}\)
⇒ \(\frac{x-300}{200}\) = 0
x = 3000 m
(iii) time taken to hit the target.
By the given data
T₁(15^th sec) D₁ (1400 m)
T₂(18^th sec) D₂(800 m)
Using two point form, the linear relationship between time and distance is
\(\frac { T-{ T }_{ 1 } }{ { T }_{ 2 }-{ T }_{ 1 } } =\frac { D-{ D }_{ 1 } }{ { D }_{ 2 }-{ D }_{ 1 } } \)
\(\Rightarrow \quad \frac { T-15 }{ 18-15 } =\frac { D-1400 }{ 800-1400 } \quad \quad \Rightarrow \quad \frac { T-15 }{ 3 } =\frac { D-1400 }{ -600 } \)
\(\Rightarrow \quad T-15=\frac { 1400-D }{ 200 } \quad \Rightarrow \quad T=\frac { 1400-D }{ 200 } +15\)
Putting D = 0 \(\Rightarrow\) \(T=\frac { 1400 }{ 200 } +15=7+15=22\quad sec\)

(i) Find the linear relationship between C and F.
By the given data
x₁ (100°C)  y₁(212°F)
x₂(0°C) y₂(32°F)
Using two point form, the linear relationship between C and F is
\(\frac { y-{ y }_{ 1 } }{ { y }_{ 2 }-{ y }_{ 1 } } =\frac { x-{ x }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } } \)
\(\frac { y-212 }{ 32-212 } =\frac { x-100 }{ 0-100 } \)
\(\Rightarrow \quad \frac { y-212 }{ -180 } =\frac { x-100 }{ -100 } \)
\(\Rightarrow \quad \frac { y-212 }{ 9 } =\frac { x-100 }{ 5 } =\frac { 5 }{ 9 } (y-212)=x-100\)
\(\Rightarrow \quad x=\frac { 5 }{ 9 } (y-212)+100\quad \Rightarrow \quad x=\frac { 5 }{ 9 } y-\frac { 5 }{ 9 } \times 212+100\)
\(\Rightarrow \quad x=\frac { 5 }{ 9 } y-118+100\quad \Rightarrow x=\frac { 5 }{ 9 } y-18\)
\(\Rightarrow \quad x=\frac { 5 }{ 9 } (y-32) \Rightarrow C=\frac { 5 }{ 9 } (F-32)\quad .....(1)\)
[\(\because \)  x represents Celsius and y represents Fahrenheit]
Which is the required relationship between C and F.
(ii) Find the value of C for 98.6°F and
Find C when F = 98.6° F
Substituting F = 98.6° in (1) we get,
C = \(\frac{5}{9}(98.6-32)=\frac{5}{9}(66.6)=\frac{333}{9}=37°\)
(iii) Find the value of F for 38°C.
Substituting C = 38° in (1) we get,
38=\(\frac{5}{9}(F-32)\)
⇒ \(\frac{342}{5}+32\) = F
⇒ F = 100.4°C

Let P(h, k) be the locus of the point and A(4, 0) B(-4, 0) are the given points.
Given PA + PB = 10
\(\sqrt { { (h-4) }^{ 2 }+{ (k-0) }^{ 2 } } +\sqrt { { (h+4) }^{ 2 }+{ k-0) }^{ 2 } } =10\)
\(\sqrt { { (h-4) }^{ 2 }+{ k }^{ 2 } } =10-\sqrt { { (h+4) }^{ 2 }+{ k }^{ 2 } } \)
Squaring both sides we get,
\({ (h-4) }^{ 2 }+{ k }^{ 2 }=100+[({ h }+4)^{ 2 }+{ k }^{ 2 }]-20\sqrt { { (h+4) }^{ 2 }+{ k }^{ 2 } } \)
\(\Rightarrow \ { h }^{ 2 }-8h+16+{ k }^{ 2 }=100+{ h }^{ 2 }+16+8h+{ k }^{ 2 }-20\sqrt { { (h+4) }^{ 2 }+{ k }^{ 2 } } \)
\(\Rightarrow \ { h }^{ 2 }-8h+16+{ k }^{ 2 }-100+{ h }^{ 2 }+16+8h+{ k }^{ 2 }-20\sqrt { { (h+4) }^{ 2 }+{ k }^{ 2 } } \)
\(\Rightarrow -16h-100=-20\sqrt { { (h+4) }^{ 2 }+{ k }^{ 2 } } \)
Dividing by -4, we get
4h + 25 = \(\\ 5\sqrt { ({ h+4) }^{ 2 }+{ k }^{ 2 } } \)
Squaring both sides we get,
\(\Rightarrow\) (4h + 25)² = 25(h²+ 8h + 16 + k²)
\(\Rightarrow\) 16h² + 625 + 200h = 25h²+ 400 + 200h + 25k²
\(\Rightarrow\)  9h² + 25k² = 225
\(\Rightarrow\) 9h²+ 25k² = 225
Dividing by 225, we get
\(\Rightarrow \ \frac { { h }^{ 2 } }{ 25 } +\frac { { k }^{ 2 } }{ 9 } =1\)
\(\therefore \) Locus of (h, k) is \(\frac{x^2}{25}+\frac{y^2}{9}=1\)

Let P(h, k) be any point on the locus
Let P(\(\alpha \), \(\beta\)) be the point.
Given P divides OQ externally in the ratio 3: 4
\(\therefore \ (h,k)=\left( \frac { 3\alpha -4(0) }{ 3-4 } ,\frac { 3\beta -4(0) }{ 3-4 } \right) \ \left[ \because (z,y)=\left( \frac { { mx }_{ 2 }-{ nx }_{ 1 } }{ m-n } ,\frac { { my }_{ 2 }-{ ny }_{ 1 } }{ m-n } \right) \right] \)
\(\therefore \ (h,k)=\left( \frac { 3\alpha }{ -1 } ,\frac { 3\beta }{ -1 } \right) \)
\(\Rightarrow \ (h,k)=\left( -3\alpha ,-3\beta ) \right) \)
Equating the like co-ordinates both sides we get
\(\Rightarrow h= -3\alpha \Rightarrow k=-3\beta \)

\(\Rightarrow a=\frac { -h }{ 3 } \Rightarrow \beta =\frac { -k }{ 3 } \)
Since Q(\(\alpha \), \(\beta\)) lies on the locus of x²+ y²+ 4x - 3y+ 7 = 0, we get \({ \left( \frac { -h }{ 3 } \right) }^{ 2 }+{ \left( \frac { -k }{ 3 } \right) }^{ 2 }+4{ \left( \frac { -h }{ 3 } \right) }-3{ \left( \frac { -k }{ 3 } \right) }+7=0\)
\(\Rightarrow \frac { { h }^{ 2 } }{ 9 } +\frac { { k }^{ 2 } }{ 9 } -\frac { 4h }{ 3 } +k=7=0\)
Multiplying by 9, throughout we get,
h²+ k²-12h + 9k + 63 = 0
\(\therefore\) Locus of (h,k) is x²+ y²-12x + 9y + 63 = 0

Let R(h, k) be the locus of the mid-point of PQ where, P is (2, -7) and Q is a point on (2x² + 9y² = 18)
Given equation is 2x² + 9y² = 18
Dividing by 18 we get,
\(\Rightarrow \quad \frac { { x }^{ 2 } }{ 9 } +\frac { { y }^{ 2 } }{ 2 } =1\)
\(\Rightarrow \quad \frac { { x }^{ 2 } }{ { 3 }^{ 2 } } +\frac { { y }^{ 2 } }{ { \left( \sqrt { 2 } \right) }^{ 2 } } =1\)
\(\Rightarrow \quad a=3\quad and\quad b=\sqrt { 2 } \)
Any point on the ellipse is (a cos \(\theta\), b sin \(\theta\))
\(\therefore\) Q is (3 cos \(\theta\), \(\sqrt { 2 } \) sin \(\theta\)
Since R is the mid-point of PQ, we get, (h,k) = \(\left( \frac { 2+3cos\theta }{ 2 } ,\frac { -7+\sqrt { 2 } sin\theta }{ 2 } \right) \)
\(\Rightarrow \quad h=\frac { 2+3\quad cos\quad \theta }{ 2 } \)
\(\Rightarrow \quad 2h=2+3cos\theta \)
\(\Rightarrow \quad 2h-2=3cos\theta \)
\(\Rightarrow \quad \frac { 2h-2 }{ 3 } =cos\quad \theta \)
\(k=\frac { -7+\sqrt { 2 } sin\theta }{ 2 } \)
\(\Rightarrow \quad 2k=-7+\sqrt { 2 } sin\theta \)
\(\Rightarrow \quad 2k+7\quad =\quad \sqrt { 2 } sin\theta \)
\(\Rightarrow \quad \frac { 2k+7 }{ \sqrt { 2 } } =sin\quad \theta \)
Squaring and adding we get,
\({ \left( \frac { 2h-2 }{ 3 } \right) }^{ 2 }+{ \left( \frac { 2k+7 }{ \sqrt { 2 } } \right) }^{ 2 }={ cos }^{ 2 }\theta +{ sin }^{ 2 }\theta \)
\(\Rightarrow \quad \frac { { 4h }^{ 2 }+4-8h }{ 9 } +\frac { { 4k }^{ 2 }+49+28k }{ 2 } =1\quad [\because { cos }^{ 2 }\theta +{ sin }^{ 2 }\theta =1]\)
\(\Rightarrow \quad 2({ 4h }^{ 2 }+4-8h)=9({ 4k }^{ 2 }+49+28k)=18\)
\(\Rightarrow \quad { 8h }^{ 2 }+8-16h+36{ k }^{ 2 }+441+252k-18=0\)
\(\Rightarrow \quad { 8h }^{ 2 }+36{ k }^{ 2 }-16h+252k+431=0\)
\(\therefore\)   Locus of (h,k) is
8x²+36y²-16x+252y+431=0

Let P(h, k) be the point on the locus and A(1, -6) B(4, -2) be the given points.
By the given condition, \(\angle APB=90°\)

\(\therefore \) \(\Delta\)  APB is a right angled triangle
\(\Rightarrow\)  AB² = PA² + PB²
\(\Rightarrow { (1-4) }^{ 2 }+(-6+{ 2) }^{ 2 }={ (h-1) }^{ 2 }+{ (k+6) }^{ 2 }+{ (h-4) }^{ 2 }+{ (k+2) }^{ 2 }\)
\(\Rightarrow 9+16={ h }^{ 2 }-2h+1+{ k }^{ 2 }+36+12k+{ h }^{ 2 }-8h+16+{ k }^{ 2 }+4k+4\)
\(\Rightarrow { 2h }^{ 2 }+2{ k }^{ 2 }-10h+16k+57-25=0\)
\( \Rightarrow { 2h }^{ 2 }+{ 2k }^{ 2 }-10h+16k+32=0\)
Dividing by 2, we get,
\(\Rightarrow { h }^{ 2 }+{ k }^{ 2 }-5h+8k+16=0\)
\(\therefore\) Locus of (h, k) is
x² + y²-5x+8y+16 = 0

Let P(h, k) be the point on the locus and A(3, 5) B (1, -1) be the given points.
By the given condition, PA² + PB² = 20

Given PA² + PB² = 20
⇒ (h - 3)² + (k - 5)² + (h - 1)²+ (k + 1)² = 20
h² - 6h + 9 + k² - 10k + 25 + h² - 2h + 1 + k² + 2k + 1 = 20
⇒ 2h² + 2k² - 8h - 8k + 36 = 20
⇒ 2h² + 2k² - 8h - 8k + 36 - 20 = 0
Dividing by 2 we get,
h² + k² - 4h - 4k + 8 = 0
ஃ Locus of (h, k) is x² + y²- 4x - 4y + 8 = 0

Let (h, k) be the mid-point on the required path.
Let the co-ordinates of A and B be A(a, 0) and B(0, b).
As the rod slides, the values of a and b change.
So a and b are two variables.
Then \(h=\frac { a+0 }{ 2 } ,and\quad k=\frac { 0+b }{ 2 } \)
\(\Rightarrow \ h=\frac { a }{ 2 } and\quad k=\frac { b }{ 2 } \)
\(\Rightarrow\)  a = 2h and b = 2k

From \(\Delta\)OAB we have
AB² = OA² + OB²
\(\Rightarrow\) a² + b² = 8²    [\(\because\) AB = 8 units]
\(\Rightarrow\)  (2h)² + (2k)² = 64
\(\Rightarrow\)  4h² + 4k² = 64
\(\Rightarrow\)    h² + k² = 16      [Dividing by 4]
\(\therefore\) Locus of (h, k) is x² + y² = 16