Question

If x = p cos , y = q sin , then find the value of q 2 x 2 + p 2 y 2 - p 2 q 2

Answer 1

GIVEN :–

$\\ \bf \: \: {\huge{.} }\: \: x = p \cos( \theta)$

$\\ \bf \: \: {\huge{.} }\: \: y = q \sin( \theta)$

TO FIND :–

• Value of q²x² + p²y² - p²q² = ?

SOLUTION :–

• Let –

$\\ \implies\bf A = {q}^{2} {x}^{2} + {p}^{2} {x}^{2} - {p}^{2} {q}^{2}$

• Now put the values –

$\\ \implies\bf A = {q}^{2} {(p \cos \theta)}^{2} + {p}^{2} {(q \sin\theta)}^{2} - {p}^{2} {q}^{2}$

$\\ \implies\bf A = {q}^{2} {(p)}^{2} {( \cos\theta)}^{2} + {p}^{2} {(q )}^{2} {(\sin\theta)}^{2} - {p}^{2} {q}^{2}$

$\\ \implies\bf A = {q}^{2} {p}^{2} \cos^{2} (\theta)+ {p}^{2} {q}^{2} \sin^{2} (\theta)- {p}^{2} {q}^{2}$

$\\ \implies\bf A = {p}^{2} {q}^{2}\cos^{2} (\theta)+ {p}^{2} {q}^{2} \sin^{2} (\theta)- {p}^{2} {q}^{2}$

$\\ \implies\bf A = {p}^{2} {q}^{2} \{\cos^{2} (\theta)+\sin^{2} (\theta) \}- {p}^{2} {q}^{2}$

• We know that –

$\\ \: \longrightarrow \: \large \pink{ \boxed{\bf \cos^{2} (\theta)+\sin^{2} (\theta) = 1}}$

• So that –

$\\ \implies\bf A = {p}^{2} {q}^{2} \{1\}- {p}^{2} {q}^{2}$

$\\ \implies\bf A = {p}^{2} {q}^{2}- {p}^{2} {q}^{2}$

$\\ \implies \large{\boxed{\bf A =0}}$

Answer 2

Answer:

GIVEN :–

\begin{gathered} \\ \bf \: \: {\huge{.} }\: \: x = p \cos( \theta) \end{gathered}

.x=pcos(θ)

\begin{gathered} \\ \bf \: \: {\huge{.} }\: \: y = q \sin( \theta) \end{gathered}

.y=qsin(θ)

TO FIND :–

• Value of q²x² + p²y² - p²q² = ?

SOLUTION :–

• Let –

\begin{gathered} \\ \implies\bf A = {q}^{2} {x}^{2} + {p}^{2} {x}^{2} - {p}^{2} {q}^{2} \end{gathered}

⟹A=q

2

x

2

+p

2

x

2

−p

2

q

2

• Now put the values –

\begin{gathered} \\ \implies\bf A = {q}^{2} {(p \cos \theta)}^{2} + {p}^{2} {(q \sin\theta)}^{2} - {p}^{2} {q}^{2} \end{gathered}

⟹A=q

2

(pcosθ)

2

+p

2

(qsinθ)

2

−p

2

q

2

\begin{gathered} \\ \implies\bf A = {q}^{2} {(p)}^{2} {( \cos\theta)}^{2} + {p}^{2} {(q )}^{2} {(\sin\theta)}^{2} - {p}^{2} {q}^{2} \end{gathered}

⟹A=q

2

(p)

2

(cosθ)

2

+p

2

(q)

2

(sinθ)

2

−p

2

q

2

\begin{gathered} \\ \implies\bf A = {q}^{2} {p}^{2} \cos^{2} (\theta)+ {p}^{2} {q}^{2} \sin^{2} (\theta)- {p}^{2} {q}^{2} \end{gathered}

⟹A=q

2

p

2

cos

2

(θ)+p

2

q

2

sin

2

(θ)−p

2

q

2

\begin{gathered} \\ \implies\bf A = {p}^{2} {q}^{2}\cos^{2} (\theta)+ {p}^{2} {q}^{2} \sin^{2} (\theta)- {p}^{2} {q}^{2} \end{gathered}

⟹A=p

2

q

2

cos

2

(θ)+p

2

q

2

sin

2

(θ)−p

2

q

2

\begin{gathered} \\ \implies\bf A = {p}^{2} {q}^{2} \{\cos^{2} (\theta)+\sin^{2} (\theta) \}- {p}^{2} {q}^{2} \end{gathered}

⟹A=p

2

q

2

{cos

2

(θ)+sin

2

(θ)}−p

2

q

2

• We know that –

\begin{gathered} \\ \: \longrightarrow \: \large \pink{ \boxed{\bf \cos^{2} (\theta)+\sin^{2} (\theta) = 1}}\end{gathered}

⟶

cos

2

(θ)+sin

2

(θ)=1

• So that –

\begin{gathered} \\ \implies\bf A = {p}^{2} {q}^{2} \{1\}- {p}^{2} {q}^{2} \end{gathered}

⟹A=p

2

q

2

{1}−p

2

q

2

\begin{gathered} \\ \implies\bf A = {p}^{2} {q}^{2}- {p}^{2} {q}^{2} \end{gathered}

⟹A=p

2

q

2

−p

2

q

2

\begin{gathered} \\ \implies \large{\boxed{\bf A =0}}\end{gathered}

⟹

A=0