Assertion(A): The minimum radius vector of the curve x2a2+y2b2=1 is of length a+bReason (R) : The minimum value of a2sec2θ+b2cosec2θ is (a+b)2

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Assertion(A): The minimum radius vector of the curve {{{a}^{2}

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Assertion(A): The minimum radius vector of the curve $\frac{a ^{2}}{x ^{2}} + \frac{b ^{2}}{y ^{2}} = 1$ is of length $a + b$
Reason (R) : The minimum value of $a^{2} sec^{2} θ + b^{2} cos e c^{2} θ$ is $(a + b)^{2}$

Both A and R are true and R is the correct explanation of A

Both A and R are true and R is not the correct explanation of A

A is true and R is false

A is false and R is true

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Assertion(A): The minimum radius vector of the curve

\frac{a ^{2}}{x ^{2}} + \frac{b ^{2}}{y ^{2}} = 1

is of length

a + b

Reason (R) : The minimum value of

a^{2} sec^{2} θ + b^{2} cos e c^{2} θ

(a + b)^{2}

9 mins ago

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9 mins ago

Any point on the curve

\frac{a ^{2}}{x ^{2}} + \frac{b ^{2}}{y ^{2}} = 1

can be given by

(a sec θ, b csc θ)

Radius vector

= r = a^{2} sec^{2} θ + b^{2} csc^{2} θ

\Rightarrow r^{2} = a^{2} sec^{2} θ + b^{2} csc^{2} θ = a^{2} (1 + tan^{2} θ) + b^{2} (1 + cot^{2} θ) = a^{2} + b^{2} + a^{2} tan^{2} θ + b^{2} cot^{2} θ

Since, A.M

\geq

G.M
Therefore,

\frac{a ^{2} tan ^{2} θ + b ^{2} cot ^{2} θ}{2} \geq a^{2} tan^{2} θ b^{2} cot^{2} θ = ab

\Rightarrow r^{2} \geq a^{2} + b^{2} + 2 ab = (a + b)^{2}

Therefore,

r \geq a + b

Ans: A

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Question Text	Assertion(A): The minimum radius vector of the curve $\frac{a ^{2}}{x ^{2}} + \frac{b ^{2}}{y ^{2}} = 1$ is of length $a + b$ Reason (R) : The minimum value of $a^{2} sec^{2} θ + b^{2} cos e c^{2} θ$ is $(a + b)^{2}$
Answer Type	Text solution:1
Upvotes	150