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# Solution for Let C Be a Curve Defined Parametrically as X = a Cos 3 θ , Y = a Sin 3 θ , 0 ≤ θ ≤ π 2 . Determine a Point P on C, Where the Tangent to C is Parallel to the Chord Joining the Points (A, 0) and (0, A) - CBSE (Science) Class 12 - Mathematics

ConceptMaximum and Minimum Values of a Function in a Closed Interval

#### Question

Let C be a curve defined parametrically as $x = a \cos^3 \theta, y = a \sin^3 \theta, 0 \leq \theta \leq \frac{\pi}{2}$ . Determine a point P on C, where the tangent to C is parallel to the chord joining the points (a, 0) and (0, a).

#### Solution

$\text { As,} x = \text { a }\cos^3 \theta$

$\Rightarrow \frac{dx}{d\theta} = - 3a \cos^2 \theta\sin\theta$

$\text { And }, y = a \sin^3 \theta$

$\Rightarrow \frac{dy}{d\theta} = 3a \sin^2 \theta\cos\theta$

$So, \frac{dy}{dx} = \frac{\left( \frac{dy}{d\theta} \right)}{\left( \frac{dx}{d\theta} \right)} = \frac{3a \sin^2 \theta\cos\theta}{- 3a \cos^2 \theta\sin\theta} = - \tan\theta$

$\text { For the tangent to be parallel to the chord joining the points } \left( a, 0 \right) \text { and } \left( 0, a \right),$

$\frac{dy}{dx} = \frac{0 - a}{a - 0}$

$\Rightarrow - \tan\theta = - 1$

$\Rightarrow \tan\theta = 1$

$\Rightarrow \theta = \frac{\pi}{4}$

$\text { Now },$

$x = a \cos^3 \frac{\pi}{4} = a \left( \frac{1}{\sqrt{2}} \right)^3 = \frac{a}{2\sqrt{2}} \text { and }$

$y = a \sin^3 \frac{\pi}{4} = a \left( \frac{1}{\sqrt{2}} \right)^3 = \frac{a}{2\sqrt{2}}$

$\text { So, the point P on the curve C is } \left( \frac{a}{2\sqrt{2}}, \frac{a}{2\sqrt{2}} \right) .$

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Solution Let C Be a Curve Defined Parametrically as X = a Cos 3 θ , Y = a Sin 3 θ , 0 ≤ θ ≤ π 2 . Determine a Point P on C, Where the Tangent to C is Parallel to the Chord Joining the Points (A, 0) and (0, A) Concept: Maximum and Minimum Values of a Function in a Closed Interval.
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