Question:
A circle passes through the ends of the latus rectum of parabola \( y^2=12x \) and has its centre at the vertex. The area inside the circle and outside the parabola in the \( 1^{st} \) quadrant is:
A circle passes through the ends of the latus rectum of parabola \( y^2=12x \) and has its centre at the vertex. The area inside the circle and outside the parabola in the \( 1^{st} \) quadrant is:
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When finding the area between a parabola and a circle, determine the intersection points first to define the limits of your integration.
Updated On: Jun 9, 2026
- \( \frac{45}{2} \sin^{-1}\left(\frac{1}{\sqrt{5}}\right) - 3 \)
- \( \frac{45}{2} \sin^{-1}\left(\frac{3}{\sqrt{5}}\right) + \frac{45}{2}\pi \)
- \( 45 \sin^{-1}\left(\frac{1}{\sqrt{5}}\right) + 6 \)
- \( 45 \sin^{-1}\left(\frac{3}{\sqrt{5}}\right) + 6\pi \)
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The Correct Option is A
Solution and Explanation
Concept:
We find the area between two curves: the circle \( x^2 + y^2 = 45 \) (derived from latus rectum ends) and the parabola \( y^2 = 12x \)[cite: 1028].
Step 1: Determine the circle equation.
Ends of latus rectum for \( y^2 = 12x \) are \( (3, 6) \) and \( (3, -6) \). Distance from vertex \((0,0)\) is \( R = \sqrt{3^2 + 6^2} = \sqrt{45} \). $$ x^2 + y^2 = 45 $$
Step 2: Find intersection point.
$$ x^2 + 12x - 45 = 0 \implies (x+15)(x-3) = 0 $$ Intersection is at \( x=3 \)[cite: 1030].
Step 3: Set up the area integral.
$$ Area = \int_{0}^{3} (\sqrt{45-x^2} - \sqrt{12x}) dx $$
Step 4: Integrate.
The integral of \( \sqrt{45-x^2} \) gives the circular sector area involving \( \sin^{-1} \), and the integral of \( \sqrt{12x} \) is \( 2\sqrt{3} \cdot \frac{2}{3} x^{3/2} \)[cite: 1037]. Evaluating at the boundaries gives the required area. $$\boxed{\frac{45}{2} \sin^{-1}\left(\frac{1}{\sqrt{5}}\right) - 3}$$
Step 1: Determine the circle equation.
Ends of latus rectum for \( y^2 = 12x \) are \( (3, 6) \) and \( (3, -6) \). Distance from vertex \((0,0)\) is \( R = \sqrt{3^2 + 6^2} = \sqrt{45} \). $$ x^2 + y^2 = 45 $$
Step 2: Find intersection point.
$$ x^2 + 12x - 45 = 0 \implies (x+15)(x-3) = 0 $$ Intersection is at \( x=3 \)[cite: 1030].
Step 3: Set up the area integral.
$$ Area = \int_{0}^{3} (\sqrt{45-x^2} - \sqrt{12x}) dx $$
Step 4: Integrate.
The integral of \( \sqrt{45-x^2} \) gives the circular sector area involving \( \sin^{-1} \), and the integral of \( \sqrt{12x} \) is \( 2\sqrt{3} \cdot \frac{2}{3} x^{3/2} \)[cite: 1037]. Evaluating at the boundaries gives the required area. $$\boxed{\frac{45}{2} \sin^{-1}\left(\frac{1}{\sqrt{5}}\right) - 3}$$
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