14. Find the area of the circle 4x2 + 4y2 = 9 which is interior to the parabola x2 = 4y

The equation given circle is4x2+4y2=9⇒x2+y2=94⇒x2+y2=(32)2i.e, centre (0,0), radius r=32since x2=4y intersect the circlewe can put x2=4y in x2+y2=(32)2(4y)+y2=94⇒y2+4y−94=0⇒4y2+16y−9=0⇒4y2+18y−2y−9=0⇒2y(2y+9)−1(2y+9)=0⇒(2y+9)(2y−1)=0⇒y=−92&y=12y=−92,x2=4×(−92)=−18 which is not possible or x2 cannot be (-)ve

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Application of Integrals Ncert Solutions Maths class 12th Maths Class 12th Application of Integrals

14. Find the area of the circle 4x² + 4y² = 9 which is interior to the parabola x² = 4y

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Vishal Baghel | Contributor-Level 10

4 months ago

The equation given circle is

$\begin{array}{l} 4 x^{2} + 4 y^{2} = 9 \\ \Rightarrow x^{2} + y^{2} = \frac{9}{4} \\ \Rightarrow x^{2} + y^{2} = {(\frac{3}{2})}^{2} \end{array}$

i.e, centre (0,0), radius $r = \frac{3}{2}$

since $x^{2} = 4 y$ intersect the circle

we can put $x^{2} = 4 y$ in $x^{2} + y^{2} = {(\frac{3}{2})}^{2}$

$\begin{array}{l} (4 y) + y^{2} = \frac{9}{4} \\ \Rightarrow y^{2} + 4 y - \frac{9}{4} = 0 \\ \Rightarrow 4 y^{2} + 16 y - 9 = 0 \\ \Rightarrow 4 y^{2} + 18 y - 2 y - 9 = 0 \end{array}$

$\begin{array}{l} \Rightarrow 2 y (2 y + 9) - 1 (2 y + 9) = 0 \\ \Rightarrow (2 y + 9) (2 y - 1) = 0 \\ \Rightarrow y = \frac{- 9}{2} & y = \frac{1}{2} \end{array}$

$y = \frac{- 9}{2}, x^{2} = 4 \times (\frac{- 9}{2}) = - 18$ which is not possible or $x^{2}$ cannot be (-)ve

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