87102
The order and degree of differential equations, \(p=\frac{\left[1+\left(\frac{d y}{d x}\right)^{2}\right]^{3 / 2}}{\frac{d^{2} y}{d x^{2}}} \text { are, respectively }\)
1 2,2
2 2,3
3 2,1
4 None of these
Explanation:
(A) : Given, differential equation- \(p=\frac{\left[1+\left(\frac{d y}{d x}\right)^{2}\right]^{3 / 2}}{\frac{d^{2} y}{d x^{2}}} \Rightarrow p\left(\frac{d^{2} y}{d x^{2}}\right)=\left[1+\left(\frac{d y}{d x}\right)^{2}\right]^{3 / 2}\) On squarrng both sides, we get \(\mathrm{p}^{2}\left(\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}\right)^{2}=\left[1+\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^{2}\right]^{3}\) Clearly, it is a second order differential equation of degree 2 . Note: The higher order derivative is in the transcendential, then we do not determined the degree of the equation.
JCECE-2016
Differential Equation
87103
The order and degree of the differential equation \(\left(1+4 \frac{d y}{d x}\right)^{2 / 3}=4 \frac{d^{2} y}{d x^{2}}\) are respectively
1 \(1, \frac{2}{3}\)
2 3,2
3 2,3
4 \(2, \frac{2}{3}\)
Explanation:
(C) : Given differential equation is \(\left(1+4 \frac{\mathrm{dy}}{\mathrm{dx}}\right)^{\frac{2}{3}}=4 \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}\) Cubic on both sides, \(\left(1+4 \frac{d y}{d x}\right)^{2}=4^{3}\left(\frac{d^{2} y}{d x^{2}}\right)^{3}\) Hence, order is 2 and degree is 3 .
VITEEE-2011]**#
Differential Equation
87104
The degree of the differential equation of all curves having normal of constant length \(\mathbf{c}\) is
1 1
2 3
3 4
4 None of these
Explanation:
(D) : Length of normal \(=\mathrm{c}\) \(y \sqrt{1+\left(\frac{d y}{d x}\right)^{2}}=c\) On squaring both sides, \(\mathrm{y}^{2}\left[1+\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^{2}\right]=\mathrm{c}^{2}\) Clearly, this is the differential equation of degree 2.
VITEEE-2010
Differential Equation
87105
The order and power of differential equation \(\frac{d^{2} y}{d x^{2}}+7 \frac{d y}{d x}+\int y d x=\sin x\) is
1 1,3
2 3,1
3 1,2
4 2, 1
Explanation:
(B) : Given differential equation \(\frac{d^{2} y}{d x^{2}}+7 \frac{d y}{d x}+\int y d x=\sin x\) Differentiating with respect to \(x\), we get- \(\frac{d^{3} y}{d x^{3}}+7 \frac{d^{2} y}{d x^{2}}+y \frac{d y}{d x}=\cos x\) Clearly order is 3 and degree is 1
UPSEE-2010
Differential Equation
87107
The length of the normal to the curve \(\mathbf{x}=\mathbf{a}(\theta+\sin \theta), y=\mathbf{a}(1-\cos \theta)\) at \(\theta=\frac{\pi}{2}\) is
1 \(2 \mathrm{a}\)
2 \(\frac{a}{2}\)
3 \(\frac{\mathrm{a}}{\sqrt{2}}\)
4 \(\sqrt{2} \mathrm{a}\)
Explanation:
(D) : Given, \(\mathrm{x}=\mathrm{a}(\theta+\sin \theta)\), and \(\mathrm{y}=\mathrm{a}(1-\cos \theta)\) Differentiating we get - \(\frac{\mathrm{dx}}{\mathrm{d} \theta}=\mathrm{a}(1+\cos \theta)\) and \(\frac{\mathrm{dy}}{\mathrm{d} \theta}=\mathrm{a}(0+\sin \theta)\) \(\frac{d y}{d x}=\frac{\frac{d y}{d x}}{\frac{d x}{d \theta}}=\frac{a \sin \theta}{a(1+\cos \theta)} \Rightarrow \frac{d y}{d x}=\frac{\sin \theta}{1+\cos \theta}\) \(\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{\theta=\frac{\pi}{2}}=\frac{\sin \frac{\pi}{2}}{1+\cos \frac{\pi}{2}}=\frac{1}{1+0}=1\) \(y=a(1-\cos \theta)\) at, \(\quad \theta=\frac{\pi}{2}\) \(\mathrm{y}=\mathrm{a}\left(1-\cos \frac{\pi}{2}\right)=\mathrm{a}(1-0) \Rightarrow \mathrm{y}=\mathrm{a}\) \(\therefore\) Length of normal \(=y \sqrt{1+\left(\frac{d y}{d x}\right)^{2}}=a \sqrt{1+1}=\sqrt{2} a\)
87102
The order and degree of differential equations, \(p=\frac{\left[1+\left(\frac{d y}{d x}\right)^{2}\right]^{3 / 2}}{\frac{d^{2} y}{d x^{2}}} \text { are, respectively }\)
1 2,2
2 2,3
3 2,1
4 None of these
Explanation:
(A) : Given, differential equation- \(p=\frac{\left[1+\left(\frac{d y}{d x}\right)^{2}\right]^{3 / 2}}{\frac{d^{2} y}{d x^{2}}} \Rightarrow p\left(\frac{d^{2} y}{d x^{2}}\right)=\left[1+\left(\frac{d y}{d x}\right)^{2}\right]^{3 / 2}\) On squarrng both sides, we get \(\mathrm{p}^{2}\left(\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}\right)^{2}=\left[1+\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^{2}\right]^{3}\) Clearly, it is a second order differential equation of degree 2 . Note: The higher order derivative is in the transcendential, then we do not determined the degree of the equation.
JCECE-2016
Differential Equation
87103
The order and degree of the differential equation \(\left(1+4 \frac{d y}{d x}\right)^{2 / 3}=4 \frac{d^{2} y}{d x^{2}}\) are respectively
1 \(1, \frac{2}{3}\)
2 3,2
3 2,3
4 \(2, \frac{2}{3}\)
Explanation:
(C) : Given differential equation is \(\left(1+4 \frac{\mathrm{dy}}{\mathrm{dx}}\right)^{\frac{2}{3}}=4 \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}\) Cubic on both sides, \(\left(1+4 \frac{d y}{d x}\right)^{2}=4^{3}\left(\frac{d^{2} y}{d x^{2}}\right)^{3}\) Hence, order is 2 and degree is 3 .
VITEEE-2011]**#
Differential Equation
87104
The degree of the differential equation of all curves having normal of constant length \(\mathbf{c}\) is
1 1
2 3
3 4
4 None of these
Explanation:
(D) : Length of normal \(=\mathrm{c}\) \(y \sqrt{1+\left(\frac{d y}{d x}\right)^{2}}=c\) On squaring both sides, \(\mathrm{y}^{2}\left[1+\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^{2}\right]=\mathrm{c}^{2}\) Clearly, this is the differential equation of degree 2.
VITEEE-2010
Differential Equation
87105
The order and power of differential equation \(\frac{d^{2} y}{d x^{2}}+7 \frac{d y}{d x}+\int y d x=\sin x\) is
1 1,3
2 3,1
3 1,2
4 2, 1
Explanation:
(B) : Given differential equation \(\frac{d^{2} y}{d x^{2}}+7 \frac{d y}{d x}+\int y d x=\sin x\) Differentiating with respect to \(x\), we get- \(\frac{d^{3} y}{d x^{3}}+7 \frac{d^{2} y}{d x^{2}}+y \frac{d y}{d x}=\cos x\) Clearly order is 3 and degree is 1
UPSEE-2010
Differential Equation
87107
The length of the normal to the curve \(\mathbf{x}=\mathbf{a}(\theta+\sin \theta), y=\mathbf{a}(1-\cos \theta)\) at \(\theta=\frac{\pi}{2}\) is
1 \(2 \mathrm{a}\)
2 \(\frac{a}{2}\)
3 \(\frac{\mathrm{a}}{\sqrt{2}}\)
4 \(\sqrt{2} \mathrm{a}\)
Explanation:
(D) : Given, \(\mathrm{x}=\mathrm{a}(\theta+\sin \theta)\), and \(\mathrm{y}=\mathrm{a}(1-\cos \theta)\) Differentiating we get - \(\frac{\mathrm{dx}}{\mathrm{d} \theta}=\mathrm{a}(1+\cos \theta)\) and \(\frac{\mathrm{dy}}{\mathrm{d} \theta}=\mathrm{a}(0+\sin \theta)\) \(\frac{d y}{d x}=\frac{\frac{d y}{d x}}{\frac{d x}{d \theta}}=\frac{a \sin \theta}{a(1+\cos \theta)} \Rightarrow \frac{d y}{d x}=\frac{\sin \theta}{1+\cos \theta}\) \(\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{\theta=\frac{\pi}{2}}=\frac{\sin \frac{\pi}{2}}{1+\cos \frac{\pi}{2}}=\frac{1}{1+0}=1\) \(y=a(1-\cos \theta)\) at, \(\quad \theta=\frac{\pi}{2}\) \(\mathrm{y}=\mathrm{a}\left(1-\cos \frac{\pi}{2}\right)=\mathrm{a}(1-0) \Rightarrow \mathrm{y}=\mathrm{a}\) \(\therefore\) Length of normal \(=y \sqrt{1+\left(\frac{d y}{d x}\right)^{2}}=a \sqrt{1+1}=\sqrt{2} a\)
87102
The order and degree of differential equations, \(p=\frac{\left[1+\left(\frac{d y}{d x}\right)^{2}\right]^{3 / 2}}{\frac{d^{2} y}{d x^{2}}} \text { are, respectively }\)
1 2,2
2 2,3
3 2,1
4 None of these
Explanation:
(A) : Given, differential equation- \(p=\frac{\left[1+\left(\frac{d y}{d x}\right)^{2}\right]^{3 / 2}}{\frac{d^{2} y}{d x^{2}}} \Rightarrow p\left(\frac{d^{2} y}{d x^{2}}\right)=\left[1+\left(\frac{d y}{d x}\right)^{2}\right]^{3 / 2}\) On squarrng both sides, we get \(\mathrm{p}^{2}\left(\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}\right)^{2}=\left[1+\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^{2}\right]^{3}\) Clearly, it is a second order differential equation of degree 2 . Note: The higher order derivative is in the transcendential, then we do not determined the degree of the equation.
JCECE-2016
Differential Equation
87103
The order and degree of the differential equation \(\left(1+4 \frac{d y}{d x}\right)^{2 / 3}=4 \frac{d^{2} y}{d x^{2}}\) are respectively
1 \(1, \frac{2}{3}\)
2 3,2
3 2,3
4 \(2, \frac{2}{3}\)
Explanation:
(C) : Given differential equation is \(\left(1+4 \frac{\mathrm{dy}}{\mathrm{dx}}\right)^{\frac{2}{3}}=4 \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}\) Cubic on both sides, \(\left(1+4 \frac{d y}{d x}\right)^{2}=4^{3}\left(\frac{d^{2} y}{d x^{2}}\right)^{3}\) Hence, order is 2 and degree is 3 .
VITEEE-2011]**#
Differential Equation
87104
The degree of the differential equation of all curves having normal of constant length \(\mathbf{c}\) is
1 1
2 3
3 4
4 None of these
Explanation:
(D) : Length of normal \(=\mathrm{c}\) \(y \sqrt{1+\left(\frac{d y}{d x}\right)^{2}}=c\) On squaring both sides, \(\mathrm{y}^{2}\left[1+\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^{2}\right]=\mathrm{c}^{2}\) Clearly, this is the differential equation of degree 2.
VITEEE-2010
Differential Equation
87105
The order and power of differential equation \(\frac{d^{2} y}{d x^{2}}+7 \frac{d y}{d x}+\int y d x=\sin x\) is
1 1,3
2 3,1
3 1,2
4 2, 1
Explanation:
(B) : Given differential equation \(\frac{d^{2} y}{d x^{2}}+7 \frac{d y}{d x}+\int y d x=\sin x\) Differentiating with respect to \(x\), we get- \(\frac{d^{3} y}{d x^{3}}+7 \frac{d^{2} y}{d x^{2}}+y \frac{d y}{d x}=\cos x\) Clearly order is 3 and degree is 1
UPSEE-2010
Differential Equation
87107
The length of the normal to the curve \(\mathbf{x}=\mathbf{a}(\theta+\sin \theta), y=\mathbf{a}(1-\cos \theta)\) at \(\theta=\frac{\pi}{2}\) is
1 \(2 \mathrm{a}\)
2 \(\frac{a}{2}\)
3 \(\frac{\mathrm{a}}{\sqrt{2}}\)
4 \(\sqrt{2} \mathrm{a}\)
Explanation:
(D) : Given, \(\mathrm{x}=\mathrm{a}(\theta+\sin \theta)\), and \(\mathrm{y}=\mathrm{a}(1-\cos \theta)\) Differentiating we get - \(\frac{\mathrm{dx}}{\mathrm{d} \theta}=\mathrm{a}(1+\cos \theta)\) and \(\frac{\mathrm{dy}}{\mathrm{d} \theta}=\mathrm{a}(0+\sin \theta)\) \(\frac{d y}{d x}=\frac{\frac{d y}{d x}}{\frac{d x}{d \theta}}=\frac{a \sin \theta}{a(1+\cos \theta)} \Rightarrow \frac{d y}{d x}=\frac{\sin \theta}{1+\cos \theta}\) \(\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{\theta=\frac{\pi}{2}}=\frac{\sin \frac{\pi}{2}}{1+\cos \frac{\pi}{2}}=\frac{1}{1+0}=1\) \(y=a(1-\cos \theta)\) at, \(\quad \theta=\frac{\pi}{2}\) \(\mathrm{y}=\mathrm{a}\left(1-\cos \frac{\pi}{2}\right)=\mathrm{a}(1-0) \Rightarrow \mathrm{y}=\mathrm{a}\) \(\therefore\) Length of normal \(=y \sqrt{1+\left(\frac{d y}{d x}\right)^{2}}=a \sqrt{1+1}=\sqrt{2} a\)
NEET Test Series from KOTA - 10 Papers In MS WORD
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Differential Equation
87102
The order and degree of differential equations, \(p=\frac{\left[1+\left(\frac{d y}{d x}\right)^{2}\right]^{3 / 2}}{\frac{d^{2} y}{d x^{2}}} \text { are, respectively }\)
1 2,2
2 2,3
3 2,1
4 None of these
Explanation:
(A) : Given, differential equation- \(p=\frac{\left[1+\left(\frac{d y}{d x}\right)^{2}\right]^{3 / 2}}{\frac{d^{2} y}{d x^{2}}} \Rightarrow p\left(\frac{d^{2} y}{d x^{2}}\right)=\left[1+\left(\frac{d y}{d x}\right)^{2}\right]^{3 / 2}\) On squarrng both sides, we get \(\mathrm{p}^{2}\left(\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}\right)^{2}=\left[1+\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^{2}\right]^{3}\) Clearly, it is a second order differential equation of degree 2 . Note: The higher order derivative is in the transcendential, then we do not determined the degree of the equation.
JCECE-2016
Differential Equation
87103
The order and degree of the differential equation \(\left(1+4 \frac{d y}{d x}\right)^{2 / 3}=4 \frac{d^{2} y}{d x^{2}}\) are respectively
1 \(1, \frac{2}{3}\)
2 3,2
3 2,3
4 \(2, \frac{2}{3}\)
Explanation:
(C) : Given differential equation is \(\left(1+4 \frac{\mathrm{dy}}{\mathrm{dx}}\right)^{\frac{2}{3}}=4 \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}\) Cubic on both sides, \(\left(1+4 \frac{d y}{d x}\right)^{2}=4^{3}\left(\frac{d^{2} y}{d x^{2}}\right)^{3}\) Hence, order is 2 and degree is 3 .
VITEEE-2011]**#
Differential Equation
87104
The degree of the differential equation of all curves having normal of constant length \(\mathbf{c}\) is
1 1
2 3
3 4
4 None of these
Explanation:
(D) : Length of normal \(=\mathrm{c}\) \(y \sqrt{1+\left(\frac{d y}{d x}\right)^{2}}=c\) On squaring both sides, \(\mathrm{y}^{2}\left[1+\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^{2}\right]=\mathrm{c}^{2}\) Clearly, this is the differential equation of degree 2.
VITEEE-2010
Differential Equation
87105
The order and power of differential equation \(\frac{d^{2} y}{d x^{2}}+7 \frac{d y}{d x}+\int y d x=\sin x\) is
1 1,3
2 3,1
3 1,2
4 2, 1
Explanation:
(B) : Given differential equation \(\frac{d^{2} y}{d x^{2}}+7 \frac{d y}{d x}+\int y d x=\sin x\) Differentiating with respect to \(x\), we get- \(\frac{d^{3} y}{d x^{3}}+7 \frac{d^{2} y}{d x^{2}}+y \frac{d y}{d x}=\cos x\) Clearly order is 3 and degree is 1
UPSEE-2010
Differential Equation
87107
The length of the normal to the curve \(\mathbf{x}=\mathbf{a}(\theta+\sin \theta), y=\mathbf{a}(1-\cos \theta)\) at \(\theta=\frac{\pi}{2}\) is
1 \(2 \mathrm{a}\)
2 \(\frac{a}{2}\)
3 \(\frac{\mathrm{a}}{\sqrt{2}}\)
4 \(\sqrt{2} \mathrm{a}\)
Explanation:
(D) : Given, \(\mathrm{x}=\mathrm{a}(\theta+\sin \theta)\), and \(\mathrm{y}=\mathrm{a}(1-\cos \theta)\) Differentiating we get - \(\frac{\mathrm{dx}}{\mathrm{d} \theta}=\mathrm{a}(1+\cos \theta)\) and \(\frac{\mathrm{dy}}{\mathrm{d} \theta}=\mathrm{a}(0+\sin \theta)\) \(\frac{d y}{d x}=\frac{\frac{d y}{d x}}{\frac{d x}{d \theta}}=\frac{a \sin \theta}{a(1+\cos \theta)} \Rightarrow \frac{d y}{d x}=\frac{\sin \theta}{1+\cos \theta}\) \(\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{\theta=\frac{\pi}{2}}=\frac{\sin \frac{\pi}{2}}{1+\cos \frac{\pi}{2}}=\frac{1}{1+0}=1\) \(y=a(1-\cos \theta)\) at, \(\quad \theta=\frac{\pi}{2}\) \(\mathrm{y}=\mathrm{a}\left(1-\cos \frac{\pi}{2}\right)=\mathrm{a}(1-0) \Rightarrow \mathrm{y}=\mathrm{a}\) \(\therefore\) Length of normal \(=y \sqrt{1+\left(\frac{d y}{d x}\right)^{2}}=a \sqrt{1+1}=\sqrt{2} a\)
87102
The order and degree of differential equations, \(p=\frac{\left[1+\left(\frac{d y}{d x}\right)^{2}\right]^{3 / 2}}{\frac{d^{2} y}{d x^{2}}} \text { are, respectively }\)
1 2,2
2 2,3
3 2,1
4 None of these
Explanation:
(A) : Given, differential equation- \(p=\frac{\left[1+\left(\frac{d y}{d x}\right)^{2}\right]^{3 / 2}}{\frac{d^{2} y}{d x^{2}}} \Rightarrow p\left(\frac{d^{2} y}{d x^{2}}\right)=\left[1+\left(\frac{d y}{d x}\right)^{2}\right]^{3 / 2}\) On squarrng both sides, we get \(\mathrm{p}^{2}\left(\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}\right)^{2}=\left[1+\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^{2}\right]^{3}\) Clearly, it is a second order differential equation of degree 2 . Note: The higher order derivative is in the transcendential, then we do not determined the degree of the equation.
JCECE-2016
Differential Equation
87103
The order and degree of the differential equation \(\left(1+4 \frac{d y}{d x}\right)^{2 / 3}=4 \frac{d^{2} y}{d x^{2}}\) are respectively
1 \(1, \frac{2}{3}\)
2 3,2
3 2,3
4 \(2, \frac{2}{3}\)
Explanation:
(C) : Given differential equation is \(\left(1+4 \frac{\mathrm{dy}}{\mathrm{dx}}\right)^{\frac{2}{3}}=4 \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}\) Cubic on both sides, \(\left(1+4 \frac{d y}{d x}\right)^{2}=4^{3}\left(\frac{d^{2} y}{d x^{2}}\right)^{3}\) Hence, order is 2 and degree is 3 .
VITEEE-2011]**#
Differential Equation
87104
The degree of the differential equation of all curves having normal of constant length \(\mathbf{c}\) is
1 1
2 3
3 4
4 None of these
Explanation:
(D) : Length of normal \(=\mathrm{c}\) \(y \sqrt{1+\left(\frac{d y}{d x}\right)^{2}}=c\) On squaring both sides, \(\mathrm{y}^{2}\left[1+\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^{2}\right]=\mathrm{c}^{2}\) Clearly, this is the differential equation of degree 2.
VITEEE-2010
Differential Equation
87105
The order and power of differential equation \(\frac{d^{2} y}{d x^{2}}+7 \frac{d y}{d x}+\int y d x=\sin x\) is
1 1,3
2 3,1
3 1,2
4 2, 1
Explanation:
(B) : Given differential equation \(\frac{d^{2} y}{d x^{2}}+7 \frac{d y}{d x}+\int y d x=\sin x\) Differentiating with respect to \(x\), we get- \(\frac{d^{3} y}{d x^{3}}+7 \frac{d^{2} y}{d x^{2}}+y \frac{d y}{d x}=\cos x\) Clearly order is 3 and degree is 1
UPSEE-2010
Differential Equation
87107
The length of the normal to the curve \(\mathbf{x}=\mathbf{a}(\theta+\sin \theta), y=\mathbf{a}(1-\cos \theta)\) at \(\theta=\frac{\pi}{2}\) is
1 \(2 \mathrm{a}\)
2 \(\frac{a}{2}\)
3 \(\frac{\mathrm{a}}{\sqrt{2}}\)
4 \(\sqrt{2} \mathrm{a}\)
Explanation:
(D) : Given, \(\mathrm{x}=\mathrm{a}(\theta+\sin \theta)\), and \(\mathrm{y}=\mathrm{a}(1-\cos \theta)\) Differentiating we get - \(\frac{\mathrm{dx}}{\mathrm{d} \theta}=\mathrm{a}(1+\cos \theta)\) and \(\frac{\mathrm{dy}}{\mathrm{d} \theta}=\mathrm{a}(0+\sin \theta)\) \(\frac{d y}{d x}=\frac{\frac{d y}{d x}}{\frac{d x}{d \theta}}=\frac{a \sin \theta}{a(1+\cos \theta)} \Rightarrow \frac{d y}{d x}=\frac{\sin \theta}{1+\cos \theta}\) \(\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{\theta=\frac{\pi}{2}}=\frac{\sin \frac{\pi}{2}}{1+\cos \frac{\pi}{2}}=\frac{1}{1+0}=1\) \(y=a(1-\cos \theta)\) at, \(\quad \theta=\frac{\pi}{2}\) \(\mathrm{y}=\mathrm{a}\left(1-\cos \frac{\pi}{2}\right)=\mathrm{a}(1-0) \Rightarrow \mathrm{y}=\mathrm{a}\) \(\therefore\) Length of normal \(=y \sqrt{1+\left(\frac{d y}{d x}\right)^{2}}=a \sqrt{1+1}=\sqrt{2} a\)