NEET Test Series from KOTA - 10 Papers In MS WORD
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Application of Derivatives
85796
If \(\theta\) is the acute angle between the curves \(x^{2}+y^{2}=4\) and \(y^{2}=3 x\) then \(\tan \theta\)
1 \(\frac{5}{\sqrt{3}}\)
2 \(\frac{\sqrt{3}}{4}\)
3 \(\frac{4}{\sqrt{3}}\)
4 \(\frac{\sqrt{3}}{5}\)
Explanation:
(A) : Given curve, \(x^{2}+y^{2}=4 \tag{i}\) And \(\quad y^{2}=3 x\) Point of contact is obtained by solving above two equations- From equation (i) and (ii)- \(x^{2}+3 x-4=0\) \(\Rightarrow \quad(x+4)(x-1)=0 \quad x=-4,1\) Neglecting \(\mathrm{x}=-4\), for \(\mathrm{x}=1\) \(y=\sqrt{3}\) \(\therefore\) Point of contact is \((1, \sqrt{3})\) For curve \(x^{2}+y^{2}=1=\frac{d y}{d x}=\left(\frac{-x}{y}\right)\) \(\therefore \quad \mathrm{m}=\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{(1, \sqrt{3})}=\frac{-1}{\sqrt{3}}\) for \(y^{2}=3 x \Rightarrow\left(\frac{d y}{d x}\right)_{(1, \sqrt{3})}=\frac{\sqrt{3}}{2}\) \(\therefore\) Angle between the curve \(=\) angle between their normal \(\therefore \tan \theta=\frac{\sqrt{3}+\frac{2}{\sqrt{3}}}{1+\sqrt{3}\left(\frac{-2}{\sqrt{3}}\right)}\) \(=\frac{-5}{\sqrt{3}}\) \(\because\) Angle is acute angle, then- \(\tan \theta=\frac{5}{\sqrt{3}}\)
Shift-I
Application of Derivatives
85797
The angle between the curves \(2 x^{2}+y^{2}=20\) and \(4 y^{2}-x^{2}=8\) at a point where they intersect in the \(4^{\text {th }}\) quadrant \(i\)
85796
If \(\theta\) is the acute angle between the curves \(x^{2}+y^{2}=4\) and \(y^{2}=3 x\) then \(\tan \theta\)
1 \(\frac{5}{\sqrt{3}}\)
2 \(\frac{\sqrt{3}}{4}\)
3 \(\frac{4}{\sqrt{3}}\)
4 \(\frac{\sqrt{3}}{5}\)
Explanation:
(A) : Given curve, \(x^{2}+y^{2}=4 \tag{i}\) And \(\quad y^{2}=3 x\) Point of contact is obtained by solving above two equations- From equation (i) and (ii)- \(x^{2}+3 x-4=0\) \(\Rightarrow \quad(x+4)(x-1)=0 \quad x=-4,1\) Neglecting \(\mathrm{x}=-4\), for \(\mathrm{x}=1\) \(y=\sqrt{3}\) \(\therefore\) Point of contact is \((1, \sqrt{3})\) For curve \(x^{2}+y^{2}=1=\frac{d y}{d x}=\left(\frac{-x}{y}\right)\) \(\therefore \quad \mathrm{m}=\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{(1, \sqrt{3})}=\frac{-1}{\sqrt{3}}\) for \(y^{2}=3 x \Rightarrow\left(\frac{d y}{d x}\right)_{(1, \sqrt{3})}=\frac{\sqrt{3}}{2}\) \(\therefore\) Angle between the curve \(=\) angle between their normal \(\therefore \tan \theta=\frac{\sqrt{3}+\frac{2}{\sqrt{3}}}{1+\sqrt{3}\left(\frac{-2}{\sqrt{3}}\right)}\) \(=\frac{-5}{\sqrt{3}}\) \(\because\) Angle is acute angle, then- \(\tan \theta=\frac{5}{\sqrt{3}}\)
Shift-I
Application of Derivatives
85797
The angle between the curves \(2 x^{2}+y^{2}=20\) and \(4 y^{2}-x^{2}=8\) at a point where they intersect in the \(4^{\text {th }}\) quadrant \(i\)
85796
If \(\theta\) is the acute angle between the curves \(x^{2}+y^{2}=4\) and \(y^{2}=3 x\) then \(\tan \theta\)
1 \(\frac{5}{\sqrt{3}}\)
2 \(\frac{\sqrt{3}}{4}\)
3 \(\frac{4}{\sqrt{3}}\)
4 \(\frac{\sqrt{3}}{5}\)
Explanation:
(A) : Given curve, \(x^{2}+y^{2}=4 \tag{i}\) And \(\quad y^{2}=3 x\) Point of contact is obtained by solving above two equations- From equation (i) and (ii)- \(x^{2}+3 x-4=0\) \(\Rightarrow \quad(x+4)(x-1)=0 \quad x=-4,1\) Neglecting \(\mathrm{x}=-4\), for \(\mathrm{x}=1\) \(y=\sqrt{3}\) \(\therefore\) Point of contact is \((1, \sqrt{3})\) For curve \(x^{2}+y^{2}=1=\frac{d y}{d x}=\left(\frac{-x}{y}\right)\) \(\therefore \quad \mathrm{m}=\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{(1, \sqrt{3})}=\frac{-1}{\sqrt{3}}\) for \(y^{2}=3 x \Rightarrow\left(\frac{d y}{d x}\right)_{(1, \sqrt{3})}=\frac{\sqrt{3}}{2}\) \(\therefore\) Angle between the curve \(=\) angle between their normal \(\therefore \tan \theta=\frac{\sqrt{3}+\frac{2}{\sqrt{3}}}{1+\sqrt{3}\left(\frac{-2}{\sqrt{3}}\right)}\) \(=\frac{-5}{\sqrt{3}}\) \(\because\) Angle is acute angle, then- \(\tan \theta=\frac{5}{\sqrt{3}}\)
Shift-I
Application of Derivatives
85797
The angle between the curves \(2 x^{2}+y^{2}=20\) and \(4 y^{2}-x^{2}=8\) at a point where they intersect in the \(4^{\text {th }}\) quadrant \(i\)
85796
If \(\theta\) is the acute angle between the curves \(x^{2}+y^{2}=4\) and \(y^{2}=3 x\) then \(\tan \theta\)
1 \(\frac{5}{\sqrt{3}}\)
2 \(\frac{\sqrt{3}}{4}\)
3 \(\frac{4}{\sqrt{3}}\)
4 \(\frac{\sqrt{3}}{5}\)
Explanation:
(A) : Given curve, \(x^{2}+y^{2}=4 \tag{i}\) And \(\quad y^{2}=3 x\) Point of contact is obtained by solving above two equations- From equation (i) and (ii)- \(x^{2}+3 x-4=0\) \(\Rightarrow \quad(x+4)(x-1)=0 \quad x=-4,1\) Neglecting \(\mathrm{x}=-4\), for \(\mathrm{x}=1\) \(y=\sqrt{3}\) \(\therefore\) Point of contact is \((1, \sqrt{3})\) For curve \(x^{2}+y^{2}=1=\frac{d y}{d x}=\left(\frac{-x}{y}\right)\) \(\therefore \quad \mathrm{m}=\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{(1, \sqrt{3})}=\frac{-1}{\sqrt{3}}\) for \(y^{2}=3 x \Rightarrow\left(\frac{d y}{d x}\right)_{(1, \sqrt{3})}=\frac{\sqrt{3}}{2}\) \(\therefore\) Angle between the curve \(=\) angle between their normal \(\therefore \tan \theta=\frac{\sqrt{3}+\frac{2}{\sqrt{3}}}{1+\sqrt{3}\left(\frac{-2}{\sqrt{3}}\right)}\) \(=\frac{-5}{\sqrt{3}}\) \(\because\) Angle is acute angle, then- \(\tan \theta=\frac{5}{\sqrt{3}}\)
Shift-I
Application of Derivatives
85797
The angle between the curves \(2 x^{2}+y^{2}=20\) and \(4 y^{2}-x^{2}=8\) at a point where they intersect in the \(4^{\text {th }}\) quadrant \(i\)
