85778
The two curves \(x^{3}-3 x y^{2}+2=0\) and \(3 x^{2} y-y^{3}=2\)
1 cut at right angle
2 cut at angle \(\frac{\pi}{4}\)
3 touch each other
4 cut at angle \(\frac{\pi}{3}\)
Explanation:
(A) : Angles between two curves is same as angle between their tangents So, first we will find slope of their tangents. First curve - \(x^{3}-3 x y^{2}+2=0\) Differentiating w.r.t. to \(\mathrm{x}\). \(3 x^{2}-3 y^{2}-6 x y \frac{d y}{d x}=0 \Rightarrow 3 x^{2}-3 y^{2}=6 x y \frac{d y}{d x}\) \(\frac{d y}{d x}=\frac{3 x^{2}-3 y^{2}}{6 x y} \Rightarrow \frac{d y}{d x}=\frac{x^{2}-y^{2}}{2 x y}\) Let, \(m_{1}=\frac{x^{2}-y^{2}}{2 x y}\) Second curve - \(3 x^{2} y-y^{3}-2=0\) Differentiating w.r.t to \(\mathrm{x}\). \(3 x^{2} \frac{d y}{d x}+6 x y-3 y^{2} \frac{d y}{d x}=0 \Rightarrow\left(3 x^{2}-3 y^{2}\right) \frac{d y}{d x}=-6 x y\) \(\frac{d y}{d x}=\frac{-6 x y}{3 x^{2}-3 y^{2}} \Rightarrow \frac{d y}{d x}=\frac{-2 x y}{x^{2}-y^{2}}\) Let, \(\mathrm{m}_{2}=\frac{-2 \mathrm{xy}}{\mathrm{x}^{2}-\mathrm{y}^{2}}\) Finding product of \(\mathrm{m}_{1}\) and \(\mathrm{m}_{2}\) \(\mathrm{m}_{1} \times \mathrm{m}_{2}=\frac{\mathrm{x}^{2}-\mathrm{y}^{2}}{2 \mathrm{xy}} \times \frac{-2 \mathrm{xy}}{\mathrm{x}^{2}-\mathrm{y}^{2}}\) \(=-1\) Since, product of the slopes is -1 \(\therefore\) Angle between tangents is \(90^{\circ}\). Thus curves cut each other at right angle.
VITEEE-2018
Application of Derivatives
85779
If \(\sin ^{-1} a\) is the acute angle between the curves \(x^{2}+y^{2}=4 x\) and \(x^{2}+y^{2}=8\) at \((2,2)\), then a
1 1
2 0
3 \(\frac{1}{\sqrt{2}}\)
4 \(\frac{\sqrt{3}}{2}\)
Explanation:
(C) : Given, \(\begin{align*} & x^{2}+y^{2}-4 x=0 \tag{i}\\ & x^{2}+y^{2}-8=0 \tag{ii} \end{align*}\) Differentiating (i) and (ii) with respect ' \(x\) ' we get \(2 x+2 y \frac{d y}{d x}-4=0\) \(2 y \frac{d y}{d x}=4-2 x \Rightarrow \frac{d y}{d x}=\frac{4-2 x}{2 y}\) \(m_{1}=\frac{d y}{d x_{(2,2)}}=0 \Rightarrow 2 x+2 y \frac{d y}{d x}=0\) \(\frac{d y}{d x}=\frac{-x}{y} \Rightarrow \Rightarrow m_{2}=\frac{d y}{d x_{(2,2)}}=-1\) We know that, \(\tan \mathrm{a}=\left|\frac{\mathrm{m}_{1}-\mathrm{m}_{2}}{1+\mathrm{m}_{1} \mathrm{~m}_{2}}\right|=\left|\frac{0-(-1)}{1-0 \times-1}\right|=1\) \(\tan \mathrm{a}=1\) \(\sin \mathrm{a}=\frac{1}{\sqrt{1+1}}=\frac{1}{\sqrt{2}}\) \(\Rightarrow \mathrm{a}=\sin ^{-1}\left(\frac{1}{\sqrt{2}}\right)\)
Application of Derivatives
85780
If the normal to the curve \(y=f(x)\) at the point \((3,4)\) makes an angle \(3 \pi / 4\) with the positive \(x\) axis, then \(\mathbf{f}^{\prime}(3)=\)
1 -1
2 \(-3 / 4\)
3 \(4 / 3\)
4 1
Explanation:
(D) : Given, \(\mathrm{y}=\mathrm{f}(\mathrm{x})\) \(\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{f}^{\prime}(\mathrm{x}), \quad \frac{\mathrm{dy}}{\mathrm{dx}(3,4)}=\mathrm{f}^{\prime}(3)\) Slope of normal \(=-\frac{1}{f^{\prime}(x)}\) But, \(\quad-\frac{1}{f^{\prime}(3)}=\tan \left(\frac{3 \pi}{4}\right)\) \(-\frac{1}{f^{\prime}(3)}=\tan \left(\frac{\pi}{2}+\frac{\pi}{4}\right)\) \(-\frac{1}{f^{\prime}(3)}=-1\) \(f^{\prime}(3)=1\)
BITSAT-2012
Application of Derivatives
85781
The line which is parallel to \(X\)-axis and crosses the curve \(y=\sqrt{x}\) at an angle of \(45^{\circ}, i\)
1 \(x=\frac{1}{4}\)
2 \(\mathrm{y}=\frac{1}{4}\)
3 \(\mathrm{y}=\frac{1}{2}\)
4 \(y=1\)
Explanation:
(C) : Given, Equation of a line parallel to \(\mathrm{X}\)-axis is \(\mathrm{y}=\mathrm{k}\). Equation of the curve is \(y=\sqrt{x}\), On solving equation of line with the equation of curve, we get \(x=k^{2}\) Thus the intersecting point is \(\left(\mathrm{k}^{2}, \mathrm{k}\right)\) It is given that the line \(\mathrm{y}\) \(=\mathrm{k}\) intersect the curve \(\mathrm{y}=\sqrt{\mathrm{x}}\) at an angle of \(\pi / 4\). This means that the slope of the tangent to \(\mathrm{y}=\sqrt{\mathrm{x}}\) at \(\left(\mathrm{k}^{2}, \mathrm{k}\right)\) is \(\tan \left( \pm \frac{\pi}{4}\right)= \pm 1\) \(\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{\left(\mathrm{k}^{2}, \mathrm{k}\right)}= \pm 1 \Rightarrow\left(\frac{1}{2 \sqrt{\mathrm{x}}}\right)_{\left(\mathrm{k}^{2}, \mathrm{k}\right)}= \pm 1\) \(\mathrm{k}= \pm \frac{1}{2}\) Hence, \(\mathrm{k}=\frac{1}{2} \Rightarrow \mathrm{y}=\frac{1}{2}\)
BITSAT-2015
Application of Derivatives
85782
The angle between a pair of tangents drawn from \(T\) to the circle \(x^{2}+y^{2}+4 x-6 y+9 \sin ^{2} \alpha+13 \cos ^{2} \alpha=0 \text { is } 2 \alpha \text {. }\) The equation of the locus of the point \(T\) is
1 \(x^{2}+y^{2}+4 x-6 y+4=0\)
2 \(x^{2}+y^{2}+4 x-6 y-9=0\)
3 \(x^{2}+y^{2}+4 x-6 y-4=0\)
4 \(x^{2}+y^{2}+4 x-6 y+9=0\)
Explanation:
(D) : We have, \(x^{2}+y^{2}+4 x-6 y+9 \sin ^{2} \alpha+13 \cos ^{2} \alpha=0\) Comparing the general equation of circle \(\mathrm{x}^{2}+\mathrm{y}^{2}+2 \mathrm{gx}+2 \mathrm{fy}+\mathrm{c}=0\) \(2 \mathrm{~g}=4 \quad 2 \mathrm{f}=-6 \quad \mathrm{c}=9 \sin ^{2} \alpha+13 \cos ^{2} \alpha\) \(g=2 \quad f=-3\) Center \(\equiv(-\mathrm{g},-\mathrm{f})\) \(\equiv(-2,3)\) Radius \(=\sqrt{\mathrm{g}^{2}+\mathrm{f}^{2}-\mathrm{c}}\) \(=\sqrt{4+9\left(1-\sin ^{2} \alpha\right)-13 \cos ^{2} \alpha}\) \(=\sqrt{4+9 \cos ^{2} \alpha-13 \cos ^{2} \alpha}\) \(=\sqrt{4-4 \cos ^{2} \alpha}=\sqrt{4\left(1-\cos ^{2} \alpha\right)}\) \(=2 \sin \alpha \text { unit }\) \(\mathrm{C}\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right) ; \mathrm{D}\left(\mathrm{x}_{2}, \mathrm{y}_{2}\right)\) \(\mathrm{CD}=\sqrt{\left(\mathrm{x}_{2}-\mathrm{x}_{1}\right)^{2}+\left(\mathrm{y}_{2}-\mathrm{y}_{1}\right)^{2}}\) unit \(\mathrm{OP}=\sqrt{(\mathrm{h}+2)^{2}+(\mathrm{k}-3)^{2}} \quad\) unit \(\sin \alpha=\frac{\mathrm{OA}}{\mathrm{OP}}\) \(\sin \alpha=\frac{2 \sin \alpha}{\sqrt{(h+2)^{2}+(k-3)^{2}}}\) \((h+2)^{2}+(k-3)^{2}=4\) \(h^{2}+4 h+4+k^{2}-6 k+9-4=0\) \(h^{2}+k^{2}+4 h-6 k+9=0\) So, the locus of \(\mathrm{T}\) is \(x^{2}+y^{2}+4 x-6 y+9=0\)
85778
The two curves \(x^{3}-3 x y^{2}+2=0\) and \(3 x^{2} y-y^{3}=2\)
1 cut at right angle
2 cut at angle \(\frac{\pi}{4}\)
3 touch each other
4 cut at angle \(\frac{\pi}{3}\)
Explanation:
(A) : Angles between two curves is same as angle between their tangents So, first we will find slope of their tangents. First curve - \(x^{3}-3 x y^{2}+2=0\) Differentiating w.r.t. to \(\mathrm{x}\). \(3 x^{2}-3 y^{2}-6 x y \frac{d y}{d x}=0 \Rightarrow 3 x^{2}-3 y^{2}=6 x y \frac{d y}{d x}\) \(\frac{d y}{d x}=\frac{3 x^{2}-3 y^{2}}{6 x y} \Rightarrow \frac{d y}{d x}=\frac{x^{2}-y^{2}}{2 x y}\) Let, \(m_{1}=\frac{x^{2}-y^{2}}{2 x y}\) Second curve - \(3 x^{2} y-y^{3}-2=0\) Differentiating w.r.t to \(\mathrm{x}\). \(3 x^{2} \frac{d y}{d x}+6 x y-3 y^{2} \frac{d y}{d x}=0 \Rightarrow\left(3 x^{2}-3 y^{2}\right) \frac{d y}{d x}=-6 x y\) \(\frac{d y}{d x}=\frac{-6 x y}{3 x^{2}-3 y^{2}} \Rightarrow \frac{d y}{d x}=\frac{-2 x y}{x^{2}-y^{2}}\) Let, \(\mathrm{m}_{2}=\frac{-2 \mathrm{xy}}{\mathrm{x}^{2}-\mathrm{y}^{2}}\) Finding product of \(\mathrm{m}_{1}\) and \(\mathrm{m}_{2}\) \(\mathrm{m}_{1} \times \mathrm{m}_{2}=\frac{\mathrm{x}^{2}-\mathrm{y}^{2}}{2 \mathrm{xy}} \times \frac{-2 \mathrm{xy}}{\mathrm{x}^{2}-\mathrm{y}^{2}}\) \(=-1\) Since, product of the slopes is -1 \(\therefore\) Angle between tangents is \(90^{\circ}\). Thus curves cut each other at right angle.
VITEEE-2018
Application of Derivatives
85779
If \(\sin ^{-1} a\) is the acute angle between the curves \(x^{2}+y^{2}=4 x\) and \(x^{2}+y^{2}=8\) at \((2,2)\), then a
1 1
2 0
3 \(\frac{1}{\sqrt{2}}\)
4 \(\frac{\sqrt{3}}{2}\)
Explanation:
(C) : Given, \(\begin{align*} & x^{2}+y^{2}-4 x=0 \tag{i}\\ & x^{2}+y^{2}-8=0 \tag{ii} \end{align*}\) Differentiating (i) and (ii) with respect ' \(x\) ' we get \(2 x+2 y \frac{d y}{d x}-4=0\) \(2 y \frac{d y}{d x}=4-2 x \Rightarrow \frac{d y}{d x}=\frac{4-2 x}{2 y}\) \(m_{1}=\frac{d y}{d x_{(2,2)}}=0 \Rightarrow 2 x+2 y \frac{d y}{d x}=0\) \(\frac{d y}{d x}=\frac{-x}{y} \Rightarrow \Rightarrow m_{2}=\frac{d y}{d x_{(2,2)}}=-1\) We know that, \(\tan \mathrm{a}=\left|\frac{\mathrm{m}_{1}-\mathrm{m}_{2}}{1+\mathrm{m}_{1} \mathrm{~m}_{2}}\right|=\left|\frac{0-(-1)}{1-0 \times-1}\right|=1\) \(\tan \mathrm{a}=1\) \(\sin \mathrm{a}=\frac{1}{\sqrt{1+1}}=\frac{1}{\sqrt{2}}\) \(\Rightarrow \mathrm{a}=\sin ^{-1}\left(\frac{1}{\sqrt{2}}\right)\)
Application of Derivatives
85780
If the normal to the curve \(y=f(x)\) at the point \((3,4)\) makes an angle \(3 \pi / 4\) with the positive \(x\) axis, then \(\mathbf{f}^{\prime}(3)=\)
1 -1
2 \(-3 / 4\)
3 \(4 / 3\)
4 1
Explanation:
(D) : Given, \(\mathrm{y}=\mathrm{f}(\mathrm{x})\) \(\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{f}^{\prime}(\mathrm{x}), \quad \frac{\mathrm{dy}}{\mathrm{dx}(3,4)}=\mathrm{f}^{\prime}(3)\) Slope of normal \(=-\frac{1}{f^{\prime}(x)}\) But, \(\quad-\frac{1}{f^{\prime}(3)}=\tan \left(\frac{3 \pi}{4}\right)\) \(-\frac{1}{f^{\prime}(3)}=\tan \left(\frac{\pi}{2}+\frac{\pi}{4}\right)\) \(-\frac{1}{f^{\prime}(3)}=-1\) \(f^{\prime}(3)=1\)
BITSAT-2012
Application of Derivatives
85781
The line which is parallel to \(X\)-axis and crosses the curve \(y=\sqrt{x}\) at an angle of \(45^{\circ}, i\)
1 \(x=\frac{1}{4}\)
2 \(\mathrm{y}=\frac{1}{4}\)
3 \(\mathrm{y}=\frac{1}{2}\)
4 \(y=1\)
Explanation:
(C) : Given, Equation of a line parallel to \(\mathrm{X}\)-axis is \(\mathrm{y}=\mathrm{k}\). Equation of the curve is \(y=\sqrt{x}\), On solving equation of line with the equation of curve, we get \(x=k^{2}\) Thus the intersecting point is \(\left(\mathrm{k}^{2}, \mathrm{k}\right)\) It is given that the line \(\mathrm{y}\) \(=\mathrm{k}\) intersect the curve \(\mathrm{y}=\sqrt{\mathrm{x}}\) at an angle of \(\pi / 4\). This means that the slope of the tangent to \(\mathrm{y}=\sqrt{\mathrm{x}}\) at \(\left(\mathrm{k}^{2}, \mathrm{k}\right)\) is \(\tan \left( \pm \frac{\pi}{4}\right)= \pm 1\) \(\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{\left(\mathrm{k}^{2}, \mathrm{k}\right)}= \pm 1 \Rightarrow\left(\frac{1}{2 \sqrt{\mathrm{x}}}\right)_{\left(\mathrm{k}^{2}, \mathrm{k}\right)}= \pm 1\) \(\mathrm{k}= \pm \frac{1}{2}\) Hence, \(\mathrm{k}=\frac{1}{2} \Rightarrow \mathrm{y}=\frac{1}{2}\)
BITSAT-2015
Application of Derivatives
85782
The angle between a pair of tangents drawn from \(T\) to the circle \(x^{2}+y^{2}+4 x-6 y+9 \sin ^{2} \alpha+13 \cos ^{2} \alpha=0 \text { is } 2 \alpha \text {. }\) The equation of the locus of the point \(T\) is
1 \(x^{2}+y^{2}+4 x-6 y+4=0\)
2 \(x^{2}+y^{2}+4 x-6 y-9=0\)
3 \(x^{2}+y^{2}+4 x-6 y-4=0\)
4 \(x^{2}+y^{2}+4 x-6 y+9=0\)
Explanation:
(D) : We have, \(x^{2}+y^{2}+4 x-6 y+9 \sin ^{2} \alpha+13 \cos ^{2} \alpha=0\) Comparing the general equation of circle \(\mathrm{x}^{2}+\mathrm{y}^{2}+2 \mathrm{gx}+2 \mathrm{fy}+\mathrm{c}=0\) \(2 \mathrm{~g}=4 \quad 2 \mathrm{f}=-6 \quad \mathrm{c}=9 \sin ^{2} \alpha+13 \cos ^{2} \alpha\) \(g=2 \quad f=-3\) Center \(\equiv(-\mathrm{g},-\mathrm{f})\) \(\equiv(-2,3)\) Radius \(=\sqrt{\mathrm{g}^{2}+\mathrm{f}^{2}-\mathrm{c}}\) \(=\sqrt{4+9\left(1-\sin ^{2} \alpha\right)-13 \cos ^{2} \alpha}\) \(=\sqrt{4+9 \cos ^{2} \alpha-13 \cos ^{2} \alpha}\) \(=\sqrt{4-4 \cos ^{2} \alpha}=\sqrt{4\left(1-\cos ^{2} \alpha\right)}\) \(=2 \sin \alpha \text { unit }\) \(\mathrm{C}\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right) ; \mathrm{D}\left(\mathrm{x}_{2}, \mathrm{y}_{2}\right)\) \(\mathrm{CD}=\sqrt{\left(\mathrm{x}_{2}-\mathrm{x}_{1}\right)^{2}+\left(\mathrm{y}_{2}-\mathrm{y}_{1}\right)^{2}}\) unit \(\mathrm{OP}=\sqrt{(\mathrm{h}+2)^{2}+(\mathrm{k}-3)^{2}} \quad\) unit \(\sin \alpha=\frac{\mathrm{OA}}{\mathrm{OP}}\) \(\sin \alpha=\frac{2 \sin \alpha}{\sqrt{(h+2)^{2}+(k-3)^{2}}}\) \((h+2)^{2}+(k-3)^{2}=4\) \(h^{2}+4 h+4+k^{2}-6 k+9-4=0\) \(h^{2}+k^{2}+4 h-6 k+9=0\) So, the locus of \(\mathrm{T}\) is \(x^{2}+y^{2}+4 x-6 y+9=0\)
85778
The two curves \(x^{3}-3 x y^{2}+2=0\) and \(3 x^{2} y-y^{3}=2\)
1 cut at right angle
2 cut at angle \(\frac{\pi}{4}\)
3 touch each other
4 cut at angle \(\frac{\pi}{3}\)
Explanation:
(A) : Angles between two curves is same as angle between their tangents So, first we will find slope of their tangents. First curve - \(x^{3}-3 x y^{2}+2=0\) Differentiating w.r.t. to \(\mathrm{x}\). \(3 x^{2}-3 y^{2}-6 x y \frac{d y}{d x}=0 \Rightarrow 3 x^{2}-3 y^{2}=6 x y \frac{d y}{d x}\) \(\frac{d y}{d x}=\frac{3 x^{2}-3 y^{2}}{6 x y} \Rightarrow \frac{d y}{d x}=\frac{x^{2}-y^{2}}{2 x y}\) Let, \(m_{1}=\frac{x^{2}-y^{2}}{2 x y}\) Second curve - \(3 x^{2} y-y^{3}-2=0\) Differentiating w.r.t to \(\mathrm{x}\). \(3 x^{2} \frac{d y}{d x}+6 x y-3 y^{2} \frac{d y}{d x}=0 \Rightarrow\left(3 x^{2}-3 y^{2}\right) \frac{d y}{d x}=-6 x y\) \(\frac{d y}{d x}=\frac{-6 x y}{3 x^{2}-3 y^{2}} \Rightarrow \frac{d y}{d x}=\frac{-2 x y}{x^{2}-y^{2}}\) Let, \(\mathrm{m}_{2}=\frac{-2 \mathrm{xy}}{\mathrm{x}^{2}-\mathrm{y}^{2}}\) Finding product of \(\mathrm{m}_{1}\) and \(\mathrm{m}_{2}\) \(\mathrm{m}_{1} \times \mathrm{m}_{2}=\frac{\mathrm{x}^{2}-\mathrm{y}^{2}}{2 \mathrm{xy}} \times \frac{-2 \mathrm{xy}}{\mathrm{x}^{2}-\mathrm{y}^{2}}\) \(=-1\) Since, product of the slopes is -1 \(\therefore\) Angle between tangents is \(90^{\circ}\). Thus curves cut each other at right angle.
VITEEE-2018
Application of Derivatives
85779
If \(\sin ^{-1} a\) is the acute angle between the curves \(x^{2}+y^{2}=4 x\) and \(x^{2}+y^{2}=8\) at \((2,2)\), then a
1 1
2 0
3 \(\frac{1}{\sqrt{2}}\)
4 \(\frac{\sqrt{3}}{2}\)
Explanation:
(C) : Given, \(\begin{align*} & x^{2}+y^{2}-4 x=0 \tag{i}\\ & x^{2}+y^{2}-8=0 \tag{ii} \end{align*}\) Differentiating (i) and (ii) with respect ' \(x\) ' we get \(2 x+2 y \frac{d y}{d x}-4=0\) \(2 y \frac{d y}{d x}=4-2 x \Rightarrow \frac{d y}{d x}=\frac{4-2 x}{2 y}\) \(m_{1}=\frac{d y}{d x_{(2,2)}}=0 \Rightarrow 2 x+2 y \frac{d y}{d x}=0\) \(\frac{d y}{d x}=\frac{-x}{y} \Rightarrow \Rightarrow m_{2}=\frac{d y}{d x_{(2,2)}}=-1\) We know that, \(\tan \mathrm{a}=\left|\frac{\mathrm{m}_{1}-\mathrm{m}_{2}}{1+\mathrm{m}_{1} \mathrm{~m}_{2}}\right|=\left|\frac{0-(-1)}{1-0 \times-1}\right|=1\) \(\tan \mathrm{a}=1\) \(\sin \mathrm{a}=\frac{1}{\sqrt{1+1}}=\frac{1}{\sqrt{2}}\) \(\Rightarrow \mathrm{a}=\sin ^{-1}\left(\frac{1}{\sqrt{2}}\right)\)
Application of Derivatives
85780
If the normal to the curve \(y=f(x)\) at the point \((3,4)\) makes an angle \(3 \pi / 4\) with the positive \(x\) axis, then \(\mathbf{f}^{\prime}(3)=\)
1 -1
2 \(-3 / 4\)
3 \(4 / 3\)
4 1
Explanation:
(D) : Given, \(\mathrm{y}=\mathrm{f}(\mathrm{x})\) \(\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{f}^{\prime}(\mathrm{x}), \quad \frac{\mathrm{dy}}{\mathrm{dx}(3,4)}=\mathrm{f}^{\prime}(3)\) Slope of normal \(=-\frac{1}{f^{\prime}(x)}\) But, \(\quad-\frac{1}{f^{\prime}(3)}=\tan \left(\frac{3 \pi}{4}\right)\) \(-\frac{1}{f^{\prime}(3)}=\tan \left(\frac{\pi}{2}+\frac{\pi}{4}\right)\) \(-\frac{1}{f^{\prime}(3)}=-1\) \(f^{\prime}(3)=1\)
BITSAT-2012
Application of Derivatives
85781
The line which is parallel to \(X\)-axis and crosses the curve \(y=\sqrt{x}\) at an angle of \(45^{\circ}, i\)
1 \(x=\frac{1}{4}\)
2 \(\mathrm{y}=\frac{1}{4}\)
3 \(\mathrm{y}=\frac{1}{2}\)
4 \(y=1\)
Explanation:
(C) : Given, Equation of a line parallel to \(\mathrm{X}\)-axis is \(\mathrm{y}=\mathrm{k}\). Equation of the curve is \(y=\sqrt{x}\), On solving equation of line with the equation of curve, we get \(x=k^{2}\) Thus the intersecting point is \(\left(\mathrm{k}^{2}, \mathrm{k}\right)\) It is given that the line \(\mathrm{y}\) \(=\mathrm{k}\) intersect the curve \(\mathrm{y}=\sqrt{\mathrm{x}}\) at an angle of \(\pi / 4\). This means that the slope of the tangent to \(\mathrm{y}=\sqrt{\mathrm{x}}\) at \(\left(\mathrm{k}^{2}, \mathrm{k}\right)\) is \(\tan \left( \pm \frac{\pi}{4}\right)= \pm 1\) \(\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{\left(\mathrm{k}^{2}, \mathrm{k}\right)}= \pm 1 \Rightarrow\left(\frac{1}{2 \sqrt{\mathrm{x}}}\right)_{\left(\mathrm{k}^{2}, \mathrm{k}\right)}= \pm 1\) \(\mathrm{k}= \pm \frac{1}{2}\) Hence, \(\mathrm{k}=\frac{1}{2} \Rightarrow \mathrm{y}=\frac{1}{2}\)
BITSAT-2015
Application of Derivatives
85782
The angle between a pair of tangents drawn from \(T\) to the circle \(x^{2}+y^{2}+4 x-6 y+9 \sin ^{2} \alpha+13 \cos ^{2} \alpha=0 \text { is } 2 \alpha \text {. }\) The equation of the locus of the point \(T\) is
1 \(x^{2}+y^{2}+4 x-6 y+4=0\)
2 \(x^{2}+y^{2}+4 x-6 y-9=0\)
3 \(x^{2}+y^{2}+4 x-6 y-4=0\)
4 \(x^{2}+y^{2}+4 x-6 y+9=0\)
Explanation:
(D) : We have, \(x^{2}+y^{2}+4 x-6 y+9 \sin ^{2} \alpha+13 \cos ^{2} \alpha=0\) Comparing the general equation of circle \(\mathrm{x}^{2}+\mathrm{y}^{2}+2 \mathrm{gx}+2 \mathrm{fy}+\mathrm{c}=0\) \(2 \mathrm{~g}=4 \quad 2 \mathrm{f}=-6 \quad \mathrm{c}=9 \sin ^{2} \alpha+13 \cos ^{2} \alpha\) \(g=2 \quad f=-3\) Center \(\equiv(-\mathrm{g},-\mathrm{f})\) \(\equiv(-2,3)\) Radius \(=\sqrt{\mathrm{g}^{2}+\mathrm{f}^{2}-\mathrm{c}}\) \(=\sqrt{4+9\left(1-\sin ^{2} \alpha\right)-13 \cos ^{2} \alpha}\) \(=\sqrt{4+9 \cos ^{2} \alpha-13 \cos ^{2} \alpha}\) \(=\sqrt{4-4 \cos ^{2} \alpha}=\sqrt{4\left(1-\cos ^{2} \alpha\right)}\) \(=2 \sin \alpha \text { unit }\) \(\mathrm{C}\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right) ; \mathrm{D}\left(\mathrm{x}_{2}, \mathrm{y}_{2}\right)\) \(\mathrm{CD}=\sqrt{\left(\mathrm{x}_{2}-\mathrm{x}_{1}\right)^{2}+\left(\mathrm{y}_{2}-\mathrm{y}_{1}\right)^{2}}\) unit \(\mathrm{OP}=\sqrt{(\mathrm{h}+2)^{2}+(\mathrm{k}-3)^{2}} \quad\) unit \(\sin \alpha=\frac{\mathrm{OA}}{\mathrm{OP}}\) \(\sin \alpha=\frac{2 \sin \alpha}{\sqrt{(h+2)^{2}+(k-3)^{2}}}\) \((h+2)^{2}+(k-3)^{2}=4\) \(h^{2}+4 h+4+k^{2}-6 k+9-4=0\) \(h^{2}+k^{2}+4 h-6 k+9=0\) So, the locus of \(\mathrm{T}\) is \(x^{2}+y^{2}+4 x-6 y+9=0\)
NEET Test Series from KOTA - 10 Papers In MS WORD
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Application of Derivatives
85778
The two curves \(x^{3}-3 x y^{2}+2=0\) and \(3 x^{2} y-y^{3}=2\)
1 cut at right angle
2 cut at angle \(\frac{\pi}{4}\)
3 touch each other
4 cut at angle \(\frac{\pi}{3}\)
Explanation:
(A) : Angles between two curves is same as angle between their tangents So, first we will find slope of their tangents. First curve - \(x^{3}-3 x y^{2}+2=0\) Differentiating w.r.t. to \(\mathrm{x}\). \(3 x^{2}-3 y^{2}-6 x y \frac{d y}{d x}=0 \Rightarrow 3 x^{2}-3 y^{2}=6 x y \frac{d y}{d x}\) \(\frac{d y}{d x}=\frac{3 x^{2}-3 y^{2}}{6 x y} \Rightarrow \frac{d y}{d x}=\frac{x^{2}-y^{2}}{2 x y}\) Let, \(m_{1}=\frac{x^{2}-y^{2}}{2 x y}\) Second curve - \(3 x^{2} y-y^{3}-2=0\) Differentiating w.r.t to \(\mathrm{x}\). \(3 x^{2} \frac{d y}{d x}+6 x y-3 y^{2} \frac{d y}{d x}=0 \Rightarrow\left(3 x^{2}-3 y^{2}\right) \frac{d y}{d x}=-6 x y\) \(\frac{d y}{d x}=\frac{-6 x y}{3 x^{2}-3 y^{2}} \Rightarrow \frac{d y}{d x}=\frac{-2 x y}{x^{2}-y^{2}}\) Let, \(\mathrm{m}_{2}=\frac{-2 \mathrm{xy}}{\mathrm{x}^{2}-\mathrm{y}^{2}}\) Finding product of \(\mathrm{m}_{1}\) and \(\mathrm{m}_{2}\) \(\mathrm{m}_{1} \times \mathrm{m}_{2}=\frac{\mathrm{x}^{2}-\mathrm{y}^{2}}{2 \mathrm{xy}} \times \frac{-2 \mathrm{xy}}{\mathrm{x}^{2}-\mathrm{y}^{2}}\) \(=-1\) Since, product of the slopes is -1 \(\therefore\) Angle between tangents is \(90^{\circ}\). Thus curves cut each other at right angle.
VITEEE-2018
Application of Derivatives
85779
If \(\sin ^{-1} a\) is the acute angle between the curves \(x^{2}+y^{2}=4 x\) and \(x^{2}+y^{2}=8\) at \((2,2)\), then a
1 1
2 0
3 \(\frac{1}{\sqrt{2}}\)
4 \(\frac{\sqrt{3}}{2}\)
Explanation:
(C) : Given, \(\begin{align*} & x^{2}+y^{2}-4 x=0 \tag{i}\\ & x^{2}+y^{2}-8=0 \tag{ii} \end{align*}\) Differentiating (i) and (ii) with respect ' \(x\) ' we get \(2 x+2 y \frac{d y}{d x}-4=0\) \(2 y \frac{d y}{d x}=4-2 x \Rightarrow \frac{d y}{d x}=\frac{4-2 x}{2 y}\) \(m_{1}=\frac{d y}{d x_{(2,2)}}=0 \Rightarrow 2 x+2 y \frac{d y}{d x}=0\) \(\frac{d y}{d x}=\frac{-x}{y} \Rightarrow \Rightarrow m_{2}=\frac{d y}{d x_{(2,2)}}=-1\) We know that, \(\tan \mathrm{a}=\left|\frac{\mathrm{m}_{1}-\mathrm{m}_{2}}{1+\mathrm{m}_{1} \mathrm{~m}_{2}}\right|=\left|\frac{0-(-1)}{1-0 \times-1}\right|=1\) \(\tan \mathrm{a}=1\) \(\sin \mathrm{a}=\frac{1}{\sqrt{1+1}}=\frac{1}{\sqrt{2}}\) \(\Rightarrow \mathrm{a}=\sin ^{-1}\left(\frac{1}{\sqrt{2}}\right)\)
Application of Derivatives
85780
If the normal to the curve \(y=f(x)\) at the point \((3,4)\) makes an angle \(3 \pi / 4\) with the positive \(x\) axis, then \(\mathbf{f}^{\prime}(3)=\)
1 -1
2 \(-3 / 4\)
3 \(4 / 3\)
4 1
Explanation:
(D) : Given, \(\mathrm{y}=\mathrm{f}(\mathrm{x})\) \(\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{f}^{\prime}(\mathrm{x}), \quad \frac{\mathrm{dy}}{\mathrm{dx}(3,4)}=\mathrm{f}^{\prime}(3)\) Slope of normal \(=-\frac{1}{f^{\prime}(x)}\) But, \(\quad-\frac{1}{f^{\prime}(3)}=\tan \left(\frac{3 \pi}{4}\right)\) \(-\frac{1}{f^{\prime}(3)}=\tan \left(\frac{\pi}{2}+\frac{\pi}{4}\right)\) \(-\frac{1}{f^{\prime}(3)}=-1\) \(f^{\prime}(3)=1\)
BITSAT-2012
Application of Derivatives
85781
The line which is parallel to \(X\)-axis and crosses the curve \(y=\sqrt{x}\) at an angle of \(45^{\circ}, i\)
1 \(x=\frac{1}{4}\)
2 \(\mathrm{y}=\frac{1}{4}\)
3 \(\mathrm{y}=\frac{1}{2}\)
4 \(y=1\)
Explanation:
(C) : Given, Equation of a line parallel to \(\mathrm{X}\)-axis is \(\mathrm{y}=\mathrm{k}\). Equation of the curve is \(y=\sqrt{x}\), On solving equation of line with the equation of curve, we get \(x=k^{2}\) Thus the intersecting point is \(\left(\mathrm{k}^{2}, \mathrm{k}\right)\) It is given that the line \(\mathrm{y}\) \(=\mathrm{k}\) intersect the curve \(\mathrm{y}=\sqrt{\mathrm{x}}\) at an angle of \(\pi / 4\). This means that the slope of the tangent to \(\mathrm{y}=\sqrt{\mathrm{x}}\) at \(\left(\mathrm{k}^{2}, \mathrm{k}\right)\) is \(\tan \left( \pm \frac{\pi}{4}\right)= \pm 1\) \(\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{\left(\mathrm{k}^{2}, \mathrm{k}\right)}= \pm 1 \Rightarrow\left(\frac{1}{2 \sqrt{\mathrm{x}}}\right)_{\left(\mathrm{k}^{2}, \mathrm{k}\right)}= \pm 1\) \(\mathrm{k}= \pm \frac{1}{2}\) Hence, \(\mathrm{k}=\frac{1}{2} \Rightarrow \mathrm{y}=\frac{1}{2}\)
BITSAT-2015
Application of Derivatives
85782
The angle between a pair of tangents drawn from \(T\) to the circle \(x^{2}+y^{2}+4 x-6 y+9 \sin ^{2} \alpha+13 \cos ^{2} \alpha=0 \text { is } 2 \alpha \text {. }\) The equation of the locus of the point \(T\) is
1 \(x^{2}+y^{2}+4 x-6 y+4=0\)
2 \(x^{2}+y^{2}+4 x-6 y-9=0\)
3 \(x^{2}+y^{2}+4 x-6 y-4=0\)
4 \(x^{2}+y^{2}+4 x-6 y+9=0\)
Explanation:
(D) : We have, \(x^{2}+y^{2}+4 x-6 y+9 \sin ^{2} \alpha+13 \cos ^{2} \alpha=0\) Comparing the general equation of circle \(\mathrm{x}^{2}+\mathrm{y}^{2}+2 \mathrm{gx}+2 \mathrm{fy}+\mathrm{c}=0\) \(2 \mathrm{~g}=4 \quad 2 \mathrm{f}=-6 \quad \mathrm{c}=9 \sin ^{2} \alpha+13 \cos ^{2} \alpha\) \(g=2 \quad f=-3\) Center \(\equiv(-\mathrm{g},-\mathrm{f})\) \(\equiv(-2,3)\) Radius \(=\sqrt{\mathrm{g}^{2}+\mathrm{f}^{2}-\mathrm{c}}\) \(=\sqrt{4+9\left(1-\sin ^{2} \alpha\right)-13 \cos ^{2} \alpha}\) \(=\sqrt{4+9 \cos ^{2} \alpha-13 \cos ^{2} \alpha}\) \(=\sqrt{4-4 \cos ^{2} \alpha}=\sqrt{4\left(1-\cos ^{2} \alpha\right)}\) \(=2 \sin \alpha \text { unit }\) \(\mathrm{C}\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right) ; \mathrm{D}\left(\mathrm{x}_{2}, \mathrm{y}_{2}\right)\) \(\mathrm{CD}=\sqrt{\left(\mathrm{x}_{2}-\mathrm{x}_{1}\right)^{2}+\left(\mathrm{y}_{2}-\mathrm{y}_{1}\right)^{2}}\) unit \(\mathrm{OP}=\sqrt{(\mathrm{h}+2)^{2}+(\mathrm{k}-3)^{2}} \quad\) unit \(\sin \alpha=\frac{\mathrm{OA}}{\mathrm{OP}}\) \(\sin \alpha=\frac{2 \sin \alpha}{\sqrt{(h+2)^{2}+(k-3)^{2}}}\) \((h+2)^{2}+(k-3)^{2}=4\) \(h^{2}+4 h+4+k^{2}-6 k+9-4=0\) \(h^{2}+k^{2}+4 h-6 k+9=0\) So, the locus of \(\mathrm{T}\) is \(x^{2}+y^{2}+4 x-6 y+9=0\)
85778
The two curves \(x^{3}-3 x y^{2}+2=0\) and \(3 x^{2} y-y^{3}=2\)
1 cut at right angle
2 cut at angle \(\frac{\pi}{4}\)
3 touch each other
4 cut at angle \(\frac{\pi}{3}\)
Explanation:
(A) : Angles between two curves is same as angle between their tangents So, first we will find slope of their tangents. First curve - \(x^{3}-3 x y^{2}+2=0\) Differentiating w.r.t. to \(\mathrm{x}\). \(3 x^{2}-3 y^{2}-6 x y \frac{d y}{d x}=0 \Rightarrow 3 x^{2}-3 y^{2}=6 x y \frac{d y}{d x}\) \(\frac{d y}{d x}=\frac{3 x^{2}-3 y^{2}}{6 x y} \Rightarrow \frac{d y}{d x}=\frac{x^{2}-y^{2}}{2 x y}\) Let, \(m_{1}=\frac{x^{2}-y^{2}}{2 x y}\) Second curve - \(3 x^{2} y-y^{3}-2=0\) Differentiating w.r.t to \(\mathrm{x}\). \(3 x^{2} \frac{d y}{d x}+6 x y-3 y^{2} \frac{d y}{d x}=0 \Rightarrow\left(3 x^{2}-3 y^{2}\right) \frac{d y}{d x}=-6 x y\) \(\frac{d y}{d x}=\frac{-6 x y}{3 x^{2}-3 y^{2}} \Rightarrow \frac{d y}{d x}=\frac{-2 x y}{x^{2}-y^{2}}\) Let, \(\mathrm{m}_{2}=\frac{-2 \mathrm{xy}}{\mathrm{x}^{2}-\mathrm{y}^{2}}\) Finding product of \(\mathrm{m}_{1}\) and \(\mathrm{m}_{2}\) \(\mathrm{m}_{1} \times \mathrm{m}_{2}=\frac{\mathrm{x}^{2}-\mathrm{y}^{2}}{2 \mathrm{xy}} \times \frac{-2 \mathrm{xy}}{\mathrm{x}^{2}-\mathrm{y}^{2}}\) \(=-1\) Since, product of the slopes is -1 \(\therefore\) Angle between tangents is \(90^{\circ}\). Thus curves cut each other at right angle.
VITEEE-2018
Application of Derivatives
85779
If \(\sin ^{-1} a\) is the acute angle between the curves \(x^{2}+y^{2}=4 x\) and \(x^{2}+y^{2}=8\) at \((2,2)\), then a
1 1
2 0
3 \(\frac{1}{\sqrt{2}}\)
4 \(\frac{\sqrt{3}}{2}\)
Explanation:
(C) : Given, \(\begin{align*} & x^{2}+y^{2}-4 x=0 \tag{i}\\ & x^{2}+y^{2}-8=0 \tag{ii} \end{align*}\) Differentiating (i) and (ii) with respect ' \(x\) ' we get \(2 x+2 y \frac{d y}{d x}-4=0\) \(2 y \frac{d y}{d x}=4-2 x \Rightarrow \frac{d y}{d x}=\frac{4-2 x}{2 y}\) \(m_{1}=\frac{d y}{d x_{(2,2)}}=0 \Rightarrow 2 x+2 y \frac{d y}{d x}=0\) \(\frac{d y}{d x}=\frac{-x}{y} \Rightarrow \Rightarrow m_{2}=\frac{d y}{d x_{(2,2)}}=-1\) We know that, \(\tan \mathrm{a}=\left|\frac{\mathrm{m}_{1}-\mathrm{m}_{2}}{1+\mathrm{m}_{1} \mathrm{~m}_{2}}\right|=\left|\frac{0-(-1)}{1-0 \times-1}\right|=1\) \(\tan \mathrm{a}=1\) \(\sin \mathrm{a}=\frac{1}{\sqrt{1+1}}=\frac{1}{\sqrt{2}}\) \(\Rightarrow \mathrm{a}=\sin ^{-1}\left(\frac{1}{\sqrt{2}}\right)\)
Application of Derivatives
85780
If the normal to the curve \(y=f(x)\) at the point \((3,4)\) makes an angle \(3 \pi / 4\) with the positive \(x\) axis, then \(\mathbf{f}^{\prime}(3)=\)
1 -1
2 \(-3 / 4\)
3 \(4 / 3\)
4 1
Explanation:
(D) : Given, \(\mathrm{y}=\mathrm{f}(\mathrm{x})\) \(\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{f}^{\prime}(\mathrm{x}), \quad \frac{\mathrm{dy}}{\mathrm{dx}(3,4)}=\mathrm{f}^{\prime}(3)\) Slope of normal \(=-\frac{1}{f^{\prime}(x)}\) But, \(\quad-\frac{1}{f^{\prime}(3)}=\tan \left(\frac{3 \pi}{4}\right)\) \(-\frac{1}{f^{\prime}(3)}=\tan \left(\frac{\pi}{2}+\frac{\pi}{4}\right)\) \(-\frac{1}{f^{\prime}(3)}=-1\) \(f^{\prime}(3)=1\)
BITSAT-2012
Application of Derivatives
85781
The line which is parallel to \(X\)-axis and crosses the curve \(y=\sqrt{x}\) at an angle of \(45^{\circ}, i\)
1 \(x=\frac{1}{4}\)
2 \(\mathrm{y}=\frac{1}{4}\)
3 \(\mathrm{y}=\frac{1}{2}\)
4 \(y=1\)
Explanation:
(C) : Given, Equation of a line parallel to \(\mathrm{X}\)-axis is \(\mathrm{y}=\mathrm{k}\). Equation of the curve is \(y=\sqrt{x}\), On solving equation of line with the equation of curve, we get \(x=k^{2}\) Thus the intersecting point is \(\left(\mathrm{k}^{2}, \mathrm{k}\right)\) It is given that the line \(\mathrm{y}\) \(=\mathrm{k}\) intersect the curve \(\mathrm{y}=\sqrt{\mathrm{x}}\) at an angle of \(\pi / 4\). This means that the slope of the tangent to \(\mathrm{y}=\sqrt{\mathrm{x}}\) at \(\left(\mathrm{k}^{2}, \mathrm{k}\right)\) is \(\tan \left( \pm \frac{\pi}{4}\right)= \pm 1\) \(\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{\left(\mathrm{k}^{2}, \mathrm{k}\right)}= \pm 1 \Rightarrow\left(\frac{1}{2 \sqrt{\mathrm{x}}}\right)_{\left(\mathrm{k}^{2}, \mathrm{k}\right)}= \pm 1\) \(\mathrm{k}= \pm \frac{1}{2}\) Hence, \(\mathrm{k}=\frac{1}{2} \Rightarrow \mathrm{y}=\frac{1}{2}\)
BITSAT-2015
Application of Derivatives
85782
The angle between a pair of tangents drawn from \(T\) to the circle \(x^{2}+y^{2}+4 x-6 y+9 \sin ^{2} \alpha+13 \cos ^{2} \alpha=0 \text { is } 2 \alpha \text {. }\) The equation of the locus of the point \(T\) is
1 \(x^{2}+y^{2}+4 x-6 y+4=0\)
2 \(x^{2}+y^{2}+4 x-6 y-9=0\)
3 \(x^{2}+y^{2}+4 x-6 y-4=0\)
4 \(x^{2}+y^{2}+4 x-6 y+9=0\)
Explanation:
(D) : We have, \(x^{2}+y^{2}+4 x-6 y+9 \sin ^{2} \alpha+13 \cos ^{2} \alpha=0\) Comparing the general equation of circle \(\mathrm{x}^{2}+\mathrm{y}^{2}+2 \mathrm{gx}+2 \mathrm{fy}+\mathrm{c}=0\) \(2 \mathrm{~g}=4 \quad 2 \mathrm{f}=-6 \quad \mathrm{c}=9 \sin ^{2} \alpha+13 \cos ^{2} \alpha\) \(g=2 \quad f=-3\) Center \(\equiv(-\mathrm{g},-\mathrm{f})\) \(\equiv(-2,3)\) Radius \(=\sqrt{\mathrm{g}^{2}+\mathrm{f}^{2}-\mathrm{c}}\) \(=\sqrt{4+9\left(1-\sin ^{2} \alpha\right)-13 \cos ^{2} \alpha}\) \(=\sqrt{4+9 \cos ^{2} \alpha-13 \cos ^{2} \alpha}\) \(=\sqrt{4-4 \cos ^{2} \alpha}=\sqrt{4\left(1-\cos ^{2} \alpha\right)}\) \(=2 \sin \alpha \text { unit }\) \(\mathrm{C}\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right) ; \mathrm{D}\left(\mathrm{x}_{2}, \mathrm{y}_{2}\right)\) \(\mathrm{CD}=\sqrt{\left(\mathrm{x}_{2}-\mathrm{x}_{1}\right)^{2}+\left(\mathrm{y}_{2}-\mathrm{y}_{1}\right)^{2}}\) unit \(\mathrm{OP}=\sqrt{(\mathrm{h}+2)^{2}+(\mathrm{k}-3)^{2}} \quad\) unit \(\sin \alpha=\frac{\mathrm{OA}}{\mathrm{OP}}\) \(\sin \alpha=\frac{2 \sin \alpha}{\sqrt{(h+2)^{2}+(k-3)^{2}}}\) \((h+2)^{2}+(k-3)^{2}=4\) \(h^{2}+4 h+4+k^{2}-6 k+9-4=0\) \(h^{2}+k^{2}+4 h-6 k+9=0\) So, the locus of \(\mathrm{T}\) is \(x^{2}+y^{2}+4 x-6 y+9=0\)
