85414
The slope of the tangent to the hyperbola \(2 x^{2}-3 y^{2}=6\) at \((3,2) i\)
1 -1
2 1
3 0
4 2
Explanation:
(B) : Given, \(2 x^{2}-3 y^{2}=6\) On differentiating both sides w.r. to \(x\), we get - \(4 x-6 y \frac{d y}{d x}=0 \Rightarrow \therefore \frac{d y}{d x}=\frac{2 x}{3 y}\) The slope of the tangent at point \((3,2)\), \(\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{(3,2)}=\frac{2(3)}{3(2)}=\frac{6}{6}\) \(\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{(3,2)}=1\)
BITSAT-2010
Application of Derivatives
85415
The minimum value of the function \(y=x^{4}-2 x^{2}+1\) in the interval \(\left[\frac{1}{2}, 2\right] i\)
1 0
2 2
3 8
4 9
Explanation:
(A) : Given \(y=x^{4}-2 x^{2}+1\) \(\frac{d y}{d x}=\frac{d}{d x}\left(x^{4}-2 x^{2}+1\right)=4 x\left(x^{2}-1\right)\) For maximum or minimum, \(\frac{\mathrm{dy}}{\mathrm{dx}}=0\) \(4 x\left(x^{2}-1\right)=0, \text { either } x=0 \text { or } x= \pm 1\) \(\mathrm{x}=0\) and \(\mathrm{x}=-1\) does not belong to \(\left[\frac{1}{2}, 2\right]\) \(\frac{d^{2} y}{d x^{2}}=12 x^{2}-4\) \(\therefore \quad\left(\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}\right)_{\mathrm{x}=1}=12(1)^{2}-4=8>0\) \(\therefore\) There is minimum value of function at \(\mathrm{x}=1\) \(\therefore\) minimum value is \(y(1)=1^{4}-2(1)^{2}+1=1-2+1=0\)
BITSAT-2009
Application of Derivatives
85416
The slope of the tangent to the curve \(y=e^{x} \cos\) \(x\) is minimum at \(x=\alpha, 0 \leq a \leq 2 \pi\), then the value of \(\alpha \mathbf{i}\)
1 0
2 \(\pi\)
3 \(2 \pi\)
4 \(3 \pi / 2\)
Explanation:
(B) : Let \(\mathrm{m}\) be the slope of the tangent to the curve: \(y=e^{x} \cos x\). Then, \(\mathrm{m}=\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{e}^{\mathrm{x}}(\cos \mathrm{x}-\sin \mathrm{x})\) On differentiation both sides w.r.t. x, we get- \(\frac{d m}{d x}=e^{x}(\cos x-\sin x)+e^{x}(-\sin x-\cos x)=-2 e^{x} \sin x\) Again on differentiation both sides w.r.t. \(x\), we get- \(\frac{\mathrm{d}^{2} \mathrm{~m}}{\mathrm{dx}^{2}}=-2 \mathrm{e}^{\mathrm{x}}(\sin \mathrm{x}+\cos \mathrm{x})\) \(\frac{\mathrm{dm}}{\mathrm{dx}}=0\) Put, \(\quad \sin x=0\) \(\mathrm{x}=0, \pi, 2 \pi\) Clearly, \(\frac{\mathrm{d}^{2} \mathrm{~m}}{\mathrm{dx}^{2}}>0\) for \(\mathrm{x}=\pi\) Thus, \(\mathrm{y}\) is minimum at \(\mathrm{x}=\pi\). Hence the value of \(\alpha=\pi\).
BITSAT-2015
Application of Derivatives
85417
Tangents are drawn from the origin to the curve \(y=\cos x\). Their points of contact lie on
1 \(x^{2} y^{2}=y^{2}-x^{2}\)
2 \(x^{2} y^{2}=x^{2}+y^{2}\)
3 \(x^{2} y^{2}=x^{2}-y^{2}\)
4 None of these
Explanation:
(C) : Let (x, y) be one of the points of contact. Then, the equation of the tangent at \(\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)\) is \(y=\cos x\) On differentiating both side w.r.t.x, we get- \(\frac{d y}{d x}=-\sin x \Rightarrow\left(\frac{d y}{d x}\right)_{\left(x_{1} y_{1}\right)}=-\sin x_{1}\) Equation of tangent at point \(\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)\) is- \(\mathrm{y}-\mathrm{y}_{1}=\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)}\left(\mathrm{x}-\mathrm{x}_{1}\right)\) \(\mathrm{y}-\mathrm{y}_{1}=-\sin \mathrm{x}_{1}\left(\mathrm{x}-\mathrm{x}_{1}\right)\) Which passes through the origin. \(\therefore \quad 0-\mathrm{y}_{1}=-\sin \mathrm{x}_{1}\left(0-\mathrm{x}_{1}\right)\) \(\mathrm{y}_{1}=-\mathrm{x}_{1} \sin \mathrm{x}_{1}\) \(\frac{\mathrm{y}_{1}}{\mathrm{x}_{1}}=-\sin \mathrm{x}_{1}\) Also, point \(\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)\) lies on \(\mathrm{y}=\cos \mathrm{x}\). \(\therefore \quad \mathrm{y}_{1}=\cos \mathrm{x}_{1}\) From equation (i) and (ii), we get- \(\sin ^{2} x_{1}+\cos ^{2} x_{1}=\frac{y_{1}^{2}}{x_{1}^{2}}+y_{1}^{2}=1\) \(x_{1}^{2}=y_{1}^{2}+y_{1}^{2} x_{1}^{2}\) Hence, the locus of \(\left(x_{1}, y_{1}\right)\) is \(\mathrm{x}^{2}=\mathrm{y}^{2}+\mathrm{y}^{2} \mathrm{x}^{2}\) \(x^{2} y^{2}=x^{2}-y^{2}\)
85414
The slope of the tangent to the hyperbola \(2 x^{2}-3 y^{2}=6\) at \((3,2) i\)
1 -1
2 1
3 0
4 2
Explanation:
(B) : Given, \(2 x^{2}-3 y^{2}=6\) On differentiating both sides w.r. to \(x\), we get - \(4 x-6 y \frac{d y}{d x}=0 \Rightarrow \therefore \frac{d y}{d x}=\frac{2 x}{3 y}\) The slope of the tangent at point \((3,2)\), \(\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{(3,2)}=\frac{2(3)}{3(2)}=\frac{6}{6}\) \(\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{(3,2)}=1\)
BITSAT-2010
Application of Derivatives
85415
The minimum value of the function \(y=x^{4}-2 x^{2}+1\) in the interval \(\left[\frac{1}{2}, 2\right] i\)
1 0
2 2
3 8
4 9
Explanation:
(A) : Given \(y=x^{4}-2 x^{2}+1\) \(\frac{d y}{d x}=\frac{d}{d x}\left(x^{4}-2 x^{2}+1\right)=4 x\left(x^{2}-1\right)\) For maximum or minimum, \(\frac{\mathrm{dy}}{\mathrm{dx}}=0\) \(4 x\left(x^{2}-1\right)=0, \text { either } x=0 \text { or } x= \pm 1\) \(\mathrm{x}=0\) and \(\mathrm{x}=-1\) does not belong to \(\left[\frac{1}{2}, 2\right]\) \(\frac{d^{2} y}{d x^{2}}=12 x^{2}-4\) \(\therefore \quad\left(\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}\right)_{\mathrm{x}=1}=12(1)^{2}-4=8>0\) \(\therefore\) There is minimum value of function at \(\mathrm{x}=1\) \(\therefore\) minimum value is \(y(1)=1^{4}-2(1)^{2}+1=1-2+1=0\)
BITSAT-2009
Application of Derivatives
85416
The slope of the tangent to the curve \(y=e^{x} \cos\) \(x\) is minimum at \(x=\alpha, 0 \leq a \leq 2 \pi\), then the value of \(\alpha \mathbf{i}\)
1 0
2 \(\pi\)
3 \(2 \pi\)
4 \(3 \pi / 2\)
Explanation:
(B) : Let \(\mathrm{m}\) be the slope of the tangent to the curve: \(y=e^{x} \cos x\). Then, \(\mathrm{m}=\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{e}^{\mathrm{x}}(\cos \mathrm{x}-\sin \mathrm{x})\) On differentiation both sides w.r.t. x, we get- \(\frac{d m}{d x}=e^{x}(\cos x-\sin x)+e^{x}(-\sin x-\cos x)=-2 e^{x} \sin x\) Again on differentiation both sides w.r.t. \(x\), we get- \(\frac{\mathrm{d}^{2} \mathrm{~m}}{\mathrm{dx}^{2}}=-2 \mathrm{e}^{\mathrm{x}}(\sin \mathrm{x}+\cos \mathrm{x})\) \(\frac{\mathrm{dm}}{\mathrm{dx}}=0\) Put, \(\quad \sin x=0\) \(\mathrm{x}=0, \pi, 2 \pi\) Clearly, \(\frac{\mathrm{d}^{2} \mathrm{~m}}{\mathrm{dx}^{2}}>0\) for \(\mathrm{x}=\pi\) Thus, \(\mathrm{y}\) is minimum at \(\mathrm{x}=\pi\). Hence the value of \(\alpha=\pi\).
BITSAT-2015
Application of Derivatives
85417
Tangents are drawn from the origin to the curve \(y=\cos x\). Their points of contact lie on
1 \(x^{2} y^{2}=y^{2}-x^{2}\)
2 \(x^{2} y^{2}=x^{2}+y^{2}\)
3 \(x^{2} y^{2}=x^{2}-y^{2}\)
4 None of these
Explanation:
(C) : Let (x, y) be one of the points of contact. Then, the equation of the tangent at \(\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)\) is \(y=\cos x\) On differentiating both side w.r.t.x, we get- \(\frac{d y}{d x}=-\sin x \Rightarrow\left(\frac{d y}{d x}\right)_{\left(x_{1} y_{1}\right)}=-\sin x_{1}\) Equation of tangent at point \(\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)\) is- \(\mathrm{y}-\mathrm{y}_{1}=\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)}\left(\mathrm{x}-\mathrm{x}_{1}\right)\) \(\mathrm{y}-\mathrm{y}_{1}=-\sin \mathrm{x}_{1}\left(\mathrm{x}-\mathrm{x}_{1}\right)\) Which passes through the origin. \(\therefore \quad 0-\mathrm{y}_{1}=-\sin \mathrm{x}_{1}\left(0-\mathrm{x}_{1}\right)\) \(\mathrm{y}_{1}=-\mathrm{x}_{1} \sin \mathrm{x}_{1}\) \(\frac{\mathrm{y}_{1}}{\mathrm{x}_{1}}=-\sin \mathrm{x}_{1}\) Also, point \(\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)\) lies on \(\mathrm{y}=\cos \mathrm{x}\). \(\therefore \quad \mathrm{y}_{1}=\cos \mathrm{x}_{1}\) From equation (i) and (ii), we get- \(\sin ^{2} x_{1}+\cos ^{2} x_{1}=\frac{y_{1}^{2}}{x_{1}^{2}}+y_{1}^{2}=1\) \(x_{1}^{2}=y_{1}^{2}+y_{1}^{2} x_{1}^{2}\) Hence, the locus of \(\left(x_{1}, y_{1}\right)\) is \(\mathrm{x}^{2}=\mathrm{y}^{2}+\mathrm{y}^{2} \mathrm{x}^{2}\) \(x^{2} y^{2}=x^{2}-y^{2}\)
85414
The slope of the tangent to the hyperbola \(2 x^{2}-3 y^{2}=6\) at \((3,2) i\)
1 -1
2 1
3 0
4 2
Explanation:
(B) : Given, \(2 x^{2}-3 y^{2}=6\) On differentiating both sides w.r. to \(x\), we get - \(4 x-6 y \frac{d y}{d x}=0 \Rightarrow \therefore \frac{d y}{d x}=\frac{2 x}{3 y}\) The slope of the tangent at point \((3,2)\), \(\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{(3,2)}=\frac{2(3)}{3(2)}=\frac{6}{6}\) \(\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{(3,2)}=1\)
BITSAT-2010
Application of Derivatives
85415
The minimum value of the function \(y=x^{4}-2 x^{2}+1\) in the interval \(\left[\frac{1}{2}, 2\right] i\)
1 0
2 2
3 8
4 9
Explanation:
(A) : Given \(y=x^{4}-2 x^{2}+1\) \(\frac{d y}{d x}=\frac{d}{d x}\left(x^{4}-2 x^{2}+1\right)=4 x\left(x^{2}-1\right)\) For maximum or minimum, \(\frac{\mathrm{dy}}{\mathrm{dx}}=0\) \(4 x\left(x^{2}-1\right)=0, \text { either } x=0 \text { or } x= \pm 1\) \(\mathrm{x}=0\) and \(\mathrm{x}=-1\) does not belong to \(\left[\frac{1}{2}, 2\right]\) \(\frac{d^{2} y}{d x^{2}}=12 x^{2}-4\) \(\therefore \quad\left(\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}\right)_{\mathrm{x}=1}=12(1)^{2}-4=8>0\) \(\therefore\) There is minimum value of function at \(\mathrm{x}=1\) \(\therefore\) minimum value is \(y(1)=1^{4}-2(1)^{2}+1=1-2+1=0\)
BITSAT-2009
Application of Derivatives
85416
The slope of the tangent to the curve \(y=e^{x} \cos\) \(x\) is minimum at \(x=\alpha, 0 \leq a \leq 2 \pi\), then the value of \(\alpha \mathbf{i}\)
1 0
2 \(\pi\)
3 \(2 \pi\)
4 \(3 \pi / 2\)
Explanation:
(B) : Let \(\mathrm{m}\) be the slope of the tangent to the curve: \(y=e^{x} \cos x\). Then, \(\mathrm{m}=\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{e}^{\mathrm{x}}(\cos \mathrm{x}-\sin \mathrm{x})\) On differentiation both sides w.r.t. x, we get- \(\frac{d m}{d x}=e^{x}(\cos x-\sin x)+e^{x}(-\sin x-\cos x)=-2 e^{x} \sin x\) Again on differentiation both sides w.r.t. \(x\), we get- \(\frac{\mathrm{d}^{2} \mathrm{~m}}{\mathrm{dx}^{2}}=-2 \mathrm{e}^{\mathrm{x}}(\sin \mathrm{x}+\cos \mathrm{x})\) \(\frac{\mathrm{dm}}{\mathrm{dx}}=0\) Put, \(\quad \sin x=0\) \(\mathrm{x}=0, \pi, 2 \pi\) Clearly, \(\frac{\mathrm{d}^{2} \mathrm{~m}}{\mathrm{dx}^{2}}>0\) for \(\mathrm{x}=\pi\) Thus, \(\mathrm{y}\) is minimum at \(\mathrm{x}=\pi\). Hence the value of \(\alpha=\pi\).
BITSAT-2015
Application of Derivatives
85417
Tangents are drawn from the origin to the curve \(y=\cos x\). Their points of contact lie on
1 \(x^{2} y^{2}=y^{2}-x^{2}\)
2 \(x^{2} y^{2}=x^{2}+y^{2}\)
3 \(x^{2} y^{2}=x^{2}-y^{2}\)
4 None of these
Explanation:
(C) : Let (x, y) be one of the points of contact. Then, the equation of the tangent at \(\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)\) is \(y=\cos x\) On differentiating both side w.r.t.x, we get- \(\frac{d y}{d x}=-\sin x \Rightarrow\left(\frac{d y}{d x}\right)_{\left(x_{1} y_{1}\right)}=-\sin x_{1}\) Equation of tangent at point \(\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)\) is- \(\mathrm{y}-\mathrm{y}_{1}=\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)}\left(\mathrm{x}-\mathrm{x}_{1}\right)\) \(\mathrm{y}-\mathrm{y}_{1}=-\sin \mathrm{x}_{1}\left(\mathrm{x}-\mathrm{x}_{1}\right)\) Which passes through the origin. \(\therefore \quad 0-\mathrm{y}_{1}=-\sin \mathrm{x}_{1}\left(0-\mathrm{x}_{1}\right)\) \(\mathrm{y}_{1}=-\mathrm{x}_{1} \sin \mathrm{x}_{1}\) \(\frac{\mathrm{y}_{1}}{\mathrm{x}_{1}}=-\sin \mathrm{x}_{1}\) Also, point \(\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)\) lies on \(\mathrm{y}=\cos \mathrm{x}\). \(\therefore \quad \mathrm{y}_{1}=\cos \mathrm{x}_{1}\) From equation (i) and (ii), we get- \(\sin ^{2} x_{1}+\cos ^{2} x_{1}=\frac{y_{1}^{2}}{x_{1}^{2}}+y_{1}^{2}=1\) \(x_{1}^{2}=y_{1}^{2}+y_{1}^{2} x_{1}^{2}\) Hence, the locus of \(\left(x_{1}, y_{1}\right)\) is \(\mathrm{x}^{2}=\mathrm{y}^{2}+\mathrm{y}^{2} \mathrm{x}^{2}\) \(x^{2} y^{2}=x^{2}-y^{2}\)
85414
The slope of the tangent to the hyperbola \(2 x^{2}-3 y^{2}=6\) at \((3,2) i\)
1 -1
2 1
3 0
4 2
Explanation:
(B) : Given, \(2 x^{2}-3 y^{2}=6\) On differentiating both sides w.r. to \(x\), we get - \(4 x-6 y \frac{d y}{d x}=0 \Rightarrow \therefore \frac{d y}{d x}=\frac{2 x}{3 y}\) The slope of the tangent at point \((3,2)\), \(\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{(3,2)}=\frac{2(3)}{3(2)}=\frac{6}{6}\) \(\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{(3,2)}=1\)
BITSAT-2010
Application of Derivatives
85415
The minimum value of the function \(y=x^{4}-2 x^{2}+1\) in the interval \(\left[\frac{1}{2}, 2\right] i\)
1 0
2 2
3 8
4 9
Explanation:
(A) : Given \(y=x^{4}-2 x^{2}+1\) \(\frac{d y}{d x}=\frac{d}{d x}\left(x^{4}-2 x^{2}+1\right)=4 x\left(x^{2}-1\right)\) For maximum or minimum, \(\frac{\mathrm{dy}}{\mathrm{dx}}=0\) \(4 x\left(x^{2}-1\right)=0, \text { either } x=0 \text { or } x= \pm 1\) \(\mathrm{x}=0\) and \(\mathrm{x}=-1\) does not belong to \(\left[\frac{1}{2}, 2\right]\) \(\frac{d^{2} y}{d x^{2}}=12 x^{2}-4\) \(\therefore \quad\left(\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}\right)_{\mathrm{x}=1}=12(1)^{2}-4=8>0\) \(\therefore\) There is minimum value of function at \(\mathrm{x}=1\) \(\therefore\) minimum value is \(y(1)=1^{4}-2(1)^{2}+1=1-2+1=0\)
BITSAT-2009
Application of Derivatives
85416
The slope of the tangent to the curve \(y=e^{x} \cos\) \(x\) is minimum at \(x=\alpha, 0 \leq a \leq 2 \pi\), then the value of \(\alpha \mathbf{i}\)
1 0
2 \(\pi\)
3 \(2 \pi\)
4 \(3 \pi / 2\)
Explanation:
(B) : Let \(\mathrm{m}\) be the slope of the tangent to the curve: \(y=e^{x} \cos x\). Then, \(\mathrm{m}=\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{e}^{\mathrm{x}}(\cos \mathrm{x}-\sin \mathrm{x})\) On differentiation both sides w.r.t. x, we get- \(\frac{d m}{d x}=e^{x}(\cos x-\sin x)+e^{x}(-\sin x-\cos x)=-2 e^{x} \sin x\) Again on differentiation both sides w.r.t. \(x\), we get- \(\frac{\mathrm{d}^{2} \mathrm{~m}}{\mathrm{dx}^{2}}=-2 \mathrm{e}^{\mathrm{x}}(\sin \mathrm{x}+\cos \mathrm{x})\) \(\frac{\mathrm{dm}}{\mathrm{dx}}=0\) Put, \(\quad \sin x=0\) \(\mathrm{x}=0, \pi, 2 \pi\) Clearly, \(\frac{\mathrm{d}^{2} \mathrm{~m}}{\mathrm{dx}^{2}}>0\) for \(\mathrm{x}=\pi\) Thus, \(\mathrm{y}\) is minimum at \(\mathrm{x}=\pi\). Hence the value of \(\alpha=\pi\).
BITSAT-2015
Application of Derivatives
85417
Tangents are drawn from the origin to the curve \(y=\cos x\). Their points of contact lie on
1 \(x^{2} y^{2}=y^{2}-x^{2}\)
2 \(x^{2} y^{2}=x^{2}+y^{2}\)
3 \(x^{2} y^{2}=x^{2}-y^{2}\)
4 None of these
Explanation:
(C) : Let (x, y) be one of the points of contact. Then, the equation of the tangent at \(\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)\) is \(y=\cos x\) On differentiating both side w.r.t.x, we get- \(\frac{d y}{d x}=-\sin x \Rightarrow\left(\frac{d y}{d x}\right)_{\left(x_{1} y_{1}\right)}=-\sin x_{1}\) Equation of tangent at point \(\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)\) is- \(\mathrm{y}-\mathrm{y}_{1}=\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)}\left(\mathrm{x}-\mathrm{x}_{1}\right)\) \(\mathrm{y}-\mathrm{y}_{1}=-\sin \mathrm{x}_{1}\left(\mathrm{x}-\mathrm{x}_{1}\right)\) Which passes through the origin. \(\therefore \quad 0-\mathrm{y}_{1}=-\sin \mathrm{x}_{1}\left(0-\mathrm{x}_{1}\right)\) \(\mathrm{y}_{1}=-\mathrm{x}_{1} \sin \mathrm{x}_{1}\) \(\frac{\mathrm{y}_{1}}{\mathrm{x}_{1}}=-\sin \mathrm{x}_{1}\) Also, point \(\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)\) lies on \(\mathrm{y}=\cos \mathrm{x}\). \(\therefore \quad \mathrm{y}_{1}=\cos \mathrm{x}_{1}\) From equation (i) and (ii), we get- \(\sin ^{2} x_{1}+\cos ^{2} x_{1}=\frac{y_{1}^{2}}{x_{1}^{2}}+y_{1}^{2}=1\) \(x_{1}^{2}=y_{1}^{2}+y_{1}^{2} x_{1}^{2}\) Hence, the locus of \(\left(x_{1}, y_{1}\right)\) is \(\mathrm{x}^{2}=\mathrm{y}^{2}+\mathrm{y}^{2} \mathrm{x}^{2}\) \(x^{2} y^{2}=x^{2}-y^{2}\)
