85397
If the line \(y=4 x-5\) touches to the curve \(y^{2}=\mathbf{a x}{ }^{3}+b\) at the point \((2,3)\) then
1 0
2 1
3 -1
4 2
Explanation:
(A) : Given, \(y^{2}=a x^{3}+b \tag{i}\) On differentiating both side w.r.t. \(x\), Now, \(2 y \frac{d y}{d x}=3 a x^{2} \Rightarrow \frac{d y}{d x}=\frac{3 a x^{2}}{2 y}\) \(\left(\frac{d y}{d x}\right)_{(2,3)}=\frac{3 a(2)^{2}}{2 \times 3} \Rightarrow\left(\frac{d y}{d x}\right)_{(2,3)}=2 a\) Slope of line \(y=4 x-5\) is 4 \(\therefore\) Slope of curve is \(2 \mathrm{a}=4\) \(a=2\) On putting the value of \(\mathrm{a}=2\) in \(\mathrm{eq}^{\mathrm{n}}(\mathrm{i})\) \(y^{2}=a x^{3}+b\) At point \((2,3)\) - \((3)^{2}=2(2)^{3}+b\) \(9=16+\mathrm{b}\) \(\mathrm{b}=-7\) Then, Now, the value of \(7 a+2 b\) \(=7(2)+2(-7)=14-14=0\)
MHT CET-2018
Application of Derivatives
85398
The point on the curve \(y=\sqrt{x-1}\), where the tangent is perpendicular to the line \(2 x+y-5=0\) is
1 \((2,-1)\)
2 \((10,3)\)
3 \((2,1)\)
4 \((5,-2)\)
Explanation:
(C) : Suppose that slope of the curve \(y=\sqrt{x-1}\) is \(\mathrm{m}_{1}\) \(\therefore \quad \mathrm{m}_{1}=\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}} \sqrt{\mathrm{x}-1}=\frac{1}{2 \sqrt{\mathrm{x}-1}}\) Slope of the line \(2 x+y-5=0\) is \(\mathrm{m}_{2}\) \(m_{2}=\frac{d y}{d x}=\frac{d}{d x}(5-2 x)=-2\) As lines are perpendicular if- \(\mathrm{m}_{1} \mathrm{~m}_{2}=-1\) So, \(\begin{aligned} \frac{1}{2 \sqrt{\mathrm{x}-1}}(-2)=-1 \\ \sqrt{\mathrm{x}-1} =1\end{aligned}\) On squaring both sides. \((\sqrt{\mathrm{x}-1})^{2}=(1)^{2}\) \(\mathrm{x}-1=1\) \(\mathrm{x}=2\) Then, \(y=\sqrt{x-1} \Rightarrow y=\sqrt{2-1}\) \(y=\sqrt{1} \Rightarrow y=1\) Hence, the required point is \((2,1)\).
MHT CET-2017
Application of Derivatives
85399
The point on the curve \(9 y^{2}-x^{3}\), where the normal to the curve make equal intercepts with the axes are
1 \(\left(4, \pm \frac{8}{3}\right)\)
2 \(\left(4,-\frac{8}{3}\right)\)
3 \(\left(4 \pm \frac{3}{8}\right)\)
4 \(\left( \pm 4, \frac{3}{8}\right)\)
Explanation:
(A) : Given the curve \(9 y^{2}=x^{3}\) Since, \(\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)\) lies on the given curve \(\therefore \quad 9 \mathrm{y}_{1}^{2}=\mathrm{x}_{1}^{3}\) On differentiating (i) on both side w.r.t \(x\), we get, \(18 y \frac{d y}{d x}=3 x^{2} \Rightarrow \frac{d y}{d x}=\frac{3 x^{2}}{18 y}\) \(\frac{d y}{d x}=\frac{x^{2}}{6 y}\) Slope of the tangent \(=\left(\frac{d y}{d x}\right)_{\left(x_{1}, y_{1}\right)}=\frac{x_{1}^{2}}{6 y_{1}}\) Slope of the normal \(=\frac{-1}{\frac{x_{1}^{2}}{6 y_{1}}}=\frac{-6 y_{1}}{x_{1}^{2}}\) \(\therefore\) Slope of the normal \(= \pm 1\) \(\begin{aligned} & \frac{-6 y}{x_1^2}= \pm 1 \\ & \frac{-6 y_1}{x_1^2}=1 \text { or } \frac{-6 y_1}{x_1^2}-1 \\ & y_1=\frac{-x_1^2}{6} \text { or } y_1=\frac{x_1^2}{6}\end{aligned}\) When, \(y_{1}=\frac{-x_{1}^{2}}{6}\) from eq \(^{\mathrm{n}}(\mathrm{i})\) \(9\left(\frac{x_{1}^{4}}{36}\right)=x_{1}^{3}\) \(x_{1}^{4}=4 x_{1}^{3}\) \(x_{1}^{4}-4 x_{1}^{3}=0\) \(x_{1}^{3}\left(x_{1}-4\right)=0\) \(x_{1}-4=0\) \(x_{1}=4,0\) Putting, \(\mathrm{x}_{1}=0\), \(9 \mathrm{y}_{1}^{2}=0\) \(\mathrm{y}_{1}=0\) putting \(\mathrm{x}_{1}=4\) in \(\left(\mathrm{eq}^{\mathrm{n}}(\mathrm{i})\right)\) \(9 y_{1}^{2}=4^{3}\) \(9 y_{1}^{2}=64\) \(y_{1}^{2}=\frac{64}{9} \Rightarrow y_{1}= \pm \frac{8}{3}\) But, the line making the equal intercepts with the coordinate axes cannot pass through the origin so, the point at \(\left(4, \pm \frac{8}{3}\right)\)
MHT CET-2012
Application of Derivatives
85400
All the points on the cure \(y=4 a\left[x+\operatorname{asin}\left(\frac{x}{a}\right)\right]\), where the tangent is parallel to the \(x\) axis lie on
1 circle
2 parabola
3 straight line
4 None of these
Explanation:
(B) : Given the curve - \(y^{2}=4 a\left[x+a \sin \left(\frac{x}{a}\right)\right] \tag{i}\) On differentiating both side we get \(2 y \frac{d y}{d x}=4 a\left[1+\cos \left(\frac{x}{a}\right)\right] \tag{ii}\) If the tangent is parallel to the \(x\)-axis, \(\frac{\mathrm{dy}}{\mathrm{dx}}=0\) Putting the value of \(\frac{d y}{d x}=0\) in equation(ii) we get - \(0=4 a\left[1+\cos \frac{x}{a}\right] \Rightarrow \cos \left(\frac{x}{a}\right)=-1\) \(\therefore \quad \sin \left(\frac{\mathrm{x}}{\mathrm{a}}\right)=0\) On putting the value in equation (i) we get - \(\mathrm{y}^{2}=4 \mathrm{a}(\mathrm{x}+0)\) \(\mathrm{y}^{2}=4 \mathrm{a}\) Hence, this is the equation of a parabola.
85397
If the line \(y=4 x-5\) touches to the curve \(y^{2}=\mathbf{a x}{ }^{3}+b\) at the point \((2,3)\) then
1 0
2 1
3 -1
4 2
Explanation:
(A) : Given, \(y^{2}=a x^{3}+b \tag{i}\) On differentiating both side w.r.t. \(x\), Now, \(2 y \frac{d y}{d x}=3 a x^{2} \Rightarrow \frac{d y}{d x}=\frac{3 a x^{2}}{2 y}\) \(\left(\frac{d y}{d x}\right)_{(2,3)}=\frac{3 a(2)^{2}}{2 \times 3} \Rightarrow\left(\frac{d y}{d x}\right)_{(2,3)}=2 a\) Slope of line \(y=4 x-5\) is 4 \(\therefore\) Slope of curve is \(2 \mathrm{a}=4\) \(a=2\) On putting the value of \(\mathrm{a}=2\) in \(\mathrm{eq}^{\mathrm{n}}(\mathrm{i})\) \(y^{2}=a x^{3}+b\) At point \((2,3)\) - \((3)^{2}=2(2)^{3}+b\) \(9=16+\mathrm{b}\) \(\mathrm{b}=-7\) Then, Now, the value of \(7 a+2 b\) \(=7(2)+2(-7)=14-14=0\)
MHT CET-2018
Application of Derivatives
85398
The point on the curve \(y=\sqrt{x-1}\), where the tangent is perpendicular to the line \(2 x+y-5=0\) is
1 \((2,-1)\)
2 \((10,3)\)
3 \((2,1)\)
4 \((5,-2)\)
Explanation:
(C) : Suppose that slope of the curve \(y=\sqrt{x-1}\) is \(\mathrm{m}_{1}\) \(\therefore \quad \mathrm{m}_{1}=\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}} \sqrt{\mathrm{x}-1}=\frac{1}{2 \sqrt{\mathrm{x}-1}}\) Slope of the line \(2 x+y-5=0\) is \(\mathrm{m}_{2}\) \(m_{2}=\frac{d y}{d x}=\frac{d}{d x}(5-2 x)=-2\) As lines are perpendicular if- \(\mathrm{m}_{1} \mathrm{~m}_{2}=-1\) So, \(\begin{aligned} \frac{1}{2 \sqrt{\mathrm{x}-1}}(-2)=-1 \\ \sqrt{\mathrm{x}-1} =1\end{aligned}\) On squaring both sides. \((\sqrt{\mathrm{x}-1})^{2}=(1)^{2}\) \(\mathrm{x}-1=1\) \(\mathrm{x}=2\) Then, \(y=\sqrt{x-1} \Rightarrow y=\sqrt{2-1}\) \(y=\sqrt{1} \Rightarrow y=1\) Hence, the required point is \((2,1)\).
MHT CET-2017
Application of Derivatives
85399
The point on the curve \(9 y^{2}-x^{3}\), where the normal to the curve make equal intercepts with the axes are
1 \(\left(4, \pm \frac{8}{3}\right)\)
2 \(\left(4,-\frac{8}{3}\right)\)
3 \(\left(4 \pm \frac{3}{8}\right)\)
4 \(\left( \pm 4, \frac{3}{8}\right)\)
Explanation:
(A) : Given the curve \(9 y^{2}=x^{3}\) Since, \(\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)\) lies on the given curve \(\therefore \quad 9 \mathrm{y}_{1}^{2}=\mathrm{x}_{1}^{3}\) On differentiating (i) on both side w.r.t \(x\), we get, \(18 y \frac{d y}{d x}=3 x^{2} \Rightarrow \frac{d y}{d x}=\frac{3 x^{2}}{18 y}\) \(\frac{d y}{d x}=\frac{x^{2}}{6 y}\) Slope of the tangent \(=\left(\frac{d y}{d x}\right)_{\left(x_{1}, y_{1}\right)}=\frac{x_{1}^{2}}{6 y_{1}}\) Slope of the normal \(=\frac{-1}{\frac{x_{1}^{2}}{6 y_{1}}}=\frac{-6 y_{1}}{x_{1}^{2}}\) \(\therefore\) Slope of the normal \(= \pm 1\) \(\begin{aligned} & \frac{-6 y}{x_1^2}= \pm 1 \\ & \frac{-6 y_1}{x_1^2}=1 \text { or } \frac{-6 y_1}{x_1^2}-1 \\ & y_1=\frac{-x_1^2}{6} \text { or } y_1=\frac{x_1^2}{6}\end{aligned}\) When, \(y_{1}=\frac{-x_{1}^{2}}{6}\) from eq \(^{\mathrm{n}}(\mathrm{i})\) \(9\left(\frac{x_{1}^{4}}{36}\right)=x_{1}^{3}\) \(x_{1}^{4}=4 x_{1}^{3}\) \(x_{1}^{4}-4 x_{1}^{3}=0\) \(x_{1}^{3}\left(x_{1}-4\right)=0\) \(x_{1}-4=0\) \(x_{1}=4,0\) Putting, \(\mathrm{x}_{1}=0\), \(9 \mathrm{y}_{1}^{2}=0\) \(\mathrm{y}_{1}=0\) putting \(\mathrm{x}_{1}=4\) in \(\left(\mathrm{eq}^{\mathrm{n}}(\mathrm{i})\right)\) \(9 y_{1}^{2}=4^{3}\) \(9 y_{1}^{2}=64\) \(y_{1}^{2}=\frac{64}{9} \Rightarrow y_{1}= \pm \frac{8}{3}\) But, the line making the equal intercepts with the coordinate axes cannot pass through the origin so, the point at \(\left(4, \pm \frac{8}{3}\right)\)
MHT CET-2012
Application of Derivatives
85400
All the points on the cure \(y=4 a\left[x+\operatorname{asin}\left(\frac{x}{a}\right)\right]\), where the tangent is parallel to the \(x\) axis lie on
1 circle
2 parabola
3 straight line
4 None of these
Explanation:
(B) : Given the curve - \(y^{2}=4 a\left[x+a \sin \left(\frac{x}{a}\right)\right] \tag{i}\) On differentiating both side we get \(2 y \frac{d y}{d x}=4 a\left[1+\cos \left(\frac{x}{a}\right)\right] \tag{ii}\) If the tangent is parallel to the \(x\)-axis, \(\frac{\mathrm{dy}}{\mathrm{dx}}=0\) Putting the value of \(\frac{d y}{d x}=0\) in equation(ii) we get - \(0=4 a\left[1+\cos \frac{x}{a}\right] \Rightarrow \cos \left(\frac{x}{a}\right)=-1\) \(\therefore \quad \sin \left(\frac{\mathrm{x}}{\mathrm{a}}\right)=0\) On putting the value in equation (i) we get - \(\mathrm{y}^{2}=4 \mathrm{a}(\mathrm{x}+0)\) \(\mathrm{y}^{2}=4 \mathrm{a}\) Hence, this is the equation of a parabola.
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Application of Derivatives
85397
If the line \(y=4 x-5\) touches to the curve \(y^{2}=\mathbf{a x}{ }^{3}+b\) at the point \((2,3)\) then
1 0
2 1
3 -1
4 2
Explanation:
(A) : Given, \(y^{2}=a x^{3}+b \tag{i}\) On differentiating both side w.r.t. \(x\), Now, \(2 y \frac{d y}{d x}=3 a x^{2} \Rightarrow \frac{d y}{d x}=\frac{3 a x^{2}}{2 y}\) \(\left(\frac{d y}{d x}\right)_{(2,3)}=\frac{3 a(2)^{2}}{2 \times 3} \Rightarrow\left(\frac{d y}{d x}\right)_{(2,3)}=2 a\) Slope of line \(y=4 x-5\) is 4 \(\therefore\) Slope of curve is \(2 \mathrm{a}=4\) \(a=2\) On putting the value of \(\mathrm{a}=2\) in \(\mathrm{eq}^{\mathrm{n}}(\mathrm{i})\) \(y^{2}=a x^{3}+b\) At point \((2,3)\) - \((3)^{2}=2(2)^{3}+b\) \(9=16+\mathrm{b}\) \(\mathrm{b}=-7\) Then, Now, the value of \(7 a+2 b\) \(=7(2)+2(-7)=14-14=0\)
MHT CET-2018
Application of Derivatives
85398
The point on the curve \(y=\sqrt{x-1}\), where the tangent is perpendicular to the line \(2 x+y-5=0\) is
1 \((2,-1)\)
2 \((10,3)\)
3 \((2,1)\)
4 \((5,-2)\)
Explanation:
(C) : Suppose that slope of the curve \(y=\sqrt{x-1}\) is \(\mathrm{m}_{1}\) \(\therefore \quad \mathrm{m}_{1}=\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}} \sqrt{\mathrm{x}-1}=\frac{1}{2 \sqrt{\mathrm{x}-1}}\) Slope of the line \(2 x+y-5=0\) is \(\mathrm{m}_{2}\) \(m_{2}=\frac{d y}{d x}=\frac{d}{d x}(5-2 x)=-2\) As lines are perpendicular if- \(\mathrm{m}_{1} \mathrm{~m}_{2}=-1\) So, \(\begin{aligned} \frac{1}{2 \sqrt{\mathrm{x}-1}}(-2)=-1 \\ \sqrt{\mathrm{x}-1} =1\end{aligned}\) On squaring both sides. \((\sqrt{\mathrm{x}-1})^{2}=(1)^{2}\) \(\mathrm{x}-1=1\) \(\mathrm{x}=2\) Then, \(y=\sqrt{x-1} \Rightarrow y=\sqrt{2-1}\) \(y=\sqrt{1} \Rightarrow y=1\) Hence, the required point is \((2,1)\).
MHT CET-2017
Application of Derivatives
85399
The point on the curve \(9 y^{2}-x^{3}\), where the normal to the curve make equal intercepts with the axes are
1 \(\left(4, \pm \frac{8}{3}\right)\)
2 \(\left(4,-\frac{8}{3}\right)\)
3 \(\left(4 \pm \frac{3}{8}\right)\)
4 \(\left( \pm 4, \frac{3}{8}\right)\)
Explanation:
(A) : Given the curve \(9 y^{2}=x^{3}\) Since, \(\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)\) lies on the given curve \(\therefore \quad 9 \mathrm{y}_{1}^{2}=\mathrm{x}_{1}^{3}\) On differentiating (i) on both side w.r.t \(x\), we get, \(18 y \frac{d y}{d x}=3 x^{2} \Rightarrow \frac{d y}{d x}=\frac{3 x^{2}}{18 y}\) \(\frac{d y}{d x}=\frac{x^{2}}{6 y}\) Slope of the tangent \(=\left(\frac{d y}{d x}\right)_{\left(x_{1}, y_{1}\right)}=\frac{x_{1}^{2}}{6 y_{1}}\) Slope of the normal \(=\frac{-1}{\frac{x_{1}^{2}}{6 y_{1}}}=\frac{-6 y_{1}}{x_{1}^{2}}\) \(\therefore\) Slope of the normal \(= \pm 1\) \(\begin{aligned} & \frac{-6 y}{x_1^2}= \pm 1 \\ & \frac{-6 y_1}{x_1^2}=1 \text { or } \frac{-6 y_1}{x_1^2}-1 \\ & y_1=\frac{-x_1^2}{6} \text { or } y_1=\frac{x_1^2}{6}\end{aligned}\) When, \(y_{1}=\frac{-x_{1}^{2}}{6}\) from eq \(^{\mathrm{n}}(\mathrm{i})\) \(9\left(\frac{x_{1}^{4}}{36}\right)=x_{1}^{3}\) \(x_{1}^{4}=4 x_{1}^{3}\) \(x_{1}^{4}-4 x_{1}^{3}=0\) \(x_{1}^{3}\left(x_{1}-4\right)=0\) \(x_{1}-4=0\) \(x_{1}=4,0\) Putting, \(\mathrm{x}_{1}=0\), \(9 \mathrm{y}_{1}^{2}=0\) \(\mathrm{y}_{1}=0\) putting \(\mathrm{x}_{1}=4\) in \(\left(\mathrm{eq}^{\mathrm{n}}(\mathrm{i})\right)\) \(9 y_{1}^{2}=4^{3}\) \(9 y_{1}^{2}=64\) \(y_{1}^{2}=\frac{64}{9} \Rightarrow y_{1}= \pm \frac{8}{3}\) But, the line making the equal intercepts with the coordinate axes cannot pass through the origin so, the point at \(\left(4, \pm \frac{8}{3}\right)\)
MHT CET-2012
Application of Derivatives
85400
All the points on the cure \(y=4 a\left[x+\operatorname{asin}\left(\frac{x}{a}\right)\right]\), where the tangent is parallel to the \(x\) axis lie on
1 circle
2 parabola
3 straight line
4 None of these
Explanation:
(B) : Given the curve - \(y^{2}=4 a\left[x+a \sin \left(\frac{x}{a}\right)\right] \tag{i}\) On differentiating both side we get \(2 y \frac{d y}{d x}=4 a\left[1+\cos \left(\frac{x}{a}\right)\right] \tag{ii}\) If the tangent is parallel to the \(x\)-axis, \(\frac{\mathrm{dy}}{\mathrm{dx}}=0\) Putting the value of \(\frac{d y}{d x}=0\) in equation(ii) we get - \(0=4 a\left[1+\cos \frac{x}{a}\right] \Rightarrow \cos \left(\frac{x}{a}\right)=-1\) \(\therefore \quad \sin \left(\frac{\mathrm{x}}{\mathrm{a}}\right)=0\) On putting the value in equation (i) we get - \(\mathrm{y}^{2}=4 \mathrm{a}(\mathrm{x}+0)\) \(\mathrm{y}^{2}=4 \mathrm{a}\) Hence, this is the equation of a parabola.
85397
If the line \(y=4 x-5\) touches to the curve \(y^{2}=\mathbf{a x}{ }^{3}+b\) at the point \((2,3)\) then
1 0
2 1
3 -1
4 2
Explanation:
(A) : Given, \(y^{2}=a x^{3}+b \tag{i}\) On differentiating both side w.r.t. \(x\), Now, \(2 y \frac{d y}{d x}=3 a x^{2} \Rightarrow \frac{d y}{d x}=\frac{3 a x^{2}}{2 y}\) \(\left(\frac{d y}{d x}\right)_{(2,3)}=\frac{3 a(2)^{2}}{2 \times 3} \Rightarrow\left(\frac{d y}{d x}\right)_{(2,3)}=2 a\) Slope of line \(y=4 x-5\) is 4 \(\therefore\) Slope of curve is \(2 \mathrm{a}=4\) \(a=2\) On putting the value of \(\mathrm{a}=2\) in \(\mathrm{eq}^{\mathrm{n}}(\mathrm{i})\) \(y^{2}=a x^{3}+b\) At point \((2,3)\) - \((3)^{2}=2(2)^{3}+b\) \(9=16+\mathrm{b}\) \(\mathrm{b}=-7\) Then, Now, the value of \(7 a+2 b\) \(=7(2)+2(-7)=14-14=0\)
MHT CET-2018
Application of Derivatives
85398
The point on the curve \(y=\sqrt{x-1}\), where the tangent is perpendicular to the line \(2 x+y-5=0\) is
1 \((2,-1)\)
2 \((10,3)\)
3 \((2,1)\)
4 \((5,-2)\)
Explanation:
(C) : Suppose that slope of the curve \(y=\sqrt{x-1}\) is \(\mathrm{m}_{1}\) \(\therefore \quad \mathrm{m}_{1}=\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}} \sqrt{\mathrm{x}-1}=\frac{1}{2 \sqrt{\mathrm{x}-1}}\) Slope of the line \(2 x+y-5=0\) is \(\mathrm{m}_{2}\) \(m_{2}=\frac{d y}{d x}=\frac{d}{d x}(5-2 x)=-2\) As lines are perpendicular if- \(\mathrm{m}_{1} \mathrm{~m}_{2}=-1\) So, \(\begin{aligned} \frac{1}{2 \sqrt{\mathrm{x}-1}}(-2)=-1 \\ \sqrt{\mathrm{x}-1} =1\end{aligned}\) On squaring both sides. \((\sqrt{\mathrm{x}-1})^{2}=(1)^{2}\) \(\mathrm{x}-1=1\) \(\mathrm{x}=2\) Then, \(y=\sqrt{x-1} \Rightarrow y=\sqrt{2-1}\) \(y=\sqrt{1} \Rightarrow y=1\) Hence, the required point is \((2,1)\).
MHT CET-2017
Application of Derivatives
85399
The point on the curve \(9 y^{2}-x^{3}\), where the normal to the curve make equal intercepts with the axes are
1 \(\left(4, \pm \frac{8}{3}\right)\)
2 \(\left(4,-\frac{8}{3}\right)\)
3 \(\left(4 \pm \frac{3}{8}\right)\)
4 \(\left( \pm 4, \frac{3}{8}\right)\)
Explanation:
(A) : Given the curve \(9 y^{2}=x^{3}\) Since, \(\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)\) lies on the given curve \(\therefore \quad 9 \mathrm{y}_{1}^{2}=\mathrm{x}_{1}^{3}\) On differentiating (i) on both side w.r.t \(x\), we get, \(18 y \frac{d y}{d x}=3 x^{2} \Rightarrow \frac{d y}{d x}=\frac{3 x^{2}}{18 y}\) \(\frac{d y}{d x}=\frac{x^{2}}{6 y}\) Slope of the tangent \(=\left(\frac{d y}{d x}\right)_{\left(x_{1}, y_{1}\right)}=\frac{x_{1}^{2}}{6 y_{1}}\) Slope of the normal \(=\frac{-1}{\frac{x_{1}^{2}}{6 y_{1}}}=\frac{-6 y_{1}}{x_{1}^{2}}\) \(\therefore\) Slope of the normal \(= \pm 1\) \(\begin{aligned} & \frac{-6 y}{x_1^2}= \pm 1 \\ & \frac{-6 y_1}{x_1^2}=1 \text { or } \frac{-6 y_1}{x_1^2}-1 \\ & y_1=\frac{-x_1^2}{6} \text { or } y_1=\frac{x_1^2}{6}\end{aligned}\) When, \(y_{1}=\frac{-x_{1}^{2}}{6}\) from eq \(^{\mathrm{n}}(\mathrm{i})\) \(9\left(\frac{x_{1}^{4}}{36}\right)=x_{1}^{3}\) \(x_{1}^{4}=4 x_{1}^{3}\) \(x_{1}^{4}-4 x_{1}^{3}=0\) \(x_{1}^{3}\left(x_{1}-4\right)=0\) \(x_{1}-4=0\) \(x_{1}=4,0\) Putting, \(\mathrm{x}_{1}=0\), \(9 \mathrm{y}_{1}^{2}=0\) \(\mathrm{y}_{1}=0\) putting \(\mathrm{x}_{1}=4\) in \(\left(\mathrm{eq}^{\mathrm{n}}(\mathrm{i})\right)\) \(9 y_{1}^{2}=4^{3}\) \(9 y_{1}^{2}=64\) \(y_{1}^{2}=\frac{64}{9} \Rightarrow y_{1}= \pm \frac{8}{3}\) But, the line making the equal intercepts with the coordinate axes cannot pass through the origin so, the point at \(\left(4, \pm \frac{8}{3}\right)\)
MHT CET-2012
Application of Derivatives
85400
All the points on the cure \(y=4 a\left[x+\operatorname{asin}\left(\frac{x}{a}\right)\right]\), where the tangent is parallel to the \(x\) axis lie on
1 circle
2 parabola
3 straight line
4 None of these
Explanation:
(B) : Given the curve - \(y^{2}=4 a\left[x+a \sin \left(\frac{x}{a}\right)\right] \tag{i}\) On differentiating both side we get \(2 y \frac{d y}{d x}=4 a\left[1+\cos \left(\frac{x}{a}\right)\right] \tag{ii}\) If the tangent is parallel to the \(x\)-axis, \(\frac{\mathrm{dy}}{\mathrm{dx}}=0\) Putting the value of \(\frac{d y}{d x}=0\) in equation(ii) we get - \(0=4 a\left[1+\cos \frac{x}{a}\right] \Rightarrow \cos \left(\frac{x}{a}\right)=-1\) \(\therefore \quad \sin \left(\frac{\mathrm{x}}{\mathrm{a}}\right)=0\) On putting the value in equation (i) we get - \(\mathrm{y}^{2}=4 \mathrm{a}(\mathrm{x}+0)\) \(\mathrm{y}^{2}=4 \mathrm{a}\) Hence, this is the equation of a parabola.