85418
What is the \(x\)-coordinate of the point on the curve \(f(x)=\sqrt{x}(7 x-6)\), where the tangent is parallel to \(\mathrm{x}\)-axis
1 \(-\frac{1}{3}\)
2 \(\frac{2}{7}\)
3 \(\frac{6}{7}\)
4 \(\frac{1}{2}\)
Explanation:
(B) :Given, \(f(x)=\sqrt{x}(7 x-6)\) \(f(x)=7 x^{3 / 2}-6 x^{1 / 2}\) On differentiation both sides w.r.t.x, we get- \(f^{\prime}(x)=7 \times \frac{3}{2} x^{1 / 2}-6 \times \frac{1}{2} x^{-1 / 2}\) When tangent is parallel to \(x\) axis \(\mathrm{f}^{\prime}(\mathrm{x})=0\) \(\frac{21}{2} x^{1 / 2}-3 x^{-1 / 2}=0 \Rightarrow \frac{21}{2} \sqrt{x}=\frac{3}{\sqrt{x}}\) \(7 \mathrm{x}=2 \Rightarrow \mathrm{x}=\frac{2}{7}\)
BITSAT-2016
Application of Derivatives
85419
For the curve \(x^{2}+4 x y+8 y^{2}=64\) the tangents are parallel to the \(x\)-axis only at the points
1 \((0,2 \sqrt{2})\) and \((0,-2 \sqrt{2})\)
2 \((8,-4)\) and \((-8,4)\)
3 \((8 \sqrt{2},-2 \sqrt{2})\) and \((-8 \sqrt{2}, 2 \sqrt{2})\)
4 \((9,0)\) and \((-8,0)\)
Explanation:
(B) : Given, \(x^{2}+4 x y+8 y^{2}=64 \tag{i}\) On differentiating both sides w.r.t to \(x\) we get - \(2 x+4\left(y+x \frac{d y}{d x}\right)+16 y \frac{d y}{d x}=0\) \(2 x+4 y+(4 x+16 y) \frac{d y}{d x}=0\) \(\frac{d y}{d x}=-\frac{(x+2 y)}{2(x+4 y)}\) Since, tangent are parallel to \(x\)-axis only. \(\begin{aligned} & \frac{d y}{d x}=0 \Rightarrow 0=-\frac{(x+2 y)}{2(x+4 y)} \\ & x+2 y=0\end{aligned}\) On putting the values of \(x\) from equation (i) in (ii), we get \(4 y^{2}-8 y^{2}+8 y^{2}=64\) \(4 y^{2}=64\) \(y^{2}=16\) \(y= \pm 4\) On putting the value \(y\) in equation (ii) \(x+2 y=0\) \(x+2(4)=0\) \(x=-8, y=4\) \(x=+8, y=-4\) Hence, required point are \((-8,4)\) and \((8,-4)\)
WB JEE-2013
Application of Derivatives
85420
If the equation of one tangent to the circle with centre at \((2,-1)\) from the origin is \(3 x+y=0\), then the equation of the other tangent through the origin is
1 \(3 x-y=0\)
2 \(x+3 y=0\)
3 \(x-3 y=0\)
4 \(x+2 y=0\)
Explanation:
(C) : Given, \(3 x+y=0\) Radius \(=\mathrm{r}\) (let) The lengths of the perpendicular from the centre \((2,-1)\) to the line \(3 x+y=0\) \(r=\frac{6-1}{\sqrt{(9+1)}}\) \(r=\frac{5}{\sqrt{10}}\) Let, \(y=m x\) be other tangent from origin, Then, \(\frac{|2 \mathrm{~m}+1|}{\sqrt{\left(1+\mathrm{m}^{2}\right)}}=\frac{5}{\sqrt{10}}\) On squaring both sides- \(\frac{4 m^{2}+1+4 m}{1+m^{2}}=\frac{25}{10}\) \(8 m^{2}+2+8 m=5+5 m^{2}\) \(3 m^{2}+8 m-3=0\) \(m=-3,1 / 3\) As, -3 is the slope of given tangent so \(1 / 3\) is that of the other tangent. Hence, \(y=\left(\frac{1}{3}\right) \mathrm{x}\) \(3 \mathrm{y}=\mathrm{x}\) \(3 \mathrm{y}-\mathrm{x}=0\) \(\mathrm{x}-3 \mathrm{y}=0\)
WB JEE-2022
Application of Derivatives
85421
Consider the curve \(y=b e^{-x / a}\) where \(a\) and \(b\) are non-zero real numbers. Then
1 \(\frac{x}{a}+\frac{y}{b}=1\) is tangent to the curve at \((0,0)\)
2 \(\frac{x}{a}+\frac{y}{b}=1\) is tangent to the curve, where the curve crosses the axis of \(y\)
3 \(\frac{x}{a}+\frac{y}{b}=1\) is tangent to the curve at \((a, 0)\)
4 \(\frac{x}{a}+\frac{y}{b}=1\) is tangent to the curve at \((2 a, 0)\)
Explanation:
(B) : Given the curve, \(y=b e^{-x / a}\) On differentiating both sides w.r.t.x, we get- \(\frac{d y}{d x}=-\frac{b}{a} e^{\frac{-x}{a}}\) \(\therefore \quad\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{(0, \mathrm{~b})}=\frac{-\mathrm{b}}{\mathrm{a}} \mathrm{e}^{-0 / \mathrm{a}}\) \(\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{(0, \mathrm{~b})}=\frac{-\mathrm{b}}{\mathrm{a}}\) \(\therefore\) Equation of tangent \(y-b=\frac{-b}{a}(x-0) \Rightarrow y-b=\frac{-b x}{a}\) \(\frac{y}{b}-1=\frac{-x}{a} \Rightarrow \frac{x}{a}+\frac{y}{b}=1\) Hence, \(\frac{\mathrm{x}}{\mathrm{a}}+\frac{\mathrm{y}}{\mathrm{b}}=1\) is tangent to the curve where the curve crosses the axis of \(y\).
WB JEE-2020
Application of Derivatives
85422
If the tangent to the curve \(y^{2}=x^{3} a t\left(m^{2}, m^{3}\right)\) is also a normal to the curve at \(\left(M^{2}, M^{2}\right)\), then the value of \(\mathrm{mM}\) is
1 \(-\frac{1}{6}\)
2 \(-\frac{2}{9}\)
3 \(-\frac{1}{3}\)
4 \(-\frac{4}{9}\)
Explanation:
(D) : Given, \(y^{2}=x^{3}\) On differentiating both sides w.r.t.x, we get- \(2 y \frac{d y}{d x}=3 x^{2}\) Slope, \(\quad \mathrm{m}=\frac{3 \mathrm{x}^{2}}{2 \mathrm{y}}\) Slope of tangent at \(\left(\mathrm{m}^{2}, \mathrm{~m}^{3}\right)\) \(\mathrm{m}_{1}=\frac{3\left(\mathrm{~m}^{2}\right)^{2}}{2 \mathrm{~m}^{3}} \Rightarrow \mathrm{m}_{1}=\frac{3 \mathrm{~m}^{4}}{2 \mathrm{~m}^{3}}\) \(\mathrm{~m}_{1}=\frac{3 \mathrm{~m}}{2}\) Slope at \(\left[\mathrm{M}^{2}, \mathrm{M}^{3}\right]\) \(\mathrm{m}_{2}=\frac{3 \mathrm{M}}{2}\) Now, \(\mathrm{m}_{1} \mathrm{~m}_{2}=-1\) On putting the value of \(\mathrm{m}_{1} \mathrm{~m}_{2}\) \(\frac{3 m}{2} \cdot \frac{3 M}{2}=-1 \Rightarrow \frac{9 m M}{4}=-1\) \(m M=\frac{-4}{9}\)
85418
What is the \(x\)-coordinate of the point on the curve \(f(x)=\sqrt{x}(7 x-6)\), where the tangent is parallel to \(\mathrm{x}\)-axis
1 \(-\frac{1}{3}\)
2 \(\frac{2}{7}\)
3 \(\frac{6}{7}\)
4 \(\frac{1}{2}\)
Explanation:
(B) :Given, \(f(x)=\sqrt{x}(7 x-6)\) \(f(x)=7 x^{3 / 2}-6 x^{1 / 2}\) On differentiation both sides w.r.t.x, we get- \(f^{\prime}(x)=7 \times \frac{3}{2} x^{1 / 2}-6 \times \frac{1}{2} x^{-1 / 2}\) When tangent is parallel to \(x\) axis \(\mathrm{f}^{\prime}(\mathrm{x})=0\) \(\frac{21}{2} x^{1 / 2}-3 x^{-1 / 2}=0 \Rightarrow \frac{21}{2} \sqrt{x}=\frac{3}{\sqrt{x}}\) \(7 \mathrm{x}=2 \Rightarrow \mathrm{x}=\frac{2}{7}\)
BITSAT-2016
Application of Derivatives
85419
For the curve \(x^{2}+4 x y+8 y^{2}=64\) the tangents are parallel to the \(x\)-axis only at the points
1 \((0,2 \sqrt{2})\) and \((0,-2 \sqrt{2})\)
2 \((8,-4)\) and \((-8,4)\)
3 \((8 \sqrt{2},-2 \sqrt{2})\) and \((-8 \sqrt{2}, 2 \sqrt{2})\)
4 \((9,0)\) and \((-8,0)\)
Explanation:
(B) : Given, \(x^{2}+4 x y+8 y^{2}=64 \tag{i}\) On differentiating both sides w.r.t to \(x\) we get - \(2 x+4\left(y+x \frac{d y}{d x}\right)+16 y \frac{d y}{d x}=0\) \(2 x+4 y+(4 x+16 y) \frac{d y}{d x}=0\) \(\frac{d y}{d x}=-\frac{(x+2 y)}{2(x+4 y)}\) Since, tangent are parallel to \(x\)-axis only. \(\begin{aligned} & \frac{d y}{d x}=0 \Rightarrow 0=-\frac{(x+2 y)}{2(x+4 y)} \\ & x+2 y=0\end{aligned}\) On putting the values of \(x\) from equation (i) in (ii), we get \(4 y^{2}-8 y^{2}+8 y^{2}=64\) \(4 y^{2}=64\) \(y^{2}=16\) \(y= \pm 4\) On putting the value \(y\) in equation (ii) \(x+2 y=0\) \(x+2(4)=0\) \(x=-8, y=4\) \(x=+8, y=-4\) Hence, required point are \((-8,4)\) and \((8,-4)\)
WB JEE-2013
Application of Derivatives
85420
If the equation of one tangent to the circle with centre at \((2,-1)\) from the origin is \(3 x+y=0\), then the equation of the other tangent through the origin is
1 \(3 x-y=0\)
2 \(x+3 y=0\)
3 \(x-3 y=0\)
4 \(x+2 y=0\)
Explanation:
(C) : Given, \(3 x+y=0\) Radius \(=\mathrm{r}\) (let) The lengths of the perpendicular from the centre \((2,-1)\) to the line \(3 x+y=0\) \(r=\frac{6-1}{\sqrt{(9+1)}}\) \(r=\frac{5}{\sqrt{10}}\) Let, \(y=m x\) be other tangent from origin, Then, \(\frac{|2 \mathrm{~m}+1|}{\sqrt{\left(1+\mathrm{m}^{2}\right)}}=\frac{5}{\sqrt{10}}\) On squaring both sides- \(\frac{4 m^{2}+1+4 m}{1+m^{2}}=\frac{25}{10}\) \(8 m^{2}+2+8 m=5+5 m^{2}\) \(3 m^{2}+8 m-3=0\) \(m=-3,1 / 3\) As, -3 is the slope of given tangent so \(1 / 3\) is that of the other tangent. Hence, \(y=\left(\frac{1}{3}\right) \mathrm{x}\) \(3 \mathrm{y}=\mathrm{x}\) \(3 \mathrm{y}-\mathrm{x}=0\) \(\mathrm{x}-3 \mathrm{y}=0\)
WB JEE-2022
Application of Derivatives
85421
Consider the curve \(y=b e^{-x / a}\) where \(a\) and \(b\) are non-zero real numbers. Then
1 \(\frac{x}{a}+\frac{y}{b}=1\) is tangent to the curve at \((0,0)\)
2 \(\frac{x}{a}+\frac{y}{b}=1\) is tangent to the curve, where the curve crosses the axis of \(y\)
3 \(\frac{x}{a}+\frac{y}{b}=1\) is tangent to the curve at \((a, 0)\)
4 \(\frac{x}{a}+\frac{y}{b}=1\) is tangent to the curve at \((2 a, 0)\)
Explanation:
(B) : Given the curve, \(y=b e^{-x / a}\) On differentiating both sides w.r.t.x, we get- \(\frac{d y}{d x}=-\frac{b}{a} e^{\frac{-x}{a}}\) \(\therefore \quad\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{(0, \mathrm{~b})}=\frac{-\mathrm{b}}{\mathrm{a}} \mathrm{e}^{-0 / \mathrm{a}}\) \(\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{(0, \mathrm{~b})}=\frac{-\mathrm{b}}{\mathrm{a}}\) \(\therefore\) Equation of tangent \(y-b=\frac{-b}{a}(x-0) \Rightarrow y-b=\frac{-b x}{a}\) \(\frac{y}{b}-1=\frac{-x}{a} \Rightarrow \frac{x}{a}+\frac{y}{b}=1\) Hence, \(\frac{\mathrm{x}}{\mathrm{a}}+\frac{\mathrm{y}}{\mathrm{b}}=1\) is tangent to the curve where the curve crosses the axis of \(y\).
WB JEE-2020
Application of Derivatives
85422
If the tangent to the curve \(y^{2}=x^{3} a t\left(m^{2}, m^{3}\right)\) is also a normal to the curve at \(\left(M^{2}, M^{2}\right)\), then the value of \(\mathrm{mM}\) is
1 \(-\frac{1}{6}\)
2 \(-\frac{2}{9}\)
3 \(-\frac{1}{3}\)
4 \(-\frac{4}{9}\)
Explanation:
(D) : Given, \(y^{2}=x^{3}\) On differentiating both sides w.r.t.x, we get- \(2 y \frac{d y}{d x}=3 x^{2}\) Slope, \(\quad \mathrm{m}=\frac{3 \mathrm{x}^{2}}{2 \mathrm{y}}\) Slope of tangent at \(\left(\mathrm{m}^{2}, \mathrm{~m}^{3}\right)\) \(\mathrm{m}_{1}=\frac{3\left(\mathrm{~m}^{2}\right)^{2}}{2 \mathrm{~m}^{3}} \Rightarrow \mathrm{m}_{1}=\frac{3 \mathrm{~m}^{4}}{2 \mathrm{~m}^{3}}\) \(\mathrm{~m}_{1}=\frac{3 \mathrm{~m}}{2}\) Slope at \(\left[\mathrm{M}^{2}, \mathrm{M}^{3}\right]\) \(\mathrm{m}_{2}=\frac{3 \mathrm{M}}{2}\) Now, \(\mathrm{m}_{1} \mathrm{~m}_{2}=-1\) On putting the value of \(\mathrm{m}_{1} \mathrm{~m}_{2}\) \(\frac{3 m}{2} \cdot \frac{3 M}{2}=-1 \Rightarrow \frac{9 m M}{4}=-1\) \(m M=\frac{-4}{9}\)
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Application of Derivatives
85418
What is the \(x\)-coordinate of the point on the curve \(f(x)=\sqrt{x}(7 x-6)\), where the tangent is parallel to \(\mathrm{x}\)-axis
1 \(-\frac{1}{3}\)
2 \(\frac{2}{7}\)
3 \(\frac{6}{7}\)
4 \(\frac{1}{2}\)
Explanation:
(B) :Given, \(f(x)=\sqrt{x}(7 x-6)\) \(f(x)=7 x^{3 / 2}-6 x^{1 / 2}\) On differentiation both sides w.r.t.x, we get- \(f^{\prime}(x)=7 \times \frac{3}{2} x^{1 / 2}-6 \times \frac{1}{2} x^{-1 / 2}\) When tangent is parallel to \(x\) axis \(\mathrm{f}^{\prime}(\mathrm{x})=0\) \(\frac{21}{2} x^{1 / 2}-3 x^{-1 / 2}=0 \Rightarrow \frac{21}{2} \sqrt{x}=\frac{3}{\sqrt{x}}\) \(7 \mathrm{x}=2 \Rightarrow \mathrm{x}=\frac{2}{7}\)
BITSAT-2016
Application of Derivatives
85419
For the curve \(x^{2}+4 x y+8 y^{2}=64\) the tangents are parallel to the \(x\)-axis only at the points
1 \((0,2 \sqrt{2})\) and \((0,-2 \sqrt{2})\)
2 \((8,-4)\) and \((-8,4)\)
3 \((8 \sqrt{2},-2 \sqrt{2})\) and \((-8 \sqrt{2}, 2 \sqrt{2})\)
4 \((9,0)\) and \((-8,0)\)
Explanation:
(B) : Given, \(x^{2}+4 x y+8 y^{2}=64 \tag{i}\) On differentiating both sides w.r.t to \(x\) we get - \(2 x+4\left(y+x \frac{d y}{d x}\right)+16 y \frac{d y}{d x}=0\) \(2 x+4 y+(4 x+16 y) \frac{d y}{d x}=0\) \(\frac{d y}{d x}=-\frac{(x+2 y)}{2(x+4 y)}\) Since, tangent are parallel to \(x\)-axis only. \(\begin{aligned} & \frac{d y}{d x}=0 \Rightarrow 0=-\frac{(x+2 y)}{2(x+4 y)} \\ & x+2 y=0\end{aligned}\) On putting the values of \(x\) from equation (i) in (ii), we get \(4 y^{2}-8 y^{2}+8 y^{2}=64\) \(4 y^{2}=64\) \(y^{2}=16\) \(y= \pm 4\) On putting the value \(y\) in equation (ii) \(x+2 y=0\) \(x+2(4)=0\) \(x=-8, y=4\) \(x=+8, y=-4\) Hence, required point are \((-8,4)\) and \((8,-4)\)
WB JEE-2013
Application of Derivatives
85420
If the equation of one tangent to the circle with centre at \((2,-1)\) from the origin is \(3 x+y=0\), then the equation of the other tangent through the origin is
1 \(3 x-y=0\)
2 \(x+3 y=0\)
3 \(x-3 y=0\)
4 \(x+2 y=0\)
Explanation:
(C) : Given, \(3 x+y=0\) Radius \(=\mathrm{r}\) (let) The lengths of the perpendicular from the centre \((2,-1)\) to the line \(3 x+y=0\) \(r=\frac{6-1}{\sqrt{(9+1)}}\) \(r=\frac{5}{\sqrt{10}}\) Let, \(y=m x\) be other tangent from origin, Then, \(\frac{|2 \mathrm{~m}+1|}{\sqrt{\left(1+\mathrm{m}^{2}\right)}}=\frac{5}{\sqrt{10}}\) On squaring both sides- \(\frac{4 m^{2}+1+4 m}{1+m^{2}}=\frac{25}{10}\) \(8 m^{2}+2+8 m=5+5 m^{2}\) \(3 m^{2}+8 m-3=0\) \(m=-3,1 / 3\) As, -3 is the slope of given tangent so \(1 / 3\) is that of the other tangent. Hence, \(y=\left(\frac{1}{3}\right) \mathrm{x}\) \(3 \mathrm{y}=\mathrm{x}\) \(3 \mathrm{y}-\mathrm{x}=0\) \(\mathrm{x}-3 \mathrm{y}=0\)
WB JEE-2022
Application of Derivatives
85421
Consider the curve \(y=b e^{-x / a}\) where \(a\) and \(b\) are non-zero real numbers. Then
1 \(\frac{x}{a}+\frac{y}{b}=1\) is tangent to the curve at \((0,0)\)
2 \(\frac{x}{a}+\frac{y}{b}=1\) is tangent to the curve, where the curve crosses the axis of \(y\)
3 \(\frac{x}{a}+\frac{y}{b}=1\) is tangent to the curve at \((a, 0)\)
4 \(\frac{x}{a}+\frac{y}{b}=1\) is tangent to the curve at \((2 a, 0)\)
Explanation:
(B) : Given the curve, \(y=b e^{-x / a}\) On differentiating both sides w.r.t.x, we get- \(\frac{d y}{d x}=-\frac{b}{a} e^{\frac{-x}{a}}\) \(\therefore \quad\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{(0, \mathrm{~b})}=\frac{-\mathrm{b}}{\mathrm{a}} \mathrm{e}^{-0 / \mathrm{a}}\) \(\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{(0, \mathrm{~b})}=\frac{-\mathrm{b}}{\mathrm{a}}\) \(\therefore\) Equation of tangent \(y-b=\frac{-b}{a}(x-0) \Rightarrow y-b=\frac{-b x}{a}\) \(\frac{y}{b}-1=\frac{-x}{a} \Rightarrow \frac{x}{a}+\frac{y}{b}=1\) Hence, \(\frac{\mathrm{x}}{\mathrm{a}}+\frac{\mathrm{y}}{\mathrm{b}}=1\) is tangent to the curve where the curve crosses the axis of \(y\).
WB JEE-2020
Application of Derivatives
85422
If the tangent to the curve \(y^{2}=x^{3} a t\left(m^{2}, m^{3}\right)\) is also a normal to the curve at \(\left(M^{2}, M^{2}\right)\), then the value of \(\mathrm{mM}\) is
1 \(-\frac{1}{6}\)
2 \(-\frac{2}{9}\)
3 \(-\frac{1}{3}\)
4 \(-\frac{4}{9}\)
Explanation:
(D) : Given, \(y^{2}=x^{3}\) On differentiating both sides w.r.t.x, we get- \(2 y \frac{d y}{d x}=3 x^{2}\) Slope, \(\quad \mathrm{m}=\frac{3 \mathrm{x}^{2}}{2 \mathrm{y}}\) Slope of tangent at \(\left(\mathrm{m}^{2}, \mathrm{~m}^{3}\right)\) \(\mathrm{m}_{1}=\frac{3\left(\mathrm{~m}^{2}\right)^{2}}{2 \mathrm{~m}^{3}} \Rightarrow \mathrm{m}_{1}=\frac{3 \mathrm{~m}^{4}}{2 \mathrm{~m}^{3}}\) \(\mathrm{~m}_{1}=\frac{3 \mathrm{~m}}{2}\) Slope at \(\left[\mathrm{M}^{2}, \mathrm{M}^{3}\right]\) \(\mathrm{m}_{2}=\frac{3 \mathrm{M}}{2}\) Now, \(\mathrm{m}_{1} \mathrm{~m}_{2}=-1\) On putting the value of \(\mathrm{m}_{1} \mathrm{~m}_{2}\) \(\frac{3 m}{2} \cdot \frac{3 M}{2}=-1 \Rightarrow \frac{9 m M}{4}=-1\) \(m M=\frac{-4}{9}\)
85418
What is the \(x\)-coordinate of the point on the curve \(f(x)=\sqrt{x}(7 x-6)\), where the tangent is parallel to \(\mathrm{x}\)-axis
1 \(-\frac{1}{3}\)
2 \(\frac{2}{7}\)
3 \(\frac{6}{7}\)
4 \(\frac{1}{2}\)
Explanation:
(B) :Given, \(f(x)=\sqrt{x}(7 x-6)\) \(f(x)=7 x^{3 / 2}-6 x^{1 / 2}\) On differentiation both sides w.r.t.x, we get- \(f^{\prime}(x)=7 \times \frac{3}{2} x^{1 / 2}-6 \times \frac{1}{2} x^{-1 / 2}\) When tangent is parallel to \(x\) axis \(\mathrm{f}^{\prime}(\mathrm{x})=0\) \(\frac{21}{2} x^{1 / 2}-3 x^{-1 / 2}=0 \Rightarrow \frac{21}{2} \sqrt{x}=\frac{3}{\sqrt{x}}\) \(7 \mathrm{x}=2 \Rightarrow \mathrm{x}=\frac{2}{7}\)
BITSAT-2016
Application of Derivatives
85419
For the curve \(x^{2}+4 x y+8 y^{2}=64\) the tangents are parallel to the \(x\)-axis only at the points
1 \((0,2 \sqrt{2})\) and \((0,-2 \sqrt{2})\)
2 \((8,-4)\) and \((-8,4)\)
3 \((8 \sqrt{2},-2 \sqrt{2})\) and \((-8 \sqrt{2}, 2 \sqrt{2})\)
4 \((9,0)\) and \((-8,0)\)
Explanation:
(B) : Given, \(x^{2}+4 x y+8 y^{2}=64 \tag{i}\) On differentiating both sides w.r.t to \(x\) we get - \(2 x+4\left(y+x \frac{d y}{d x}\right)+16 y \frac{d y}{d x}=0\) \(2 x+4 y+(4 x+16 y) \frac{d y}{d x}=0\) \(\frac{d y}{d x}=-\frac{(x+2 y)}{2(x+4 y)}\) Since, tangent are parallel to \(x\)-axis only. \(\begin{aligned} & \frac{d y}{d x}=0 \Rightarrow 0=-\frac{(x+2 y)}{2(x+4 y)} \\ & x+2 y=0\end{aligned}\) On putting the values of \(x\) from equation (i) in (ii), we get \(4 y^{2}-8 y^{2}+8 y^{2}=64\) \(4 y^{2}=64\) \(y^{2}=16\) \(y= \pm 4\) On putting the value \(y\) in equation (ii) \(x+2 y=0\) \(x+2(4)=0\) \(x=-8, y=4\) \(x=+8, y=-4\) Hence, required point are \((-8,4)\) and \((8,-4)\)
WB JEE-2013
Application of Derivatives
85420
If the equation of one tangent to the circle with centre at \((2,-1)\) from the origin is \(3 x+y=0\), then the equation of the other tangent through the origin is
1 \(3 x-y=0\)
2 \(x+3 y=0\)
3 \(x-3 y=0\)
4 \(x+2 y=0\)
Explanation:
(C) : Given, \(3 x+y=0\) Radius \(=\mathrm{r}\) (let) The lengths of the perpendicular from the centre \((2,-1)\) to the line \(3 x+y=0\) \(r=\frac{6-1}{\sqrt{(9+1)}}\) \(r=\frac{5}{\sqrt{10}}\) Let, \(y=m x\) be other tangent from origin, Then, \(\frac{|2 \mathrm{~m}+1|}{\sqrt{\left(1+\mathrm{m}^{2}\right)}}=\frac{5}{\sqrt{10}}\) On squaring both sides- \(\frac{4 m^{2}+1+4 m}{1+m^{2}}=\frac{25}{10}\) \(8 m^{2}+2+8 m=5+5 m^{2}\) \(3 m^{2}+8 m-3=0\) \(m=-3,1 / 3\) As, -3 is the slope of given tangent so \(1 / 3\) is that of the other tangent. Hence, \(y=\left(\frac{1}{3}\right) \mathrm{x}\) \(3 \mathrm{y}=\mathrm{x}\) \(3 \mathrm{y}-\mathrm{x}=0\) \(\mathrm{x}-3 \mathrm{y}=0\)
WB JEE-2022
Application of Derivatives
85421
Consider the curve \(y=b e^{-x / a}\) where \(a\) and \(b\) are non-zero real numbers. Then
1 \(\frac{x}{a}+\frac{y}{b}=1\) is tangent to the curve at \((0,0)\)
2 \(\frac{x}{a}+\frac{y}{b}=1\) is tangent to the curve, where the curve crosses the axis of \(y\)
3 \(\frac{x}{a}+\frac{y}{b}=1\) is tangent to the curve at \((a, 0)\)
4 \(\frac{x}{a}+\frac{y}{b}=1\) is tangent to the curve at \((2 a, 0)\)
Explanation:
(B) : Given the curve, \(y=b e^{-x / a}\) On differentiating both sides w.r.t.x, we get- \(\frac{d y}{d x}=-\frac{b}{a} e^{\frac{-x}{a}}\) \(\therefore \quad\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{(0, \mathrm{~b})}=\frac{-\mathrm{b}}{\mathrm{a}} \mathrm{e}^{-0 / \mathrm{a}}\) \(\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{(0, \mathrm{~b})}=\frac{-\mathrm{b}}{\mathrm{a}}\) \(\therefore\) Equation of tangent \(y-b=\frac{-b}{a}(x-0) \Rightarrow y-b=\frac{-b x}{a}\) \(\frac{y}{b}-1=\frac{-x}{a} \Rightarrow \frac{x}{a}+\frac{y}{b}=1\) Hence, \(\frac{\mathrm{x}}{\mathrm{a}}+\frac{\mathrm{y}}{\mathrm{b}}=1\) is tangent to the curve where the curve crosses the axis of \(y\).
WB JEE-2020
Application of Derivatives
85422
If the tangent to the curve \(y^{2}=x^{3} a t\left(m^{2}, m^{3}\right)\) is also a normal to the curve at \(\left(M^{2}, M^{2}\right)\), then the value of \(\mathrm{mM}\) is
1 \(-\frac{1}{6}\)
2 \(-\frac{2}{9}\)
3 \(-\frac{1}{3}\)
4 \(-\frac{4}{9}\)
Explanation:
(D) : Given, \(y^{2}=x^{3}\) On differentiating both sides w.r.t.x, we get- \(2 y \frac{d y}{d x}=3 x^{2}\) Slope, \(\quad \mathrm{m}=\frac{3 \mathrm{x}^{2}}{2 \mathrm{y}}\) Slope of tangent at \(\left(\mathrm{m}^{2}, \mathrm{~m}^{3}\right)\) \(\mathrm{m}_{1}=\frac{3\left(\mathrm{~m}^{2}\right)^{2}}{2 \mathrm{~m}^{3}} \Rightarrow \mathrm{m}_{1}=\frac{3 \mathrm{~m}^{4}}{2 \mathrm{~m}^{3}}\) \(\mathrm{~m}_{1}=\frac{3 \mathrm{~m}}{2}\) Slope at \(\left[\mathrm{M}^{2}, \mathrm{M}^{3}\right]\) \(\mathrm{m}_{2}=\frac{3 \mathrm{M}}{2}\) Now, \(\mathrm{m}_{1} \mathrm{~m}_{2}=-1\) On putting the value of \(\mathrm{m}_{1} \mathrm{~m}_{2}\) \(\frac{3 m}{2} \cdot \frac{3 M}{2}=-1 \Rightarrow \frac{9 m M}{4}=-1\) \(m M=\frac{-4}{9}\)
85418
What is the \(x\)-coordinate of the point on the curve \(f(x)=\sqrt{x}(7 x-6)\), where the tangent is parallel to \(\mathrm{x}\)-axis
1 \(-\frac{1}{3}\)
2 \(\frac{2}{7}\)
3 \(\frac{6}{7}\)
4 \(\frac{1}{2}\)
Explanation:
(B) :Given, \(f(x)=\sqrt{x}(7 x-6)\) \(f(x)=7 x^{3 / 2}-6 x^{1 / 2}\) On differentiation both sides w.r.t.x, we get- \(f^{\prime}(x)=7 \times \frac{3}{2} x^{1 / 2}-6 \times \frac{1}{2} x^{-1 / 2}\) When tangent is parallel to \(x\) axis \(\mathrm{f}^{\prime}(\mathrm{x})=0\) \(\frac{21}{2} x^{1 / 2}-3 x^{-1 / 2}=0 \Rightarrow \frac{21}{2} \sqrt{x}=\frac{3}{\sqrt{x}}\) \(7 \mathrm{x}=2 \Rightarrow \mathrm{x}=\frac{2}{7}\)
BITSAT-2016
Application of Derivatives
85419
For the curve \(x^{2}+4 x y+8 y^{2}=64\) the tangents are parallel to the \(x\)-axis only at the points
1 \((0,2 \sqrt{2})\) and \((0,-2 \sqrt{2})\)
2 \((8,-4)\) and \((-8,4)\)
3 \((8 \sqrt{2},-2 \sqrt{2})\) and \((-8 \sqrt{2}, 2 \sqrt{2})\)
4 \((9,0)\) and \((-8,0)\)
Explanation:
(B) : Given, \(x^{2}+4 x y+8 y^{2}=64 \tag{i}\) On differentiating both sides w.r.t to \(x\) we get - \(2 x+4\left(y+x \frac{d y}{d x}\right)+16 y \frac{d y}{d x}=0\) \(2 x+4 y+(4 x+16 y) \frac{d y}{d x}=0\) \(\frac{d y}{d x}=-\frac{(x+2 y)}{2(x+4 y)}\) Since, tangent are parallel to \(x\)-axis only. \(\begin{aligned} & \frac{d y}{d x}=0 \Rightarrow 0=-\frac{(x+2 y)}{2(x+4 y)} \\ & x+2 y=0\end{aligned}\) On putting the values of \(x\) from equation (i) in (ii), we get \(4 y^{2}-8 y^{2}+8 y^{2}=64\) \(4 y^{2}=64\) \(y^{2}=16\) \(y= \pm 4\) On putting the value \(y\) in equation (ii) \(x+2 y=0\) \(x+2(4)=0\) \(x=-8, y=4\) \(x=+8, y=-4\) Hence, required point are \((-8,4)\) and \((8,-4)\)
WB JEE-2013
Application of Derivatives
85420
If the equation of one tangent to the circle with centre at \((2,-1)\) from the origin is \(3 x+y=0\), then the equation of the other tangent through the origin is
1 \(3 x-y=0\)
2 \(x+3 y=0\)
3 \(x-3 y=0\)
4 \(x+2 y=0\)
Explanation:
(C) : Given, \(3 x+y=0\) Radius \(=\mathrm{r}\) (let) The lengths of the perpendicular from the centre \((2,-1)\) to the line \(3 x+y=0\) \(r=\frac{6-1}{\sqrt{(9+1)}}\) \(r=\frac{5}{\sqrt{10}}\) Let, \(y=m x\) be other tangent from origin, Then, \(\frac{|2 \mathrm{~m}+1|}{\sqrt{\left(1+\mathrm{m}^{2}\right)}}=\frac{5}{\sqrt{10}}\) On squaring both sides- \(\frac{4 m^{2}+1+4 m}{1+m^{2}}=\frac{25}{10}\) \(8 m^{2}+2+8 m=5+5 m^{2}\) \(3 m^{2}+8 m-3=0\) \(m=-3,1 / 3\) As, -3 is the slope of given tangent so \(1 / 3\) is that of the other tangent. Hence, \(y=\left(\frac{1}{3}\right) \mathrm{x}\) \(3 \mathrm{y}=\mathrm{x}\) \(3 \mathrm{y}-\mathrm{x}=0\) \(\mathrm{x}-3 \mathrm{y}=0\)
WB JEE-2022
Application of Derivatives
85421
Consider the curve \(y=b e^{-x / a}\) where \(a\) and \(b\) are non-zero real numbers. Then
1 \(\frac{x}{a}+\frac{y}{b}=1\) is tangent to the curve at \((0,0)\)
2 \(\frac{x}{a}+\frac{y}{b}=1\) is tangent to the curve, where the curve crosses the axis of \(y\)
3 \(\frac{x}{a}+\frac{y}{b}=1\) is tangent to the curve at \((a, 0)\)
4 \(\frac{x}{a}+\frac{y}{b}=1\) is tangent to the curve at \((2 a, 0)\)
Explanation:
(B) : Given the curve, \(y=b e^{-x / a}\) On differentiating both sides w.r.t.x, we get- \(\frac{d y}{d x}=-\frac{b}{a} e^{\frac{-x}{a}}\) \(\therefore \quad\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{(0, \mathrm{~b})}=\frac{-\mathrm{b}}{\mathrm{a}} \mathrm{e}^{-0 / \mathrm{a}}\) \(\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{(0, \mathrm{~b})}=\frac{-\mathrm{b}}{\mathrm{a}}\) \(\therefore\) Equation of tangent \(y-b=\frac{-b}{a}(x-0) \Rightarrow y-b=\frac{-b x}{a}\) \(\frac{y}{b}-1=\frac{-x}{a} \Rightarrow \frac{x}{a}+\frac{y}{b}=1\) Hence, \(\frac{\mathrm{x}}{\mathrm{a}}+\frac{\mathrm{y}}{\mathrm{b}}=1\) is tangent to the curve where the curve crosses the axis of \(y\).
WB JEE-2020
Application of Derivatives
85422
If the tangent to the curve \(y^{2}=x^{3} a t\left(m^{2}, m^{3}\right)\) is also a normal to the curve at \(\left(M^{2}, M^{2}\right)\), then the value of \(\mathrm{mM}\) is
1 \(-\frac{1}{6}\)
2 \(-\frac{2}{9}\)
3 \(-\frac{1}{3}\)
4 \(-\frac{4}{9}\)
Explanation:
(D) : Given, \(y^{2}=x^{3}\) On differentiating both sides w.r.t.x, we get- \(2 y \frac{d y}{d x}=3 x^{2}\) Slope, \(\quad \mathrm{m}=\frac{3 \mathrm{x}^{2}}{2 \mathrm{y}}\) Slope of tangent at \(\left(\mathrm{m}^{2}, \mathrm{~m}^{3}\right)\) \(\mathrm{m}_{1}=\frac{3\left(\mathrm{~m}^{2}\right)^{2}}{2 \mathrm{~m}^{3}} \Rightarrow \mathrm{m}_{1}=\frac{3 \mathrm{~m}^{4}}{2 \mathrm{~m}^{3}}\) \(\mathrm{~m}_{1}=\frac{3 \mathrm{~m}}{2}\) Slope at \(\left[\mathrm{M}^{2}, \mathrm{M}^{3}\right]\) \(\mathrm{m}_{2}=\frac{3 \mathrm{M}}{2}\) Now, \(\mathrm{m}_{1} \mathrm{~m}_{2}=-1\) On putting the value of \(\mathrm{m}_{1} \mathrm{~m}_{2}\) \(\frac{3 m}{2} \cdot \frac{3 M}{2}=-1 \Rightarrow \frac{9 m M}{4}=-1\) \(m M=\frac{-4}{9}\)