85405
Find the equations of the tangent and the normal to the curve \(y=\frac{x^{3}}{4-x}\) at \((2,4)\).
1 \(8 x+y-12=0 ; x+8 y+34=0\)
2 \(8 x-y-12=0 ; x-8 y-34=0\)
3 \(8 x+4 y+12=0 ; x-8 y+34=0\)
4 \(8 x-y-12=0 ; x+8 y-34=0\)
Explanation:
(D) : Given the curve, \(y=\frac{x^{3}}{4-x} \tag{i}\) Differentiating (i) w.r.t.x, we get \(\frac{d y}{d x}=\frac{(4-x)\left(3 x^{2}\right)-x^{3}(-1)}{(4-x)^{2}}=\frac{12 x^{2}-2 x^{3}}{(4-x)^{2}}\) \(\left(\frac{d y}{d x}\right)_{x=2}=\frac{12(2)^{2}-2(2)^{3}}{(4-2)^{2}}=\frac{32}{4}=8\) The slope of tangent at \((2,4)\) is 8 . \(\therefore\) The equation of tangent is \(y-4=8(x-2)\) \(y-4=8 x-16\) \(8 x-y-12=0\) Now the slope of normal at \(\mathrm{P}(2,4)\) is \(\frac{-1}{8}\). \(\therefore\) The equation of normal is \(y-4=\frac{-1}{8}(x-2)\) \(x+8 y-34=0\)
COMEDK-2017
Application of Derivatives
85406
Find the point on the curve \(y=x^{2}\) where the slope of the tangent is equal to the \(x\)-coordinate of the point.
1 \((1,1)\)
2 \((-1,-1)\)
3 \((1,2)\)
4 \((0,0)\)
Explanation:
(D) : Let the required point be \(\mathrm{P}\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)\). The given curve is \(y=x^{2}\) ..... (i) Differentiating equation (i) w.r.t.x, we get- \(\frac{\mathrm{dy}}{\mathrm{dx}}=2 \mathrm{x}\) \(\text { Slope of tangent at } \mathrm{P}\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)=2 \mathrm{x}_{1} \text {. }\) But the slope of tangent at, \(\mathrm{P}\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)=\mathrm{x}\)-coordinate of \(\mathrm{P}\) \(2 x_{1}=x_{1}\) \(2 x_{1}-x_{1}=0 \Rightarrow x_{1}=0\) As \(\mathrm{P}\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)\) lies on the curve, we get \(\mathrm{y}_{1}=\mathrm{x}_{1}^{2}\) \(\mathrm{y}_{1}=0\) Hence, the required point is \((0,0)\).
COMEDK-2017
Application of Derivatives
85407
Find the equation of the tangent to the curve \(y=x^{2}-5 x+6\) at \((2,0)\)
1 \(x-y-2=0\)
2 \(x+y-2=0\)
3 \(x-y+2=0\)
4 \(x+y+2=0\)
Explanation:
(B) : The given curve is, \(y=x^{2}-5 x+6 \tag{i}\) Differentiating equation (i) w.r.t.x, we get \(\frac{d y}{d x}=2 x-5\) \(\left(\frac{d y}{d x}\right)_{(2,0)}=2(2)-5=-1\) The slope of tangent at \((2,0)\) is -1 . \(\therefore\) The equation of tangent is \((y-0)=-1(x-2)\) \(y=-x+2\) \(x+y-2=0\)
COMEDK-2018
Application of Derivatives
85408
Find the equations of the tangent and the normal to the curve \(y=x^{3}\) at \((2,8)\)
1 \(y=12 x-16 ; x+12 y-98=0\)
2 \(y=12 x-6 ; x-12 y+98=0\)
3 \(y=12 x+16 ; x-12 y-98=0\)
4 \(y=12 x-16 ; x+12 y+98=0\)
Explanation:
(A) : The given curve is, \(y=x^{3} \tag{i}\) Differentiating equation (i) w.r.t. \(x\), we get- \(\frac{d y}{d x}=3 x^{2}\) \(\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{(2,8)}=3(2)^{2}=12\) \(\therefore\) The slope of tangent at \(\mathrm{P}(2,8)\) is 12 . \(\therefore\) The equation of tangent is, \((y-8)=12(x-2)\) \(y=12 x-16\) Now the slope of normal at \(\mathrm{P}(2,8)\) is \(\frac{-1}{12}\) \(\therefore\) The equation of normal is \((y-8)=\frac{-1}{12}(x-2)\) \(x+12 y-98=0\)
COMEDK-2020
Application of Derivatives
85409
If the tangent at each point of the curve \(y=\) \(\frac{2}{3} x^{3}-2 a x^{2}+2 x+5\) makes an acute angle with the positive direction of \(x\)-axis, then
1 \(\mathrm{a} \geq 1\)
2 \(-1 \leq \mathrm{a} \leq 1\)
3 \(\mathrm{a} \leq-1\)
4 none of these
Explanation:
(B) : The given curve is, \(y=\frac{2}{3} x^{3}-2 a x^{2}+2 x+5\) Differentiating w.r.t. \(\mathrm{x}\), we get \(\frac{d y}{d x}=2 x^{2}-4 a x+2\) Since, tangent is inclined at an acute angle with the positive direction of \(\mathrm{x}\)-axis, \(\therefore \quad \frac{d y}{d x} \geq 0 \Rightarrow 2 x^{2}-4 a x+2 \geq 0 \text { for all } x \in R\) \(\Rightarrow(-4 a)^{2}-4(2)(2) \leq 0 \quad\left[\because b^{2}-4 a c \leq 0\right. \text { when }\) \(\left.\quad a x^{2}+b x+c \geq 0 \text { and } a>0\right]\) \(\Rightarrow 16 a^{2}-16 \leq 0 \Rightarrow a^{2} \leq 1 \Rightarrow-1 \leq a \leq 1\)
85405
Find the equations of the tangent and the normal to the curve \(y=\frac{x^{3}}{4-x}\) at \((2,4)\).
1 \(8 x+y-12=0 ; x+8 y+34=0\)
2 \(8 x-y-12=0 ; x-8 y-34=0\)
3 \(8 x+4 y+12=0 ; x-8 y+34=0\)
4 \(8 x-y-12=0 ; x+8 y-34=0\)
Explanation:
(D) : Given the curve, \(y=\frac{x^{3}}{4-x} \tag{i}\) Differentiating (i) w.r.t.x, we get \(\frac{d y}{d x}=\frac{(4-x)\left(3 x^{2}\right)-x^{3}(-1)}{(4-x)^{2}}=\frac{12 x^{2}-2 x^{3}}{(4-x)^{2}}\) \(\left(\frac{d y}{d x}\right)_{x=2}=\frac{12(2)^{2}-2(2)^{3}}{(4-2)^{2}}=\frac{32}{4}=8\) The slope of tangent at \((2,4)\) is 8 . \(\therefore\) The equation of tangent is \(y-4=8(x-2)\) \(y-4=8 x-16\) \(8 x-y-12=0\) Now the slope of normal at \(\mathrm{P}(2,4)\) is \(\frac{-1}{8}\). \(\therefore\) The equation of normal is \(y-4=\frac{-1}{8}(x-2)\) \(x+8 y-34=0\)
COMEDK-2017
Application of Derivatives
85406
Find the point on the curve \(y=x^{2}\) where the slope of the tangent is equal to the \(x\)-coordinate of the point.
1 \((1,1)\)
2 \((-1,-1)\)
3 \((1,2)\)
4 \((0,0)\)
Explanation:
(D) : Let the required point be \(\mathrm{P}\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)\). The given curve is \(y=x^{2}\) ..... (i) Differentiating equation (i) w.r.t.x, we get- \(\frac{\mathrm{dy}}{\mathrm{dx}}=2 \mathrm{x}\) \(\text { Slope of tangent at } \mathrm{P}\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)=2 \mathrm{x}_{1} \text {. }\) But the slope of tangent at, \(\mathrm{P}\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)=\mathrm{x}\)-coordinate of \(\mathrm{P}\) \(2 x_{1}=x_{1}\) \(2 x_{1}-x_{1}=0 \Rightarrow x_{1}=0\) As \(\mathrm{P}\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)\) lies on the curve, we get \(\mathrm{y}_{1}=\mathrm{x}_{1}^{2}\) \(\mathrm{y}_{1}=0\) Hence, the required point is \((0,0)\).
COMEDK-2017
Application of Derivatives
85407
Find the equation of the tangent to the curve \(y=x^{2}-5 x+6\) at \((2,0)\)
1 \(x-y-2=0\)
2 \(x+y-2=0\)
3 \(x-y+2=0\)
4 \(x+y+2=0\)
Explanation:
(B) : The given curve is, \(y=x^{2}-5 x+6 \tag{i}\) Differentiating equation (i) w.r.t.x, we get \(\frac{d y}{d x}=2 x-5\) \(\left(\frac{d y}{d x}\right)_{(2,0)}=2(2)-5=-1\) The slope of tangent at \((2,0)\) is -1 . \(\therefore\) The equation of tangent is \((y-0)=-1(x-2)\) \(y=-x+2\) \(x+y-2=0\)
COMEDK-2018
Application of Derivatives
85408
Find the equations of the tangent and the normal to the curve \(y=x^{3}\) at \((2,8)\)
1 \(y=12 x-16 ; x+12 y-98=0\)
2 \(y=12 x-6 ; x-12 y+98=0\)
3 \(y=12 x+16 ; x-12 y-98=0\)
4 \(y=12 x-16 ; x+12 y+98=0\)
Explanation:
(A) : The given curve is, \(y=x^{3} \tag{i}\) Differentiating equation (i) w.r.t. \(x\), we get- \(\frac{d y}{d x}=3 x^{2}\) \(\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{(2,8)}=3(2)^{2}=12\) \(\therefore\) The slope of tangent at \(\mathrm{P}(2,8)\) is 12 . \(\therefore\) The equation of tangent is, \((y-8)=12(x-2)\) \(y=12 x-16\) Now the slope of normal at \(\mathrm{P}(2,8)\) is \(\frac{-1}{12}\) \(\therefore\) The equation of normal is \((y-8)=\frac{-1}{12}(x-2)\) \(x+12 y-98=0\)
COMEDK-2020
Application of Derivatives
85409
If the tangent at each point of the curve \(y=\) \(\frac{2}{3} x^{3}-2 a x^{2}+2 x+5\) makes an acute angle with the positive direction of \(x\)-axis, then
1 \(\mathrm{a} \geq 1\)
2 \(-1 \leq \mathrm{a} \leq 1\)
3 \(\mathrm{a} \leq-1\)
4 none of these
Explanation:
(B) : The given curve is, \(y=\frac{2}{3} x^{3}-2 a x^{2}+2 x+5\) Differentiating w.r.t. \(\mathrm{x}\), we get \(\frac{d y}{d x}=2 x^{2}-4 a x+2\) Since, tangent is inclined at an acute angle with the positive direction of \(\mathrm{x}\)-axis, \(\therefore \quad \frac{d y}{d x} \geq 0 \Rightarrow 2 x^{2}-4 a x+2 \geq 0 \text { for all } x \in R\) \(\Rightarrow(-4 a)^{2}-4(2)(2) \leq 0 \quad\left[\because b^{2}-4 a c \leq 0\right. \text { when }\) \(\left.\quad a x^{2}+b x+c \geq 0 \text { and } a>0\right]\) \(\Rightarrow 16 a^{2}-16 \leq 0 \Rightarrow a^{2} \leq 1 \Rightarrow-1 \leq a \leq 1\)
85405
Find the equations of the tangent and the normal to the curve \(y=\frac{x^{3}}{4-x}\) at \((2,4)\).
1 \(8 x+y-12=0 ; x+8 y+34=0\)
2 \(8 x-y-12=0 ; x-8 y-34=0\)
3 \(8 x+4 y+12=0 ; x-8 y+34=0\)
4 \(8 x-y-12=0 ; x+8 y-34=0\)
Explanation:
(D) : Given the curve, \(y=\frac{x^{3}}{4-x} \tag{i}\) Differentiating (i) w.r.t.x, we get \(\frac{d y}{d x}=\frac{(4-x)\left(3 x^{2}\right)-x^{3}(-1)}{(4-x)^{2}}=\frac{12 x^{2}-2 x^{3}}{(4-x)^{2}}\) \(\left(\frac{d y}{d x}\right)_{x=2}=\frac{12(2)^{2}-2(2)^{3}}{(4-2)^{2}}=\frac{32}{4}=8\) The slope of tangent at \((2,4)\) is 8 . \(\therefore\) The equation of tangent is \(y-4=8(x-2)\) \(y-4=8 x-16\) \(8 x-y-12=0\) Now the slope of normal at \(\mathrm{P}(2,4)\) is \(\frac{-1}{8}\). \(\therefore\) The equation of normal is \(y-4=\frac{-1}{8}(x-2)\) \(x+8 y-34=0\)
COMEDK-2017
Application of Derivatives
85406
Find the point on the curve \(y=x^{2}\) where the slope of the tangent is equal to the \(x\)-coordinate of the point.
1 \((1,1)\)
2 \((-1,-1)\)
3 \((1,2)\)
4 \((0,0)\)
Explanation:
(D) : Let the required point be \(\mathrm{P}\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)\). The given curve is \(y=x^{2}\) ..... (i) Differentiating equation (i) w.r.t.x, we get- \(\frac{\mathrm{dy}}{\mathrm{dx}}=2 \mathrm{x}\) \(\text { Slope of tangent at } \mathrm{P}\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)=2 \mathrm{x}_{1} \text {. }\) But the slope of tangent at, \(\mathrm{P}\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)=\mathrm{x}\)-coordinate of \(\mathrm{P}\) \(2 x_{1}=x_{1}\) \(2 x_{1}-x_{1}=0 \Rightarrow x_{1}=0\) As \(\mathrm{P}\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)\) lies on the curve, we get \(\mathrm{y}_{1}=\mathrm{x}_{1}^{2}\) \(\mathrm{y}_{1}=0\) Hence, the required point is \((0,0)\).
COMEDK-2017
Application of Derivatives
85407
Find the equation of the tangent to the curve \(y=x^{2}-5 x+6\) at \((2,0)\)
1 \(x-y-2=0\)
2 \(x+y-2=0\)
3 \(x-y+2=0\)
4 \(x+y+2=0\)
Explanation:
(B) : The given curve is, \(y=x^{2}-5 x+6 \tag{i}\) Differentiating equation (i) w.r.t.x, we get \(\frac{d y}{d x}=2 x-5\) \(\left(\frac{d y}{d x}\right)_{(2,0)}=2(2)-5=-1\) The slope of tangent at \((2,0)\) is -1 . \(\therefore\) The equation of tangent is \((y-0)=-1(x-2)\) \(y=-x+2\) \(x+y-2=0\)
COMEDK-2018
Application of Derivatives
85408
Find the equations of the tangent and the normal to the curve \(y=x^{3}\) at \((2,8)\)
1 \(y=12 x-16 ; x+12 y-98=0\)
2 \(y=12 x-6 ; x-12 y+98=0\)
3 \(y=12 x+16 ; x-12 y-98=0\)
4 \(y=12 x-16 ; x+12 y+98=0\)
Explanation:
(A) : The given curve is, \(y=x^{3} \tag{i}\) Differentiating equation (i) w.r.t. \(x\), we get- \(\frac{d y}{d x}=3 x^{2}\) \(\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{(2,8)}=3(2)^{2}=12\) \(\therefore\) The slope of tangent at \(\mathrm{P}(2,8)\) is 12 . \(\therefore\) The equation of tangent is, \((y-8)=12(x-2)\) \(y=12 x-16\) Now the slope of normal at \(\mathrm{P}(2,8)\) is \(\frac{-1}{12}\) \(\therefore\) The equation of normal is \((y-8)=\frac{-1}{12}(x-2)\) \(x+12 y-98=0\)
COMEDK-2020
Application of Derivatives
85409
If the tangent at each point of the curve \(y=\) \(\frac{2}{3} x^{3}-2 a x^{2}+2 x+5\) makes an acute angle with the positive direction of \(x\)-axis, then
1 \(\mathrm{a} \geq 1\)
2 \(-1 \leq \mathrm{a} \leq 1\)
3 \(\mathrm{a} \leq-1\)
4 none of these
Explanation:
(B) : The given curve is, \(y=\frac{2}{3} x^{3}-2 a x^{2}+2 x+5\) Differentiating w.r.t. \(\mathrm{x}\), we get \(\frac{d y}{d x}=2 x^{2}-4 a x+2\) Since, tangent is inclined at an acute angle with the positive direction of \(\mathrm{x}\)-axis, \(\therefore \quad \frac{d y}{d x} \geq 0 \Rightarrow 2 x^{2}-4 a x+2 \geq 0 \text { for all } x \in R\) \(\Rightarrow(-4 a)^{2}-4(2)(2) \leq 0 \quad\left[\because b^{2}-4 a c \leq 0\right. \text { when }\) \(\left.\quad a x^{2}+b x+c \geq 0 \text { and } a>0\right]\) \(\Rightarrow 16 a^{2}-16 \leq 0 \Rightarrow a^{2} \leq 1 \Rightarrow-1 \leq a \leq 1\)
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Application of Derivatives
85405
Find the equations of the tangent and the normal to the curve \(y=\frac{x^{3}}{4-x}\) at \((2,4)\).
1 \(8 x+y-12=0 ; x+8 y+34=0\)
2 \(8 x-y-12=0 ; x-8 y-34=0\)
3 \(8 x+4 y+12=0 ; x-8 y+34=0\)
4 \(8 x-y-12=0 ; x+8 y-34=0\)
Explanation:
(D) : Given the curve, \(y=\frac{x^{3}}{4-x} \tag{i}\) Differentiating (i) w.r.t.x, we get \(\frac{d y}{d x}=\frac{(4-x)\left(3 x^{2}\right)-x^{3}(-1)}{(4-x)^{2}}=\frac{12 x^{2}-2 x^{3}}{(4-x)^{2}}\) \(\left(\frac{d y}{d x}\right)_{x=2}=\frac{12(2)^{2}-2(2)^{3}}{(4-2)^{2}}=\frac{32}{4}=8\) The slope of tangent at \((2,4)\) is 8 . \(\therefore\) The equation of tangent is \(y-4=8(x-2)\) \(y-4=8 x-16\) \(8 x-y-12=0\) Now the slope of normal at \(\mathrm{P}(2,4)\) is \(\frac{-1}{8}\). \(\therefore\) The equation of normal is \(y-4=\frac{-1}{8}(x-2)\) \(x+8 y-34=0\)
COMEDK-2017
Application of Derivatives
85406
Find the point on the curve \(y=x^{2}\) where the slope of the tangent is equal to the \(x\)-coordinate of the point.
1 \((1,1)\)
2 \((-1,-1)\)
3 \((1,2)\)
4 \((0,0)\)
Explanation:
(D) : Let the required point be \(\mathrm{P}\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)\). The given curve is \(y=x^{2}\) ..... (i) Differentiating equation (i) w.r.t.x, we get- \(\frac{\mathrm{dy}}{\mathrm{dx}}=2 \mathrm{x}\) \(\text { Slope of tangent at } \mathrm{P}\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)=2 \mathrm{x}_{1} \text {. }\) But the slope of tangent at, \(\mathrm{P}\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)=\mathrm{x}\)-coordinate of \(\mathrm{P}\) \(2 x_{1}=x_{1}\) \(2 x_{1}-x_{1}=0 \Rightarrow x_{1}=0\) As \(\mathrm{P}\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)\) lies on the curve, we get \(\mathrm{y}_{1}=\mathrm{x}_{1}^{2}\) \(\mathrm{y}_{1}=0\) Hence, the required point is \((0,0)\).
COMEDK-2017
Application of Derivatives
85407
Find the equation of the tangent to the curve \(y=x^{2}-5 x+6\) at \((2,0)\)
1 \(x-y-2=0\)
2 \(x+y-2=0\)
3 \(x-y+2=0\)
4 \(x+y+2=0\)
Explanation:
(B) : The given curve is, \(y=x^{2}-5 x+6 \tag{i}\) Differentiating equation (i) w.r.t.x, we get \(\frac{d y}{d x}=2 x-5\) \(\left(\frac{d y}{d x}\right)_{(2,0)}=2(2)-5=-1\) The slope of tangent at \((2,0)\) is -1 . \(\therefore\) The equation of tangent is \((y-0)=-1(x-2)\) \(y=-x+2\) \(x+y-2=0\)
COMEDK-2018
Application of Derivatives
85408
Find the equations of the tangent and the normal to the curve \(y=x^{3}\) at \((2,8)\)
1 \(y=12 x-16 ; x+12 y-98=0\)
2 \(y=12 x-6 ; x-12 y+98=0\)
3 \(y=12 x+16 ; x-12 y-98=0\)
4 \(y=12 x-16 ; x+12 y+98=0\)
Explanation:
(A) : The given curve is, \(y=x^{3} \tag{i}\) Differentiating equation (i) w.r.t. \(x\), we get- \(\frac{d y}{d x}=3 x^{2}\) \(\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{(2,8)}=3(2)^{2}=12\) \(\therefore\) The slope of tangent at \(\mathrm{P}(2,8)\) is 12 . \(\therefore\) The equation of tangent is, \((y-8)=12(x-2)\) \(y=12 x-16\) Now the slope of normal at \(\mathrm{P}(2,8)\) is \(\frac{-1}{12}\) \(\therefore\) The equation of normal is \((y-8)=\frac{-1}{12}(x-2)\) \(x+12 y-98=0\)
COMEDK-2020
Application of Derivatives
85409
If the tangent at each point of the curve \(y=\) \(\frac{2}{3} x^{3}-2 a x^{2}+2 x+5\) makes an acute angle with the positive direction of \(x\)-axis, then
1 \(\mathrm{a} \geq 1\)
2 \(-1 \leq \mathrm{a} \leq 1\)
3 \(\mathrm{a} \leq-1\)
4 none of these
Explanation:
(B) : The given curve is, \(y=\frac{2}{3} x^{3}-2 a x^{2}+2 x+5\) Differentiating w.r.t. \(\mathrm{x}\), we get \(\frac{d y}{d x}=2 x^{2}-4 a x+2\) Since, tangent is inclined at an acute angle with the positive direction of \(\mathrm{x}\)-axis, \(\therefore \quad \frac{d y}{d x} \geq 0 \Rightarrow 2 x^{2}-4 a x+2 \geq 0 \text { for all } x \in R\) \(\Rightarrow(-4 a)^{2}-4(2)(2) \leq 0 \quad\left[\because b^{2}-4 a c \leq 0\right. \text { when }\) \(\left.\quad a x^{2}+b x+c \geq 0 \text { and } a>0\right]\) \(\Rightarrow 16 a^{2}-16 \leq 0 \Rightarrow a^{2} \leq 1 \Rightarrow-1 \leq a \leq 1\)
85405
Find the equations of the tangent and the normal to the curve \(y=\frac{x^{3}}{4-x}\) at \((2,4)\).
1 \(8 x+y-12=0 ; x+8 y+34=0\)
2 \(8 x-y-12=0 ; x-8 y-34=0\)
3 \(8 x+4 y+12=0 ; x-8 y+34=0\)
4 \(8 x-y-12=0 ; x+8 y-34=0\)
Explanation:
(D) : Given the curve, \(y=\frac{x^{3}}{4-x} \tag{i}\) Differentiating (i) w.r.t.x, we get \(\frac{d y}{d x}=\frac{(4-x)\left(3 x^{2}\right)-x^{3}(-1)}{(4-x)^{2}}=\frac{12 x^{2}-2 x^{3}}{(4-x)^{2}}\) \(\left(\frac{d y}{d x}\right)_{x=2}=\frac{12(2)^{2}-2(2)^{3}}{(4-2)^{2}}=\frac{32}{4}=8\) The slope of tangent at \((2,4)\) is 8 . \(\therefore\) The equation of tangent is \(y-4=8(x-2)\) \(y-4=8 x-16\) \(8 x-y-12=0\) Now the slope of normal at \(\mathrm{P}(2,4)\) is \(\frac{-1}{8}\). \(\therefore\) The equation of normal is \(y-4=\frac{-1}{8}(x-2)\) \(x+8 y-34=0\)
COMEDK-2017
Application of Derivatives
85406
Find the point on the curve \(y=x^{2}\) where the slope of the tangent is equal to the \(x\)-coordinate of the point.
1 \((1,1)\)
2 \((-1,-1)\)
3 \((1,2)\)
4 \((0,0)\)
Explanation:
(D) : Let the required point be \(\mathrm{P}\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)\). The given curve is \(y=x^{2}\) ..... (i) Differentiating equation (i) w.r.t.x, we get- \(\frac{\mathrm{dy}}{\mathrm{dx}}=2 \mathrm{x}\) \(\text { Slope of tangent at } \mathrm{P}\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)=2 \mathrm{x}_{1} \text {. }\) But the slope of tangent at, \(\mathrm{P}\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)=\mathrm{x}\)-coordinate of \(\mathrm{P}\) \(2 x_{1}=x_{1}\) \(2 x_{1}-x_{1}=0 \Rightarrow x_{1}=0\) As \(\mathrm{P}\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)\) lies on the curve, we get \(\mathrm{y}_{1}=\mathrm{x}_{1}^{2}\) \(\mathrm{y}_{1}=0\) Hence, the required point is \((0,0)\).
COMEDK-2017
Application of Derivatives
85407
Find the equation of the tangent to the curve \(y=x^{2}-5 x+6\) at \((2,0)\)
1 \(x-y-2=0\)
2 \(x+y-2=0\)
3 \(x-y+2=0\)
4 \(x+y+2=0\)
Explanation:
(B) : The given curve is, \(y=x^{2}-5 x+6 \tag{i}\) Differentiating equation (i) w.r.t.x, we get \(\frac{d y}{d x}=2 x-5\) \(\left(\frac{d y}{d x}\right)_{(2,0)}=2(2)-5=-1\) The slope of tangent at \((2,0)\) is -1 . \(\therefore\) The equation of tangent is \((y-0)=-1(x-2)\) \(y=-x+2\) \(x+y-2=0\)
COMEDK-2018
Application of Derivatives
85408
Find the equations of the tangent and the normal to the curve \(y=x^{3}\) at \((2,8)\)
1 \(y=12 x-16 ; x+12 y-98=0\)
2 \(y=12 x-6 ; x-12 y+98=0\)
3 \(y=12 x+16 ; x-12 y-98=0\)
4 \(y=12 x-16 ; x+12 y+98=0\)
Explanation:
(A) : The given curve is, \(y=x^{3} \tag{i}\) Differentiating equation (i) w.r.t. \(x\), we get- \(\frac{d y}{d x}=3 x^{2}\) \(\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{(2,8)}=3(2)^{2}=12\) \(\therefore\) The slope of tangent at \(\mathrm{P}(2,8)\) is 12 . \(\therefore\) The equation of tangent is, \((y-8)=12(x-2)\) \(y=12 x-16\) Now the slope of normal at \(\mathrm{P}(2,8)\) is \(\frac{-1}{12}\) \(\therefore\) The equation of normal is \((y-8)=\frac{-1}{12}(x-2)\) \(x+12 y-98=0\)
COMEDK-2020
Application of Derivatives
85409
If the tangent at each point of the curve \(y=\) \(\frac{2}{3} x^{3}-2 a x^{2}+2 x+5\) makes an acute angle with the positive direction of \(x\)-axis, then
1 \(\mathrm{a} \geq 1\)
2 \(-1 \leq \mathrm{a} \leq 1\)
3 \(\mathrm{a} \leq-1\)
4 none of these
Explanation:
(B) : The given curve is, \(y=\frac{2}{3} x^{3}-2 a x^{2}+2 x+5\) Differentiating w.r.t. \(\mathrm{x}\), we get \(\frac{d y}{d x}=2 x^{2}-4 a x+2\) Since, tangent is inclined at an acute angle with the positive direction of \(\mathrm{x}\)-axis, \(\therefore \quad \frac{d y}{d x} \geq 0 \Rightarrow 2 x^{2}-4 a x+2 \geq 0 \text { for all } x \in R\) \(\Rightarrow(-4 a)^{2}-4(2)(2) \leq 0 \quad\left[\because b^{2}-4 a c \leq 0\right. \text { when }\) \(\left.\quad a x^{2}+b x+c \geq 0 \text { and } a>0\right]\) \(\Rightarrow 16 a^{2}-16 \leq 0 \Rightarrow a^{2} \leq 1 \Rightarrow-1 \leq a \leq 1\)