85410
If the curves \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{4}=1\) and \(y^{3}=16 x\) intersect at right angles, then
1 \(4 a^{2}=3\)
2 \(\mathrm{a}^{2}=1\)
3 \(3 \mathrm{a}^{2}=4\)
4 \(9 \mathrm{a}^{2}=4\)
Explanation:
(C) : Given curves, \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{4}=1 \tag{i}\) \(y^{3}=16 x\) Differentiating (i) and (ii) w.r.t. \(x\), we get \(\frac{2 x}{a^{2}}+\frac{2 y}{4} \frac{d y}{d x}=0 \text { and } 3 y^{2} \frac{d y}{d x}=16 \tag{ii}\) \(\Rightarrow \frac{d y}{d x}=\frac{-2 x}{a^{2}}\left(\frac{4}{2 y}\right) \text { and } \frac{d y}{d x}=\frac{16}{3 y^{2}}\) Since, curves intersect at right angles. \(\therefore \quad \frac{-2 x}{a^{2}}\left(\frac{4}{2 y}\right) \cdot\left(\frac{16}{3 y^{2}}\right)=-1\) \(\frac{64}{3}\left(\frac{x}{a^{2} y^{3}}\right)=1 \Rightarrow \frac{64}{3}\left(\frac{x}{a^{2} 16 x}\right)=1 \quad\left(\text { From eq }^{n}(\text { ii) })\right.\) \(\frac{4}{3 a^{2}}=1 \Rightarrow 3 a^{2}=4\)
SRM JEE-2013
Application of Derivatives
85411
If the curves \(y^{2}=6 x, 9 x^{2}+b y^{2}=16\) cut each other at right angles, then the value of \(b\) is
1 2
2 4
3 \(9 / 2\)
4 none of these
Explanation:
(C) : The given equation of curve is, \(y^{2}=6 x \quad \ldots .\). (i) and \(9 x^{2}+b^{2}=16 \quad \ldots .\). (ii) Differentiating equation (i) and (ii) w.r.t. \(x\), we get \(2 y \frac{d y}{d x}=6 \text { and } 18 x+2 b y \frac{d y}{d x}=0\) \(m_{1}=\frac{d y}{d x}=\frac{6}{2 y}=\frac{3}{y}\) \(\text { And } \quad m_{2}=\frac{d y}{d x}=\frac{-18 x}{2 b y}=\frac{-9 x}{b y}\) Since, curves cut each other at right angles. \(\therefore \quad \mathrm{m}_{1} \cdot \mathrm{m}_{2}=-1\) \(\frac{3}{y} \cdot\left(\frac{-9 x}{b y}\right)=-1 \Rightarrow \frac{-27 x}{b y^{2}}=-1\) \(\frac{-27 x}{b \cdot 6 x}=-1 \quad \text { [From equation (i)] }\) \(b=\frac{9}{2}\)
SRM JEE-2014
Application of Derivatives
85412
The equation of the tangent to the curve \(y=b e^{-x / a}\) at the point where it crosses the \(y\) axis is
1 \(\frac{x}{a}-\frac{y}{b}=1\)
2 ax + by \(=1\)
3 ax - by \(=1\)
4 \(\frac{x}{a}+\frac{y}{b}=1\)
Explanation:
(D) : \(y=b e^{-x / a}\) meets the \(y\)-axis at \((0, b)\) \(\therefore \quad \frac{\mathrm{dy}}{\mathrm{dx}}=b \mathrm{e}^{-\mathrm{x} / \mathrm{a}}\left(-\frac{1}{\mathrm{a}}\right)\) At \((0, b)\), \(\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{be}^{0}\left(-\frac{1}{\mathrm{a}}\right)=-\frac{\mathrm{b}}{\mathrm{a}}\) Therefore, required tangent is, \(y-b=-\frac{b}{a}(x-0) \text { or } \frac{x}{a}+\frac{y}{b}=1\)
BITSAT-2020
Application of Derivatives
85413
If at any point \(S\) of the curve \(b y^{2}=(x+a)^{3}\), the relation between subnormal \(\mathrm{SN}\) and subtangent \(S T\) be \(p(S N)=q(S T)^{2}\) then \(p / q\) is equal to
1 \(8 b / 27\)
2 \(8 \mathrm{a} / 27\)
3 \(\mathrm{b} / \mathrm{a}\)
4 None of these
Explanation:
(A) : Given curve, \(b y^{2}=(x+a)^{3}\) On differentiating both sides w.r.t. \(\mathrm{x}\) we get - Therefore, \(2 b y\left(\frac{d y}{d x}\right)=3(x+a)^{2} \Rightarrow\left(\frac{d y}{d x}\right)=\frac{3(x+a)^{2}}{2 b y}\) The subnormal is, \(y\left(\frac{d y}{d x}\right)=\frac{3}{2 b}(x+a)^{2}\) Therefore, the subtangent is, \(\frac{\mathrm{y}}{\frac{\mathrm{dy}}{\mathrm{dx}}}\) \(=\frac{y}{\frac{3(x+a)^{2}}{2 b y}}=\frac{2 b y^{2}}{3(x+a)^{2}}=\frac{2(x+a)^{3}}{3(x+a)^{2}}=\frac{2}{3}(x+a)\) \((\text { Sub tangent })^{2}=\frac{4}{9}(x+a)^{2}\) Also, \(\frac{(\text { subtangent })^{2}}{\text { subnormal }}=\frac{\frac{4}{9}(x+a)^{2}}{\frac{3}{2 b}(x+a)^{2}}\) \(\frac{\mathrm{p}}{\mathrm{q}}=\frac{(\text { subtangent })^{2}}{\text { subnormal }}=\frac{8 \mathrm{~b}}{27}\)
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Application of Derivatives
85410
If the curves \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{4}=1\) and \(y^{3}=16 x\) intersect at right angles, then
1 \(4 a^{2}=3\)
2 \(\mathrm{a}^{2}=1\)
3 \(3 \mathrm{a}^{2}=4\)
4 \(9 \mathrm{a}^{2}=4\)
Explanation:
(C) : Given curves, \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{4}=1 \tag{i}\) \(y^{3}=16 x\) Differentiating (i) and (ii) w.r.t. \(x\), we get \(\frac{2 x}{a^{2}}+\frac{2 y}{4} \frac{d y}{d x}=0 \text { and } 3 y^{2} \frac{d y}{d x}=16 \tag{ii}\) \(\Rightarrow \frac{d y}{d x}=\frac{-2 x}{a^{2}}\left(\frac{4}{2 y}\right) \text { and } \frac{d y}{d x}=\frac{16}{3 y^{2}}\) Since, curves intersect at right angles. \(\therefore \quad \frac{-2 x}{a^{2}}\left(\frac{4}{2 y}\right) \cdot\left(\frac{16}{3 y^{2}}\right)=-1\) \(\frac{64}{3}\left(\frac{x}{a^{2} y^{3}}\right)=1 \Rightarrow \frac{64}{3}\left(\frac{x}{a^{2} 16 x}\right)=1 \quad\left(\text { From eq }^{n}(\text { ii) })\right.\) \(\frac{4}{3 a^{2}}=1 \Rightarrow 3 a^{2}=4\)
SRM JEE-2013
Application of Derivatives
85411
If the curves \(y^{2}=6 x, 9 x^{2}+b y^{2}=16\) cut each other at right angles, then the value of \(b\) is
1 2
2 4
3 \(9 / 2\)
4 none of these
Explanation:
(C) : The given equation of curve is, \(y^{2}=6 x \quad \ldots .\). (i) and \(9 x^{2}+b^{2}=16 \quad \ldots .\). (ii) Differentiating equation (i) and (ii) w.r.t. \(x\), we get \(2 y \frac{d y}{d x}=6 \text { and } 18 x+2 b y \frac{d y}{d x}=0\) \(m_{1}=\frac{d y}{d x}=\frac{6}{2 y}=\frac{3}{y}\) \(\text { And } \quad m_{2}=\frac{d y}{d x}=\frac{-18 x}{2 b y}=\frac{-9 x}{b y}\) Since, curves cut each other at right angles. \(\therefore \quad \mathrm{m}_{1} \cdot \mathrm{m}_{2}=-1\) \(\frac{3}{y} \cdot\left(\frac{-9 x}{b y}\right)=-1 \Rightarrow \frac{-27 x}{b y^{2}}=-1\) \(\frac{-27 x}{b \cdot 6 x}=-1 \quad \text { [From equation (i)] }\) \(b=\frac{9}{2}\)
SRM JEE-2014
Application of Derivatives
85412
The equation of the tangent to the curve \(y=b e^{-x / a}\) at the point where it crosses the \(y\) axis is
1 \(\frac{x}{a}-\frac{y}{b}=1\)
2 ax + by \(=1\)
3 ax - by \(=1\)
4 \(\frac{x}{a}+\frac{y}{b}=1\)
Explanation:
(D) : \(y=b e^{-x / a}\) meets the \(y\)-axis at \((0, b)\) \(\therefore \quad \frac{\mathrm{dy}}{\mathrm{dx}}=b \mathrm{e}^{-\mathrm{x} / \mathrm{a}}\left(-\frac{1}{\mathrm{a}}\right)\) At \((0, b)\), \(\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{be}^{0}\left(-\frac{1}{\mathrm{a}}\right)=-\frac{\mathrm{b}}{\mathrm{a}}\) Therefore, required tangent is, \(y-b=-\frac{b}{a}(x-0) \text { or } \frac{x}{a}+\frac{y}{b}=1\)
BITSAT-2020
Application of Derivatives
85413
If at any point \(S\) of the curve \(b y^{2}=(x+a)^{3}\), the relation between subnormal \(\mathrm{SN}\) and subtangent \(S T\) be \(p(S N)=q(S T)^{2}\) then \(p / q\) is equal to
1 \(8 b / 27\)
2 \(8 \mathrm{a} / 27\)
3 \(\mathrm{b} / \mathrm{a}\)
4 None of these
Explanation:
(A) : Given curve, \(b y^{2}=(x+a)^{3}\) On differentiating both sides w.r.t. \(\mathrm{x}\) we get - Therefore, \(2 b y\left(\frac{d y}{d x}\right)=3(x+a)^{2} \Rightarrow\left(\frac{d y}{d x}\right)=\frac{3(x+a)^{2}}{2 b y}\) The subnormal is, \(y\left(\frac{d y}{d x}\right)=\frac{3}{2 b}(x+a)^{2}\) Therefore, the subtangent is, \(\frac{\mathrm{y}}{\frac{\mathrm{dy}}{\mathrm{dx}}}\) \(=\frac{y}{\frac{3(x+a)^{2}}{2 b y}}=\frac{2 b y^{2}}{3(x+a)^{2}}=\frac{2(x+a)^{3}}{3(x+a)^{2}}=\frac{2}{3}(x+a)\) \((\text { Sub tangent })^{2}=\frac{4}{9}(x+a)^{2}\) Also, \(\frac{(\text { subtangent })^{2}}{\text { subnormal }}=\frac{\frac{4}{9}(x+a)^{2}}{\frac{3}{2 b}(x+a)^{2}}\) \(\frac{\mathrm{p}}{\mathrm{q}}=\frac{(\text { subtangent })^{2}}{\text { subnormal }}=\frac{8 \mathrm{~b}}{27}\)
85410
If the curves \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{4}=1\) and \(y^{3}=16 x\) intersect at right angles, then
1 \(4 a^{2}=3\)
2 \(\mathrm{a}^{2}=1\)
3 \(3 \mathrm{a}^{2}=4\)
4 \(9 \mathrm{a}^{2}=4\)
Explanation:
(C) : Given curves, \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{4}=1 \tag{i}\) \(y^{3}=16 x\) Differentiating (i) and (ii) w.r.t. \(x\), we get \(\frac{2 x}{a^{2}}+\frac{2 y}{4} \frac{d y}{d x}=0 \text { and } 3 y^{2} \frac{d y}{d x}=16 \tag{ii}\) \(\Rightarrow \frac{d y}{d x}=\frac{-2 x}{a^{2}}\left(\frac{4}{2 y}\right) \text { and } \frac{d y}{d x}=\frac{16}{3 y^{2}}\) Since, curves intersect at right angles. \(\therefore \quad \frac{-2 x}{a^{2}}\left(\frac{4}{2 y}\right) \cdot\left(\frac{16}{3 y^{2}}\right)=-1\) \(\frac{64}{3}\left(\frac{x}{a^{2} y^{3}}\right)=1 \Rightarrow \frac{64}{3}\left(\frac{x}{a^{2} 16 x}\right)=1 \quad\left(\text { From eq }^{n}(\text { ii) })\right.\) \(\frac{4}{3 a^{2}}=1 \Rightarrow 3 a^{2}=4\)
SRM JEE-2013
Application of Derivatives
85411
If the curves \(y^{2}=6 x, 9 x^{2}+b y^{2}=16\) cut each other at right angles, then the value of \(b\) is
1 2
2 4
3 \(9 / 2\)
4 none of these
Explanation:
(C) : The given equation of curve is, \(y^{2}=6 x \quad \ldots .\). (i) and \(9 x^{2}+b^{2}=16 \quad \ldots .\). (ii) Differentiating equation (i) and (ii) w.r.t. \(x\), we get \(2 y \frac{d y}{d x}=6 \text { and } 18 x+2 b y \frac{d y}{d x}=0\) \(m_{1}=\frac{d y}{d x}=\frac{6}{2 y}=\frac{3}{y}\) \(\text { And } \quad m_{2}=\frac{d y}{d x}=\frac{-18 x}{2 b y}=\frac{-9 x}{b y}\) Since, curves cut each other at right angles. \(\therefore \quad \mathrm{m}_{1} \cdot \mathrm{m}_{2}=-1\) \(\frac{3}{y} \cdot\left(\frac{-9 x}{b y}\right)=-1 \Rightarrow \frac{-27 x}{b y^{2}}=-1\) \(\frac{-27 x}{b \cdot 6 x}=-1 \quad \text { [From equation (i)] }\) \(b=\frac{9}{2}\)
SRM JEE-2014
Application of Derivatives
85412
The equation of the tangent to the curve \(y=b e^{-x / a}\) at the point where it crosses the \(y\) axis is
1 \(\frac{x}{a}-\frac{y}{b}=1\)
2 ax + by \(=1\)
3 ax - by \(=1\)
4 \(\frac{x}{a}+\frac{y}{b}=1\)
Explanation:
(D) : \(y=b e^{-x / a}\) meets the \(y\)-axis at \((0, b)\) \(\therefore \quad \frac{\mathrm{dy}}{\mathrm{dx}}=b \mathrm{e}^{-\mathrm{x} / \mathrm{a}}\left(-\frac{1}{\mathrm{a}}\right)\) At \((0, b)\), \(\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{be}^{0}\left(-\frac{1}{\mathrm{a}}\right)=-\frac{\mathrm{b}}{\mathrm{a}}\) Therefore, required tangent is, \(y-b=-\frac{b}{a}(x-0) \text { or } \frac{x}{a}+\frac{y}{b}=1\)
BITSAT-2020
Application of Derivatives
85413
If at any point \(S\) of the curve \(b y^{2}=(x+a)^{3}\), the relation between subnormal \(\mathrm{SN}\) and subtangent \(S T\) be \(p(S N)=q(S T)^{2}\) then \(p / q\) is equal to
1 \(8 b / 27\)
2 \(8 \mathrm{a} / 27\)
3 \(\mathrm{b} / \mathrm{a}\)
4 None of these
Explanation:
(A) : Given curve, \(b y^{2}=(x+a)^{3}\) On differentiating both sides w.r.t. \(\mathrm{x}\) we get - Therefore, \(2 b y\left(\frac{d y}{d x}\right)=3(x+a)^{2} \Rightarrow\left(\frac{d y}{d x}\right)=\frac{3(x+a)^{2}}{2 b y}\) The subnormal is, \(y\left(\frac{d y}{d x}\right)=\frac{3}{2 b}(x+a)^{2}\) Therefore, the subtangent is, \(\frac{\mathrm{y}}{\frac{\mathrm{dy}}{\mathrm{dx}}}\) \(=\frac{y}{\frac{3(x+a)^{2}}{2 b y}}=\frac{2 b y^{2}}{3(x+a)^{2}}=\frac{2(x+a)^{3}}{3(x+a)^{2}}=\frac{2}{3}(x+a)\) \((\text { Sub tangent })^{2}=\frac{4}{9}(x+a)^{2}\) Also, \(\frac{(\text { subtangent })^{2}}{\text { subnormal }}=\frac{\frac{4}{9}(x+a)^{2}}{\frac{3}{2 b}(x+a)^{2}}\) \(\frac{\mathrm{p}}{\mathrm{q}}=\frac{(\text { subtangent })^{2}}{\text { subnormal }}=\frac{8 \mathrm{~b}}{27}\)
85410
If the curves \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{4}=1\) and \(y^{3}=16 x\) intersect at right angles, then
1 \(4 a^{2}=3\)
2 \(\mathrm{a}^{2}=1\)
3 \(3 \mathrm{a}^{2}=4\)
4 \(9 \mathrm{a}^{2}=4\)
Explanation:
(C) : Given curves, \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{4}=1 \tag{i}\) \(y^{3}=16 x\) Differentiating (i) and (ii) w.r.t. \(x\), we get \(\frac{2 x}{a^{2}}+\frac{2 y}{4} \frac{d y}{d x}=0 \text { and } 3 y^{2} \frac{d y}{d x}=16 \tag{ii}\) \(\Rightarrow \frac{d y}{d x}=\frac{-2 x}{a^{2}}\left(\frac{4}{2 y}\right) \text { and } \frac{d y}{d x}=\frac{16}{3 y^{2}}\) Since, curves intersect at right angles. \(\therefore \quad \frac{-2 x}{a^{2}}\left(\frac{4}{2 y}\right) \cdot\left(\frac{16}{3 y^{2}}\right)=-1\) \(\frac{64}{3}\left(\frac{x}{a^{2} y^{3}}\right)=1 \Rightarrow \frac{64}{3}\left(\frac{x}{a^{2} 16 x}\right)=1 \quad\left(\text { From eq }^{n}(\text { ii) })\right.\) \(\frac{4}{3 a^{2}}=1 \Rightarrow 3 a^{2}=4\)
SRM JEE-2013
Application of Derivatives
85411
If the curves \(y^{2}=6 x, 9 x^{2}+b y^{2}=16\) cut each other at right angles, then the value of \(b\) is
1 2
2 4
3 \(9 / 2\)
4 none of these
Explanation:
(C) : The given equation of curve is, \(y^{2}=6 x \quad \ldots .\). (i) and \(9 x^{2}+b^{2}=16 \quad \ldots .\). (ii) Differentiating equation (i) and (ii) w.r.t. \(x\), we get \(2 y \frac{d y}{d x}=6 \text { and } 18 x+2 b y \frac{d y}{d x}=0\) \(m_{1}=\frac{d y}{d x}=\frac{6}{2 y}=\frac{3}{y}\) \(\text { And } \quad m_{2}=\frac{d y}{d x}=\frac{-18 x}{2 b y}=\frac{-9 x}{b y}\) Since, curves cut each other at right angles. \(\therefore \quad \mathrm{m}_{1} \cdot \mathrm{m}_{2}=-1\) \(\frac{3}{y} \cdot\left(\frac{-9 x}{b y}\right)=-1 \Rightarrow \frac{-27 x}{b y^{2}}=-1\) \(\frac{-27 x}{b \cdot 6 x}=-1 \quad \text { [From equation (i)] }\) \(b=\frac{9}{2}\)
SRM JEE-2014
Application of Derivatives
85412
The equation of the tangent to the curve \(y=b e^{-x / a}\) at the point where it crosses the \(y\) axis is
1 \(\frac{x}{a}-\frac{y}{b}=1\)
2 ax + by \(=1\)
3 ax - by \(=1\)
4 \(\frac{x}{a}+\frac{y}{b}=1\)
Explanation:
(D) : \(y=b e^{-x / a}\) meets the \(y\)-axis at \((0, b)\) \(\therefore \quad \frac{\mathrm{dy}}{\mathrm{dx}}=b \mathrm{e}^{-\mathrm{x} / \mathrm{a}}\left(-\frac{1}{\mathrm{a}}\right)\) At \((0, b)\), \(\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{be}^{0}\left(-\frac{1}{\mathrm{a}}\right)=-\frac{\mathrm{b}}{\mathrm{a}}\) Therefore, required tangent is, \(y-b=-\frac{b}{a}(x-0) \text { or } \frac{x}{a}+\frac{y}{b}=1\)
BITSAT-2020
Application of Derivatives
85413
If at any point \(S\) of the curve \(b y^{2}=(x+a)^{3}\), the relation between subnormal \(\mathrm{SN}\) and subtangent \(S T\) be \(p(S N)=q(S T)^{2}\) then \(p / q\) is equal to
1 \(8 b / 27\)
2 \(8 \mathrm{a} / 27\)
3 \(\mathrm{b} / \mathrm{a}\)
4 None of these
Explanation:
(A) : Given curve, \(b y^{2}=(x+a)^{3}\) On differentiating both sides w.r.t. \(\mathrm{x}\) we get - Therefore, \(2 b y\left(\frac{d y}{d x}\right)=3(x+a)^{2} \Rightarrow\left(\frac{d y}{d x}\right)=\frac{3(x+a)^{2}}{2 b y}\) The subnormal is, \(y\left(\frac{d y}{d x}\right)=\frac{3}{2 b}(x+a)^{2}\) Therefore, the subtangent is, \(\frac{\mathrm{y}}{\frac{\mathrm{dy}}{\mathrm{dx}}}\) \(=\frac{y}{\frac{3(x+a)^{2}}{2 b y}}=\frac{2 b y^{2}}{3(x+a)^{2}}=\frac{2(x+a)^{3}}{3(x+a)^{2}}=\frac{2}{3}(x+a)\) \((\text { Sub tangent })^{2}=\frac{4}{9}(x+a)^{2}\) Also, \(\frac{(\text { subtangent })^{2}}{\text { subnormal }}=\frac{\frac{4}{9}(x+a)^{2}}{\frac{3}{2 b}(x+a)^{2}}\) \(\frac{\mathrm{p}}{\mathrm{q}}=\frac{(\text { subtangent })^{2}}{\text { subnormal }}=\frac{8 \mathrm{~b}}{27}\)