QBManager

Application of Derivatives

85816 The number of points on the curve \(y=54 x^{5}-\) \(135 x^{4}-70 x^{3}+180 x^{2}+210 x\) at which the normal lines are parallel to \(x+90 y+2=0\) is :

1 2

2 4

3 0

4 3

Shift-I

Application of Derivatives

85817 The longest distance of the point \((a, 0)\) from the curve \(2 x^{2}+y^{2}=2 x\) is

1 \(1+\mathrm{a}\)

2 \(|1-\mathrm{a}|\)

3 \(\sqrt{1-2 \mathrm{a}+2 \mathrm{a}^{2}}\)

4 \(\sqrt{1-2 \mathrm{a}+3 \mathrm{a}^{2}}\)

AP EAMCET-2010

Application of Derivatives

85818 The polar equation of a curve centred at \((1,0)\) with radius 1 is

1 \(r=\cos \theta\)

2 \(\mathrm{r}=2\)

3 \(\mathrm{r}=\sin 2 \theta\)

4 \(\mathrm{r}=2 \cos \theta\)

AMU-2004

Application of Derivatives

85819 For \(\theta \in I R\), when the point \((x, y)=(\tan \theta+\sin \theta\), \(\boldsymbol{\operatorname { t a n }} \theta-\boldsymbol{\operatorname { s i n }} \theta\) ) is defined, it lies on the curve:

1 \(x^{2}+y^{2}=2 x y\)

2 \(x^{2}-y^{2}=8 x y\)

3 \(\left(x^{2}-y^{2}\right)^{2}=16 x y\)

4 \(\left(x^{2}+y^{2}\right)^{2}=4 x y\)

AMU-2004

Application of Derivatives

85820 If the curves \(x^{2}+p y^{2}=1\) and \(q x^{2}+y^{2}=1\) are orthogonal to each other, then

1 \(\mathrm{p}-\mathrm{q}=2\)

2 \(\frac{1}{\mathrm{p}}-\frac{1}{\mathrm{q}}=2\)

3 \(\frac{1}{\mathrm{p}}+\frac{1}{\mathrm{q}}=-2\)

4 \(\frac{1}{\mathrm{p}}+\frac{1}{\mathrm{q}}=2\)

Explanation:

(D) : Given, the curves \(\mathrm{x}^{2}+\mathrm{py}^{2}=1\) and \(\mathrm{qx}^{2}+\mathrm{y}^{2}=1\) Then, from \(\mathrm{x}^{2}+p y^{2}=1 \Rightarrow \mathrm{x}^{2}=1-\mathrm{py}^{2}\) put in \(\mathrm{qx}^{2}+\mathrm{y}^{2}=1\) \(\therefore \quad \mathrm{q}\left(1-\mathrm{py}^{2}\right)+\mathrm{y}^{2}=1\)
\(q_{2}-p q y^{2}+y^{2}=1\)
\(\mathrm{y}_{2}-\mathrm{pqy}^{2}=1-\mathrm{q}\)
\(y^{2}(1-p q)=1-q\)
\(y^{2}=\frac{1-q}{1-p q} \Rightarrow y= \pm \sqrt{\frac{1-q}{1-p q}}\)
Then, \(\quad x^{2}=1-p\left(\frac{1-q}{1-p q}\right)\)
\(x^{2}=1-p\left(\frac{1-q}{1-p q}\right) \Rightarrow x^{2}=\frac{1-p q-p+p q}{1-p q}\)
\(x^{2}=\frac{1-p}{1-p q} \Rightarrow x= \pm \sqrt{\frac{1-p}{1-p q}}\)
Then, intersection point \((\mathrm{x}, \mathrm{y})\)
\(=\left( \pm \sqrt{\frac{1-\mathrm{p}}{1-\mathrm{pq}}}, \pm \sqrt{\frac{1-\mathrm{q}}{1-\mathrm{pq}}}\right)\)
\(\therefore \quad \mathrm{x}^{2}+\mathrm{py}^{2}=1 \Rightarrow 2 \mathrm{x}+2 \mathrm{py} \frac{\mathrm{dy}}{\mathrm{dx}}=0\)
2 py \(\frac{d y}{d x}=-2 x \Rightarrow\) py \(\frac{d y}{d x}=-x\)
\(\frac{d y}{d x}=\frac{-x}{p y}\)
\(\frac{d y}{d x}\) at \(\left(\sqrt{\frac{1-p}{1-p q}}, \sqrt{\frac{1-q}{1-p q}}\right)=\frac{\sqrt{\frac{1-p}{1-p q}}}{p \times \sqrt{\frac{1-q}{1-p q}}}=m_{1}\)
\(\mathrm{m}_{1}=\frac{-\sqrt{1-\mathrm{p}}}{\mathrm{p} \sqrt{1-\mathrm{q}}}\)
And,\(\quad \mathrm{qx}^{2}+\mathrm{y}^{2}=1 \Rightarrow 2 x q+2 y \frac{d y}{d x}=0\)
\(2 y \frac{d y}{d x}=-2 x q \Rightarrow y \frac{d y}{d x}=-x q\)
\(\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{-\mathrm{xq}}{\mathrm{y}}\)
\(\frac{d y}{d x}\) at \(\left(\sqrt{\frac{1-p}{1-p q}}, \sqrt{\frac{1-q}{1-p q}}\right)=\frac{-q \sqrt{\frac{1-p}{1-p q}}}{\sqrt{\frac{1-q}{1-p q}}}=m_{2}\)
\(\mathrm{m}_{2}=\frac{-\mathrm{q} \sqrt{1-\mathrm{q}}}{\sqrt{1-\mathrm{q}}}\)
Since, given \(\mathrm{m}_{1}\) and \(\mathrm{m}_{2}\) are orthogonal then
\(\mathrm{m}_{1} \cdot \mathrm{m}_{2}=-1\left(\because \theta=\frac{\pi}{2}\right)\)
\(\frac{-\sqrt{1-\mathrm{p}}}{\mathrm{p} \sqrt{1-q}} \times \frac{-\mathrm{q} \sqrt{1-\mathrm{p}}}{\sqrt{1-q}}=-1\)
\(\frac{\mathrm{q}(1-\mathrm{p})}{\mathrm{p}(1-\mathrm{q})}=-1\)
\(\mathrm{q}(1-\mathrm{p})=-\mathrm{p}(1-\mathrm{q})\)
Squaring both side, we get-
\(q^{2}(1-p)^{2}=p^{2}(1-q)^{2}\)
\(q^{2}\left(1+p^{2}-2 p\right)=p^{2}\left(1+q^{2}-2 q\right)\)
\(q_{2}^{2}+p^{2} q^{2}-2 p q^{2}=p^{2}+p^{2} q^{2}-2 p^{2} q\)
\(q^{2}-2 p q^{2}=p^{2}-2 p^{2} q\)
\(2 p^{2} q-2 p q^{2}=p^{2}-q^{2}\)
\(2 p q(p-q)=(p-q)(p+q)\)
\(2 p q=p+q\)
\(\frac{p}{p q}+\frac{q}{p q}=2\)
So, \(\quad \frac{1}{\mathrm{p}}+\frac{1}{\mathrm{q}}=2\)

AP EAMCET-2013

Application of Derivatives

85816 The number of points on the curve \(y=54 x^{5}-\) \(135 x^{4}-70 x^{3}+180 x^{2}+210 x\) at which the normal lines are parallel to \(x+90 y+2=0\) is :

1 2

2 4

3 0

4 3

Shift-I

Application of Derivatives

85817 The longest distance of the point \((a, 0)\) from the curve \(2 x^{2}+y^{2}=2 x\) is

1 \(1+\mathrm{a}\)

2 \(|1-\mathrm{a}|\)

3 \(\sqrt{1-2 \mathrm{a}+2 \mathrm{a}^{2}}\)

4 \(\sqrt{1-2 \mathrm{a}+3 \mathrm{a}^{2}}\)

AP EAMCET-2010

Application of Derivatives

85818 The polar equation of a curve centred at \((1,0)\) with radius 1 is

1 \(r=\cos \theta\)

2 \(\mathrm{r}=2\)

3 \(\mathrm{r}=\sin 2 \theta\)

4 \(\mathrm{r}=2 \cos \theta\)

AMU-2004

Application of Derivatives

85819 For \(\theta \in I R\), when the point \((x, y)=(\tan \theta+\sin \theta\), \(\boldsymbol{\operatorname { t a n }} \theta-\boldsymbol{\operatorname { s i n }} \theta\) ) is defined, it lies on the curve:

1 \(x^{2}+y^{2}=2 x y\)

2 \(x^{2}-y^{2}=8 x y\)

3 \(\left(x^{2}-y^{2}\right)^{2}=16 x y\)

4 \(\left(x^{2}+y^{2}\right)^{2}=4 x y\)

AMU-2004

Application of Derivatives

85820 If the curves \(x^{2}+p y^{2}=1\) and \(q x^{2}+y^{2}=1\) are orthogonal to each other, then

1 \(\mathrm{p}-\mathrm{q}=2\)

2 \(\frac{1}{\mathrm{p}}-\frac{1}{\mathrm{q}}=2\)

3 \(\frac{1}{\mathrm{p}}+\frac{1}{\mathrm{q}}=-2\)

4 \(\frac{1}{\mathrm{p}}+\frac{1}{\mathrm{q}}=2\)

Explanation:

(D) : Given, the curves \(\mathrm{x}^{2}+\mathrm{py}^{2}=1\) and \(\mathrm{qx}^{2}+\mathrm{y}^{2}=1\) Then, from \(\mathrm{x}^{2}+p y^{2}=1 \Rightarrow \mathrm{x}^{2}=1-\mathrm{py}^{2}\) put in \(\mathrm{qx}^{2}+\mathrm{y}^{2}=1\) \(\therefore \quad \mathrm{q}\left(1-\mathrm{py}^{2}\right)+\mathrm{y}^{2}=1\)
\(q_{2}-p q y^{2}+y^{2}=1\)
\(\mathrm{y}_{2}-\mathrm{pqy}^{2}=1-\mathrm{q}\)
\(y^{2}(1-p q)=1-q\)
\(y^{2}=\frac{1-q}{1-p q} \Rightarrow y= \pm \sqrt{\frac{1-q}{1-p q}}\)
Then, \(\quad x^{2}=1-p\left(\frac{1-q}{1-p q}\right)\)
\(x^{2}=1-p\left(\frac{1-q}{1-p q}\right) \Rightarrow x^{2}=\frac{1-p q-p+p q}{1-p q}\)
\(x^{2}=\frac{1-p}{1-p q} \Rightarrow x= \pm \sqrt{\frac{1-p}{1-p q}}\)
Then, intersection point \((\mathrm{x}, \mathrm{y})\)
\(=\left( \pm \sqrt{\frac{1-\mathrm{p}}{1-\mathrm{pq}}}, \pm \sqrt{\frac{1-\mathrm{q}}{1-\mathrm{pq}}}\right)\)
\(\therefore \quad \mathrm{x}^{2}+\mathrm{py}^{2}=1 \Rightarrow 2 \mathrm{x}+2 \mathrm{py} \frac{\mathrm{dy}}{\mathrm{dx}}=0\)
2 py \(\frac{d y}{d x}=-2 x \Rightarrow\) py \(\frac{d y}{d x}=-x\)
\(\frac{d y}{d x}=\frac{-x}{p y}\)
\(\frac{d y}{d x}\) at \(\left(\sqrt{\frac{1-p}{1-p q}}, \sqrt{\frac{1-q}{1-p q}}\right)=\frac{\sqrt{\frac{1-p}{1-p q}}}{p \times \sqrt{\frac{1-q}{1-p q}}}=m_{1}\)
\(\mathrm{m}_{1}=\frac{-\sqrt{1-\mathrm{p}}}{\mathrm{p} \sqrt{1-\mathrm{q}}}\)
And,\(\quad \mathrm{qx}^{2}+\mathrm{y}^{2}=1 \Rightarrow 2 x q+2 y \frac{d y}{d x}=0\)
\(2 y \frac{d y}{d x}=-2 x q \Rightarrow y \frac{d y}{d x}=-x q\)
\(\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{-\mathrm{xq}}{\mathrm{y}}\)
\(\frac{d y}{d x}\) at \(\left(\sqrt{\frac{1-p}{1-p q}}, \sqrt{\frac{1-q}{1-p q}}\right)=\frac{-q \sqrt{\frac{1-p}{1-p q}}}{\sqrt{\frac{1-q}{1-p q}}}=m_{2}\)
\(\mathrm{m}_{2}=\frac{-\mathrm{q} \sqrt{1-\mathrm{q}}}{\sqrt{1-\mathrm{q}}}\)
Since, given \(\mathrm{m}_{1}\) and \(\mathrm{m}_{2}\) are orthogonal then
\(\mathrm{m}_{1} \cdot \mathrm{m}_{2}=-1\left(\because \theta=\frac{\pi}{2}\right)\)
\(\frac{-\sqrt{1-\mathrm{p}}}{\mathrm{p} \sqrt{1-q}} \times \frac{-\mathrm{q} \sqrt{1-\mathrm{p}}}{\sqrt{1-q}}=-1\)
\(\frac{\mathrm{q}(1-\mathrm{p})}{\mathrm{p}(1-\mathrm{q})}=-1\)
\(\mathrm{q}(1-\mathrm{p})=-\mathrm{p}(1-\mathrm{q})\)
Squaring both side, we get-
\(q^{2}(1-p)^{2}=p^{2}(1-q)^{2}\)
\(q^{2}\left(1+p^{2}-2 p\right)=p^{2}\left(1+q^{2}-2 q\right)\)
\(q_{2}^{2}+p^{2} q^{2}-2 p q^{2}=p^{2}+p^{2} q^{2}-2 p^{2} q\)
\(q^{2}-2 p q^{2}=p^{2}-2 p^{2} q\)
\(2 p^{2} q-2 p q^{2}=p^{2}-q^{2}\)
\(2 p q(p-q)=(p-q)(p+q)\)
\(2 p q=p+q\)
\(\frac{p}{p q}+\frac{q}{p q}=2\)
So, \(\quad \frac{1}{\mathrm{p}}+\frac{1}{\mathrm{q}}=2\)

AP EAMCET-2013

Application of Derivatives

85816 The number of points on the curve \(y=54 x^{5}-\) \(135 x^{4}-70 x^{3}+180 x^{2}+210 x\) at which the normal lines are parallel to \(x+90 y+2=0\) is :

1 2

2 4

3 0

4 3

Shift-I

Application of Derivatives

85817 The longest distance of the point \((a, 0)\) from the curve \(2 x^{2}+y^{2}=2 x\) is

1 \(1+\mathrm{a}\)

2 \(|1-\mathrm{a}|\)

3 \(\sqrt{1-2 \mathrm{a}+2 \mathrm{a}^{2}}\)

4 \(\sqrt{1-2 \mathrm{a}+3 \mathrm{a}^{2}}\)

AP EAMCET-2010

Application of Derivatives

85818 The polar equation of a curve centred at \((1,0)\) with radius 1 is

1 \(r=\cos \theta\)

2 \(\mathrm{r}=2\)

3 \(\mathrm{r}=\sin 2 \theta\)

4 \(\mathrm{r}=2 \cos \theta\)

AMU-2004

Application of Derivatives

85819 For \(\theta \in I R\), when the point \((x, y)=(\tan \theta+\sin \theta\), \(\boldsymbol{\operatorname { t a n }} \theta-\boldsymbol{\operatorname { s i n }} \theta\) ) is defined, it lies on the curve:

1 \(x^{2}+y^{2}=2 x y\)

2 \(x^{2}-y^{2}=8 x y\)

3 \(\left(x^{2}-y^{2}\right)^{2}=16 x y\)

4 \(\left(x^{2}+y^{2}\right)^{2}=4 x y\)

AMU-2004

Application of Derivatives

85820 If the curves \(x^{2}+p y^{2}=1\) and \(q x^{2}+y^{2}=1\) are orthogonal to each other, then

1 \(\mathrm{p}-\mathrm{q}=2\)

2 \(\frac{1}{\mathrm{p}}-\frac{1}{\mathrm{q}}=2\)

3 \(\frac{1}{\mathrm{p}}+\frac{1}{\mathrm{q}}=-2\)

4 \(\frac{1}{\mathrm{p}}+\frac{1}{\mathrm{q}}=2\)

Explanation:

(D) : Given, the curves \(\mathrm{x}^{2}+\mathrm{py}^{2}=1\) and \(\mathrm{qx}^{2}+\mathrm{y}^{2}=1\) Then, from \(\mathrm{x}^{2}+p y^{2}=1 \Rightarrow \mathrm{x}^{2}=1-\mathrm{py}^{2}\) put in \(\mathrm{qx}^{2}+\mathrm{y}^{2}=1\) \(\therefore \quad \mathrm{q}\left(1-\mathrm{py}^{2}\right)+\mathrm{y}^{2}=1\)
\(q_{2}-p q y^{2}+y^{2}=1\)
\(\mathrm{y}_{2}-\mathrm{pqy}^{2}=1-\mathrm{q}\)
\(y^{2}(1-p q)=1-q\)
\(y^{2}=\frac{1-q}{1-p q} \Rightarrow y= \pm \sqrt{\frac{1-q}{1-p q}}\)
Then, \(\quad x^{2}=1-p\left(\frac{1-q}{1-p q}\right)\)
\(x^{2}=1-p\left(\frac{1-q}{1-p q}\right) \Rightarrow x^{2}=\frac{1-p q-p+p q}{1-p q}\)
\(x^{2}=\frac{1-p}{1-p q} \Rightarrow x= \pm \sqrt{\frac{1-p}{1-p q}}\)
Then, intersection point \((\mathrm{x}, \mathrm{y})\)
\(=\left( \pm \sqrt{\frac{1-\mathrm{p}}{1-\mathrm{pq}}}, \pm \sqrt{\frac{1-\mathrm{q}}{1-\mathrm{pq}}}\right)\)
\(\therefore \quad \mathrm{x}^{2}+\mathrm{py}^{2}=1 \Rightarrow 2 \mathrm{x}+2 \mathrm{py} \frac{\mathrm{dy}}{\mathrm{dx}}=0\)
2 py \(\frac{d y}{d x}=-2 x \Rightarrow\) py \(\frac{d y}{d x}=-x\)
\(\frac{d y}{d x}=\frac{-x}{p y}\)
\(\frac{d y}{d x}\) at \(\left(\sqrt{\frac{1-p}{1-p q}}, \sqrt{\frac{1-q}{1-p q}}\right)=\frac{\sqrt{\frac{1-p}{1-p q}}}{p \times \sqrt{\frac{1-q}{1-p q}}}=m_{1}\)
\(\mathrm{m}_{1}=\frac{-\sqrt{1-\mathrm{p}}}{\mathrm{p} \sqrt{1-\mathrm{q}}}\)
And,\(\quad \mathrm{qx}^{2}+\mathrm{y}^{2}=1 \Rightarrow 2 x q+2 y \frac{d y}{d x}=0\)
\(2 y \frac{d y}{d x}=-2 x q \Rightarrow y \frac{d y}{d x}=-x q\)
\(\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{-\mathrm{xq}}{\mathrm{y}}\)
\(\frac{d y}{d x}\) at \(\left(\sqrt{\frac{1-p}{1-p q}}, \sqrt{\frac{1-q}{1-p q}}\right)=\frac{-q \sqrt{\frac{1-p}{1-p q}}}{\sqrt{\frac{1-q}{1-p q}}}=m_{2}\)
\(\mathrm{m}_{2}=\frac{-\mathrm{q} \sqrt{1-\mathrm{q}}}{\sqrt{1-\mathrm{q}}}\)
Since, given \(\mathrm{m}_{1}\) and \(\mathrm{m}_{2}\) are orthogonal then
\(\mathrm{m}_{1} \cdot \mathrm{m}_{2}=-1\left(\because \theta=\frac{\pi}{2}\right)\)
\(\frac{-\sqrt{1-\mathrm{p}}}{\mathrm{p} \sqrt{1-q}} \times \frac{-\mathrm{q} \sqrt{1-\mathrm{p}}}{\sqrt{1-q}}=-1\)
\(\frac{\mathrm{q}(1-\mathrm{p})}{\mathrm{p}(1-\mathrm{q})}=-1\)
\(\mathrm{q}(1-\mathrm{p})=-\mathrm{p}(1-\mathrm{q})\)
Squaring both side, we get-
\(q^{2}(1-p)^{2}=p^{2}(1-q)^{2}\)
\(q^{2}\left(1+p^{2}-2 p\right)=p^{2}\left(1+q^{2}-2 q\right)\)
\(q_{2}^{2}+p^{2} q^{2}-2 p q^{2}=p^{2}+p^{2} q^{2}-2 p^{2} q\)
\(q^{2}-2 p q^{2}=p^{2}-2 p^{2} q\)
\(2 p^{2} q-2 p q^{2}=p^{2}-q^{2}\)
\(2 p q(p-q)=(p-q)(p+q)\)
\(2 p q=p+q\)
\(\frac{p}{p q}+\frac{q}{p q}=2\)
So, \(\quad \frac{1}{\mathrm{p}}+\frac{1}{\mathrm{q}}=2\)

AP EAMCET-2013

Application of Derivatives

85816 The number of points on the curve \(y=54 x^{5}-\) \(135 x^{4}-70 x^{3}+180 x^{2}+210 x\) at which the normal lines are parallel to \(x+90 y+2=0\) is :

1 2

2 4

3 0

4 3

Shift-I

Application of Derivatives

85817 The longest distance of the point \((a, 0)\) from the curve \(2 x^{2}+y^{2}=2 x\) is

1 \(1+\mathrm{a}\)

2 \(|1-\mathrm{a}|\)

3 \(\sqrt{1-2 \mathrm{a}+2 \mathrm{a}^{2}}\)

4 \(\sqrt{1-2 \mathrm{a}+3 \mathrm{a}^{2}}\)

AP EAMCET-2010

Application of Derivatives

85818 The polar equation of a curve centred at \((1,0)\) with radius 1 is

1 \(r=\cos \theta\)

2 \(\mathrm{r}=2\)

3 \(\mathrm{r}=\sin 2 \theta\)

4 \(\mathrm{r}=2 \cos \theta\)

AMU-2004

Application of Derivatives

85819 For \(\theta \in I R\), when the point \((x, y)=(\tan \theta+\sin \theta\), \(\boldsymbol{\operatorname { t a n }} \theta-\boldsymbol{\operatorname { s i n }} \theta\) ) is defined, it lies on the curve:

1 \(x^{2}+y^{2}=2 x y\)

2 \(x^{2}-y^{2}=8 x y\)

3 \(\left(x^{2}-y^{2}\right)^{2}=16 x y\)

4 \(\left(x^{2}+y^{2}\right)^{2}=4 x y\)

AMU-2004

Application of Derivatives

85820 If the curves \(x^{2}+p y^{2}=1\) and \(q x^{2}+y^{2}=1\) are orthogonal to each other, then

1 \(\mathrm{p}-\mathrm{q}=2\)

2 \(\frac{1}{\mathrm{p}}-\frac{1}{\mathrm{q}}=2\)

3 \(\frac{1}{\mathrm{p}}+\frac{1}{\mathrm{q}}=-2\)

4 \(\frac{1}{\mathrm{p}}+\frac{1}{\mathrm{q}}=2\)

Explanation:

(D) : Given, the curves \(\mathrm{x}^{2}+\mathrm{py}^{2}=1\) and \(\mathrm{qx}^{2}+\mathrm{y}^{2}=1\) Then, from \(\mathrm{x}^{2}+p y^{2}=1 \Rightarrow \mathrm{x}^{2}=1-\mathrm{py}^{2}\) put in \(\mathrm{qx}^{2}+\mathrm{y}^{2}=1\) \(\therefore \quad \mathrm{q}\left(1-\mathrm{py}^{2}\right)+\mathrm{y}^{2}=1\)
\(q_{2}-p q y^{2}+y^{2}=1\)
\(\mathrm{y}_{2}-\mathrm{pqy}^{2}=1-\mathrm{q}\)
\(y^{2}(1-p q)=1-q\)
\(y^{2}=\frac{1-q}{1-p q} \Rightarrow y= \pm \sqrt{\frac{1-q}{1-p q}}\)
Then, \(\quad x^{2}=1-p\left(\frac{1-q}{1-p q}\right)\)
\(x^{2}=1-p\left(\frac{1-q}{1-p q}\right) \Rightarrow x^{2}=\frac{1-p q-p+p q}{1-p q}\)
\(x^{2}=\frac{1-p}{1-p q} \Rightarrow x= \pm \sqrt{\frac{1-p}{1-p q}}\)
Then, intersection point \((\mathrm{x}, \mathrm{y})\)
\(=\left( \pm \sqrt{\frac{1-\mathrm{p}}{1-\mathrm{pq}}}, \pm \sqrt{\frac{1-\mathrm{q}}{1-\mathrm{pq}}}\right)\)
\(\therefore \quad \mathrm{x}^{2}+\mathrm{py}^{2}=1 \Rightarrow 2 \mathrm{x}+2 \mathrm{py} \frac{\mathrm{dy}}{\mathrm{dx}}=0\)
2 py \(\frac{d y}{d x}=-2 x \Rightarrow\) py \(\frac{d y}{d x}=-x\)
\(\frac{d y}{d x}=\frac{-x}{p y}\)
\(\frac{d y}{d x}\) at \(\left(\sqrt{\frac{1-p}{1-p q}}, \sqrt{\frac{1-q}{1-p q}}\right)=\frac{\sqrt{\frac{1-p}{1-p q}}}{p \times \sqrt{\frac{1-q}{1-p q}}}=m_{1}\)
\(\mathrm{m}_{1}=\frac{-\sqrt{1-\mathrm{p}}}{\mathrm{p} \sqrt{1-\mathrm{q}}}\)
And,\(\quad \mathrm{qx}^{2}+\mathrm{y}^{2}=1 \Rightarrow 2 x q+2 y \frac{d y}{d x}=0\)
\(2 y \frac{d y}{d x}=-2 x q \Rightarrow y \frac{d y}{d x}=-x q\)
\(\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{-\mathrm{xq}}{\mathrm{y}}\)
\(\frac{d y}{d x}\) at \(\left(\sqrt{\frac{1-p}{1-p q}}, \sqrt{\frac{1-q}{1-p q}}\right)=\frac{-q \sqrt{\frac{1-p}{1-p q}}}{\sqrt{\frac{1-q}{1-p q}}}=m_{2}\)
\(\mathrm{m}_{2}=\frac{-\mathrm{q} \sqrt{1-\mathrm{q}}}{\sqrt{1-\mathrm{q}}}\)
Since, given \(\mathrm{m}_{1}\) and \(\mathrm{m}_{2}\) are orthogonal then
\(\mathrm{m}_{1} \cdot \mathrm{m}_{2}=-1\left(\because \theta=\frac{\pi}{2}\right)\)
\(\frac{-\sqrt{1-\mathrm{p}}}{\mathrm{p} \sqrt{1-q}} \times \frac{-\mathrm{q} \sqrt{1-\mathrm{p}}}{\sqrt{1-q}}=-1\)
\(\frac{\mathrm{q}(1-\mathrm{p})}{\mathrm{p}(1-\mathrm{q})}=-1\)
\(\mathrm{q}(1-\mathrm{p})=-\mathrm{p}(1-\mathrm{q})\)
Squaring both side, we get-
\(q^{2}(1-p)^{2}=p^{2}(1-q)^{2}\)
\(q^{2}\left(1+p^{2}-2 p\right)=p^{2}\left(1+q^{2}-2 q\right)\)
\(q_{2}^{2}+p^{2} q^{2}-2 p q^{2}=p^{2}+p^{2} q^{2}-2 p^{2} q\)
\(q^{2}-2 p q^{2}=p^{2}-2 p^{2} q\)
\(2 p^{2} q-2 p q^{2}=p^{2}-q^{2}\)
\(2 p q(p-q)=(p-q)(p+q)\)
\(2 p q=p+q\)
\(\frac{p}{p q}+\frac{q}{p q}=2\)
So, \(\quad \frac{1}{\mathrm{p}}+\frac{1}{\mathrm{q}}=2\)

AP EAMCET-2013

Application of Derivatives

85816 The number of points on the curve \(y=54 x^{5}-\) \(135 x^{4}-70 x^{3}+180 x^{2}+210 x\) at which the normal lines are parallel to \(x+90 y+2=0\) is :

1 2

2 4

3 0

4 3

Shift-I

Application of Derivatives

85817 The longest distance of the point \((a, 0)\) from the curve \(2 x^{2}+y^{2}=2 x\) is

1 \(1+\mathrm{a}\)

2 \(|1-\mathrm{a}|\)

3 \(\sqrt{1-2 \mathrm{a}+2 \mathrm{a}^{2}}\)

4 \(\sqrt{1-2 \mathrm{a}+3 \mathrm{a}^{2}}\)

AP EAMCET-2010

Application of Derivatives

85818 The polar equation of a curve centred at \((1,0)\) with radius 1 is

1 \(r=\cos \theta\)

2 \(\mathrm{r}=2\)

3 \(\mathrm{r}=\sin 2 \theta\)

4 \(\mathrm{r}=2 \cos \theta\)

AMU-2004

Application of Derivatives

85819 For \(\theta \in I R\), when the point \((x, y)=(\tan \theta+\sin \theta\), \(\boldsymbol{\operatorname { t a n }} \theta-\boldsymbol{\operatorname { s i n }} \theta\) ) is defined, it lies on the curve:

1 \(x^{2}+y^{2}=2 x y\)

2 \(x^{2}-y^{2}=8 x y\)

3 \(\left(x^{2}-y^{2}\right)^{2}=16 x y\)

4 \(\left(x^{2}+y^{2}\right)^{2}=4 x y\)

AMU-2004

Application of Derivatives

85820 If the curves \(x^{2}+p y^{2}=1\) and \(q x^{2}+y^{2}=1\) are orthogonal to each other, then

1 \(\mathrm{p}-\mathrm{q}=2\)

2 \(\frac{1}{\mathrm{p}}-\frac{1}{\mathrm{q}}=2\)

3 \(\frac{1}{\mathrm{p}}+\frac{1}{\mathrm{q}}=-2\)

4 \(\frac{1}{\mathrm{p}}+\frac{1}{\mathrm{q}}=2\)

Explanation:

(D) : Given, the curves \(\mathrm{x}^{2}+\mathrm{py}^{2}=1\) and \(\mathrm{qx}^{2}+\mathrm{y}^{2}=1\) Then, from \(\mathrm{x}^{2}+p y^{2}=1 \Rightarrow \mathrm{x}^{2}=1-\mathrm{py}^{2}\) put in \(\mathrm{qx}^{2}+\mathrm{y}^{2}=1\) \(\therefore \quad \mathrm{q}\left(1-\mathrm{py}^{2}\right)+\mathrm{y}^{2}=1\)
\(q_{2}-p q y^{2}+y^{2}=1\)
\(\mathrm{y}_{2}-\mathrm{pqy}^{2}=1-\mathrm{q}\)
\(y^{2}(1-p q)=1-q\)
\(y^{2}=\frac{1-q}{1-p q} \Rightarrow y= \pm \sqrt{\frac{1-q}{1-p q}}\)
Then, \(\quad x^{2}=1-p\left(\frac{1-q}{1-p q}\right)\)
\(x^{2}=1-p\left(\frac{1-q}{1-p q}\right) \Rightarrow x^{2}=\frac{1-p q-p+p q}{1-p q}\)
\(x^{2}=\frac{1-p}{1-p q} \Rightarrow x= \pm \sqrt{\frac{1-p}{1-p q}}\)
Then, intersection point \((\mathrm{x}, \mathrm{y})\)
\(=\left( \pm \sqrt{\frac{1-\mathrm{p}}{1-\mathrm{pq}}}, \pm \sqrt{\frac{1-\mathrm{q}}{1-\mathrm{pq}}}\right)\)
\(\therefore \quad \mathrm{x}^{2}+\mathrm{py}^{2}=1 \Rightarrow 2 \mathrm{x}+2 \mathrm{py} \frac{\mathrm{dy}}{\mathrm{dx}}=0\)
2 py \(\frac{d y}{d x}=-2 x \Rightarrow\) py \(\frac{d y}{d x}=-x\)
\(\frac{d y}{d x}=\frac{-x}{p y}\)
\(\frac{d y}{d x}\) at \(\left(\sqrt{\frac{1-p}{1-p q}}, \sqrt{\frac{1-q}{1-p q}}\right)=\frac{\sqrt{\frac{1-p}{1-p q}}}{p \times \sqrt{\frac{1-q}{1-p q}}}=m_{1}\)
\(\mathrm{m}_{1}=\frac{-\sqrt{1-\mathrm{p}}}{\mathrm{p} \sqrt{1-\mathrm{q}}}\)
And,\(\quad \mathrm{qx}^{2}+\mathrm{y}^{2}=1 \Rightarrow 2 x q+2 y \frac{d y}{d x}=0\)
\(2 y \frac{d y}{d x}=-2 x q \Rightarrow y \frac{d y}{d x}=-x q\)
\(\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{-\mathrm{xq}}{\mathrm{y}}\)
\(\frac{d y}{d x}\) at \(\left(\sqrt{\frac{1-p}{1-p q}}, \sqrt{\frac{1-q}{1-p q}}\right)=\frac{-q \sqrt{\frac{1-p}{1-p q}}}{\sqrt{\frac{1-q}{1-p q}}}=m_{2}\)
\(\mathrm{m}_{2}=\frac{-\mathrm{q} \sqrt{1-\mathrm{q}}}{\sqrt{1-\mathrm{q}}}\)
Since, given \(\mathrm{m}_{1}\) and \(\mathrm{m}_{2}\) are orthogonal then
\(\mathrm{m}_{1} \cdot \mathrm{m}_{2}=-1\left(\because \theta=\frac{\pi}{2}\right)\)
\(\frac{-\sqrt{1-\mathrm{p}}}{\mathrm{p} \sqrt{1-q}} \times \frac{-\mathrm{q} \sqrt{1-\mathrm{p}}}{\sqrt{1-q}}=-1\)
\(\frac{\mathrm{q}(1-\mathrm{p})}{\mathrm{p}(1-\mathrm{q})}=-1\)
\(\mathrm{q}(1-\mathrm{p})=-\mathrm{p}(1-\mathrm{q})\)
Squaring both side, we get-
\(q^{2}(1-p)^{2}=p^{2}(1-q)^{2}\)
\(q^{2}\left(1+p^{2}-2 p\right)=p^{2}\left(1+q^{2}-2 q\right)\)
\(q_{2}^{2}+p^{2} q^{2}-2 p q^{2}=p^{2}+p^{2} q^{2}-2 p^{2} q\)
\(q^{2}-2 p q^{2}=p^{2}-2 p^{2} q\)
\(2 p^{2} q-2 p q^{2}=p^{2}-q^{2}\)
\(2 p q(p-q)=(p-q)(p+q)\)
\(2 p q=p+q\)
\(\frac{p}{p q}+\frac{q}{p q}=2\)
So, \(\quad \frac{1}{\mathrm{p}}+\frac{1}{\mathrm{q}}=2\)

AP EAMCET-2013