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  • please solve rd sharma class 12 chapter 15 Tangents and Normals exercise 15.3 , question 1sub question 1 maths textbook solution

please solve rd sharma class 12 chapter 15 Tangents and Normals exercise 15.3 , question 1 sub question 1 maths textbook solution

Answers (2)

\theta=\tan ^{-1}\left(\frac{3}{4}\right) \text { and } \theta=\frac{\pi}{2}                

Hint – The angle of intersection of curves is \tan \theta=\left|\frac{m_{1}-m_{2}}{1+m_{1} m_{2}}\right|

m1=slope of first curve.   m2=slope of second curve.

Given -y^{2}=x \ldots \ldots(1) \& x^{2}=y \ldots \ldots(2)

First curve is y2=x

Differentiating above with respect to x,

As we know, \frac{d}{d x}\left(x^{n}\right)=^{n} x^{n-1}, \frac{d}{d x}(\text { constants })=0

\begin{aligned} &=2 y \frac{d}{d x}=1 \\ &=m_{1}=\frac{d y}{d x}=\frac{1}{2 y} \ldots \ldots \text { (3) } \end{aligned}

 

Second curve is x^{2}=y

 

Differentiating above with respect to x,

\begin{aligned} &=2 x=\frac{d y}{d x} \\ &=m_{2}=\frac{d y}{d x}=2 x \ldots \ldots(4) \end{aligned}

 

Substituting (1) in (2),we get

 

\begin{aligned} &=x^{2}=y \\ &=\left(y^{2}\right)^{2}=y \\ &=y^{4}-y=0 \end{aligned}

\begin{aligned} &=y\left(y^{3}-1\right)=0 \\ &=y=0 \text { or } y^{3}-1=0 \\ &=y=0 \text { or } y^{3}=1 \\ &=y=0 \text { or } y=1 \end{aligned}

Substituting  y=0 or y=1 in (1)

 x=y2

When, y = 0, x = 0

             y = 1, x = 1

Substituting the values of (y = 0, x = 0),(y = 1 ,  x = 1) for m1& m2 , we get,

When, y = 0

\begin{aligned} &=m_{1}=\frac{d y}{d x}=\frac{1}{2(0)}=\frac{1}{0} \quad::\left\{\frac{1}{0}=\infty\right\} \\ &=m_{1}=\infty \end{aligned}

 

When, y = 1

 

=m_{1}=\frac{d y}{d x}=\frac{1}{2(1)}=\frac{1}{2}

Value of m1 is \infty and \frac{1}{2}

When x = 0
\begin{aligned} &=m_{2}=\frac{d y}{d x}=2 x \\ &=m_{2}=\frac{d y}{d x}=2(0)=0 \end{aligned}

When x = 1

\begin{aligned} &=m_{2}=\frac{d y}{d x}=2 x \\ &=m_{2}=\frac{d y}{d x}=2(1)=2 \end{aligned}

Values of m2 is 0 and 2.

As we know, Angle of intersection of two curves is given by \tan \theta=\left|\frac{m_{1}-m_{2}}{1+m_{1} m_{2}}\right|

When m1 is \infty and m2 is 0
\begin{aligned} &\operatorname{tan} \theta=\left(\frac{\infty-0}{1+\infty \times 0}\right) \\ &\operatorname{tan} \theta=\infty \end{aligned}

\theta=\tan ^{-1}(\infty) As we know \tan \left(\frac{\pi}{2}\right)=\infty 

\theta=\frac{\pi}{2}                                                            

When m_{1}=\frac{1}{2}and m_{2}=2
\begin{aligned} &\operatorname{Tan} \theta=(1 / 2-2) /(1+1 / 2 \times 2)=\left(\frac{-3 / 2}{2}\right) \\ &\operatorname{Tan} \theta=\left(\frac{-3}{2} \times \frac{1}{2}\right)=\left(\frac{-3}{4}\right) \\ &=\theta=\tan ^{-1}\left(\frac{3}{4}\right) \end{aligned}

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\theta=\tan ^{-1}\left(\frac{3}{4}\right) \text { and } \theta=\frac{\pi}{2}

Hint – The angle of intersection of curves is \tan \theta=\left|\frac{m_{1}-m_{2}}{1+m_{1} m_{2}}\right|

m1=slope of first curve.   m2=slope of second curve.

Given -y^{2}=x \ldots \ldots(1)\;\; \&\;\; x^{2}=y \ldots \ldots(2)

First curve is y^{2}=x

Differentiating above with respect to x,
As we know,  \frac{d}{d x}\left(x^{n}\right)={ }^{n} x^{n-1}, \frac{d}{d x}(\text { constants })=0

\begin{aligned} &=2 y \frac{d}{d x}=1 \\ &=m_{1}=\frac{d y}{d x}=\frac{1}{2 y} \ldots \ldots \text { (3) } \end{aligned}

Second curve is x^{2}=y

Differentiating above with respect to x,

\begin{aligned} &=2 \mathrm{x}=\frac{d y}{d x} \\ &=m_{2}=\frac{d y}{d x}=2 x \ldots \ldots \text { (4) } \end{aligned}

Substituting (1) in (2),we get

\begin{aligned} &=x^{2}=y \\ &=\left(y^{2}\right)^{2}=y \\ &=y^{4}-y=0 \end{aligned}

\begin{aligned} &=y\left(y^{3}-1\right)=0 \\ &=y=0 \text { or } y^{3}-1=0 \\ &=y=0 \text { or } y^{3}=1 \\ &=y=0 \text { or } y=1 \end{aligned}

Substituting  y=0 or y=1 in (1)

x=y^{2}

When, y = 0, x = 0

             y = 1, x = 1

Substituting the values of (y = 0, x = 0),(y = 1 ,  x = 1) for m1& m2 , we get,

When, y = 0
\begin{aligned} &=m_{1}=\frac{d y}{d x}=\frac{1}{2(0)}=\frac{1}{0} \quad::\left\{\frac{1}{0}=\infty\right\} \\ &=m_{1}=\infty \end{aligned}

When, y = 1

=m_{1}=\frac{d y}{d x}=\frac{1}{2(1)}=\frac{1}{2}

Value of m1 is \infty  and \frac{1}{2}

When x = 0

\begin{aligned} &=m_{2}=\frac{d y}{d x}=2 x \\ &=m_{2}=\frac{d y}{d x}=2(0)=0 \end{aligned}

 

When x = 1

\begin{aligned} &=m_{2}=\frac{d y}{d x}=2 x \\ &=m_{2}=\frac{d y}{d x}=2(1)=2 \end{aligned}

 

 

Values of m2 is 0 and 2.

 

As we know, Angle of intersection of two curves is given by \operatorname{tan} \theta=\left|\frac{m_{1}-m_{2}}{1+m_{1} m_{2}}\right|

When m1 is \infty and m2 is 0

\begin{aligned} &\operatorname{Tan} \theta=\left(\frac{\infty-0}{1+\infty \times 0}\right) \\ &\operatorname{Tan} \theta=\infty \end{aligned}

 

\theta=\tan ^{-1}(\infty) As we know \tan \left(\frac{\pi}{2}\right)=\infty

 

\infty=\frac{\pi}{2}                                                    

When m_{1}=\frac{1}{2}   and  m_{2}=\frac{1}{2}

\begin{aligned} &\operatorname{Tan} \theta=(1 / 2-2) /(1+1 / 2 \times 2)=\left(\frac{-3 / 2}{2}\right) \\ &\operatorname{Tan} \theta=\left(\frac{-3}{2} \times \frac{1}{2}\right)=\left(\frac{-3}{4}\right) \\ &=\theta=\tan ^{-1}\left(\frac{3}{4}\right) \end{aligned}

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