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Provide solution for rd sharma maths class 12 chapter 15 Tangents and Normals exercise 15.3 question 2 sub question 1

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m_{1} \times m_{2}=-1  Hence, two curves intersect orthogonally.

Hint - Two curves intersects orthogonally if m_{1} \times m_{2}=-1, where m1 and m2 are the slopes of two curves.

Given –
\begin{aligned} &y=x^{3}---(1) \\ &6 y=7-x^{2}---(2) \end{aligned}

Substituting  y= x3 in eq (2), we get

\begin{aligned} &=6\left(x^{3}\right)=7-x^{2} \\ &=6 x^{3}+x^{2}-7=0 \end{aligned}

Since \begin{aligned} &f(x)=6x^{3}+x^{2}-7=0 \\ \end{aligned}, we have to find f(x)=0, so that x is a factor of f(x).

When x = 1

\begin{aligned} f(1) &=6(1)^{3}+(1)^{2}-7 \\ &=6+1-7 \\ &=0 \end{aligned}

Hence, x = 1 is a factor of f(x)

Substituting x = 1 in  y= x3 , we get
\begin{aligned} &y=(1)^{3} \\ &y=1 \end{aligned}

The point of intersection of two curves is (1, 1)

First curve is \begin{aligned} &y=x^{3} \\ \end{aligned}

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

=m_{1}=\frac{d y}{d x}=3 x^{2}

Second curve is 6y=7-x^{2}

Differentiating above with respect to x,
\begin{aligned} &=\frac{6 d y}{d x}=0-2 x \\ &=\frac{d y}{d x}=\frac{-2 x}{6} \\ &=m_{2}=\frac{d y}{d x}=\frac{-x}{3} \end{aligned}

Now, put (1,1) in m1 & m2 , we get,

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

When m1=3 and  m2=\begin{aligned} \frac{-1}{3} \end{aligned}

Two curves intersects orthogonally if m_{1} \times m_{2}=-1

=3\times \frac{-1}{3}=-1

Hence, two curves y=x^{3} & 6y=7-x^{2}  intersect orthogonally.

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