Need solution for RD Sharm Maths Class 12 Chapter 15 Tangents and Normals Excercise Fill in the blanks Question 19

$\left ( 3,34 \right )$

Hint:

When both curves touch then slope of both curves should be same.

Given:

Given curves,

$y=4 x^{2}+2 x-8$  and  $y=x^{3}-x+13$  touch each other.

To find:

We have to find the point where the given curves touch each.

Solution:

The curves, $y=4 x^{2}+2 x-8$  and  $y=x^{3}-x+13$

For first curve say $\left(\frac{d y}{d x}\right)_{1}$

$\therefore \left(\frac{d y}{d x}\right)_{1}=8x+2$

For second curve  $\left(\frac{d y}{d x}\right)_{2}=3 x^{2}-1$

When both curves touch the slope both curves should be same

\begin{aligned} &\therefore \quad 8 x+2=3 x^{2}-1 \\ &\Rightarrow \quad 3 x^{2}-8 x-3=0 \end{aligned}

Solving the quadratic equation, we get

$\begin{array}{ll} \Rightarrow \quad & (3 x+1)(x-3)=0 \\ \end{array}$

$\Rightarrow \quad x=\frac{-1}{3}, x=3$

Now consider, $x=3$

For first curve,

$y\left ( 3 \right )=4\left ( 3 \right )^{2}+2\left ( 3 \right )-8=34$

For second curve,

$y\left ( 3 \right )=\left ( 3 \right )^{3}-3+10=34$   …(here there should be 10 inplace of 13)

Thus at  $\left ( 3,34 \right )$  both curves touch

Now consider,  $x=\frac{-1}{3}$

For first curve,

$y\left(\frac{-1}{3}\right)=4\left(\frac{-1}{3}\right)^{2}+2\left(\frac{-1}{3}\right)-5=\frac{-47}{9}$

For second curve,

$y\left(\frac{-1}{3}\right)=\left(\frac{-1}{3}\right)^{3}-\left(\frac{-1}{3}\right)+13=\frac{8}{27}+13$

Thus at  $x=\frac{-1}{3}$  both curves do not meet.

But their tangent are parallel

Hence the only point where both curves touch is  $\left ( 3,34 \right )$