Name: Local Maxima and Minima of a Function
Uploaded: 2019-07-19
Duration: 14 min 17 s
Description: Do you know what is minima or maxima of a function? This video also explains how to find local maxima and minima by differentiation in a detailed manner.

Question

Prove that the curves

y = x^{2} - 3 x + 1

and

x (y + 3) = 4

intersect at right angles at the point

(2, - 1)

.

Accepted Answer

We know that if curves intersect at right angle, then $m_{1} \cdot m_{2} = - 1$

Where $m_{1}$ and $m_{2}$ are slopes of tangents at point of intersection on the respective curves.

Now, given curves are

$y = x^{2} - 3 x + 1$ $. . . . . . . . . . . . (1)$

and $x (y + 3) = 4 . . . . . . . . . . . . (2)$

On putting values of $y$ from equation $(1)$ in equation $(2)$

$\Rightarrow x (x^{2} - 3 x + 1 + 3) = 4$

$\Rightarrow x^{3} - 3 x^{2} + 4 x - 4 = 0$

$\Rightarrow (x - 2) (x^{2} - x + 2) = 0$

But $x^{2} - x + 2 \neq 0$ [discriminant $< 0$ ]

$\Rightarrow (x - 2) = 0$

$\Rightarrow x = 2$

Put $x = 2$ in equation $(1)$

$y = 2^{2} - 3 (2) + 1$

$\Rightarrow y = - 1$

Hence, point of intersection is $(2, - 1)$

Now from equation $(1)$

$y = x^{2} - 3 x + 1$

Differentiating both side with respect to $x$

$\frac{d y}{d x} = 2 x - 3$ [ $\frac{d}{dx} x^{n} = n . x^{n - 1}$ ]

At point $(2, - 1)$

$\frac{d y}{d x} = 4 - 3 = 1 . . . . . . . . . . (1)$

$\Rightarrow m_{1} = 1 . . . . . . . . . (3)$

From equation $(2)$

$y + 3 = \frac{4}{x}$

Differentiating both side with respect to $x$

$\frac{d y}{d x} = \frac{- 4}{x^{2}}$

At point $(2, - 1)$

$\frac{d y}{d x} = \frac{- 4}{4} = - 1$

$\Rightarrow m_{2} = - 1 . . . . . . . . . . (4)$

From equation $(3)$ and $(4)$

$m_{1} \cdot m_{2} = 1 (- 1)$

$\Rightarrow m_{1} . m_{2} = - 1$

Hence, curve intersect at right angles.

Prove that the curves $y = x^{2} - 3 x + 1$ and $x (y + 3) = 4$ intersect at right angles at the point $(2, - 1)$ .

Important Questions on Applications of Derivatives

Learn and Prepare for any exam you want !

Prove that the curves y=x2-3x+1 and xy+3=4 intersect at right angles at the point 2,-1.

Important Questions on Applications of Derivatives

Learn and Prepare for any exam you want !

Prove that the curves $y = x^{2} - 3 x + 1$ and $x (y + 3) = 4$ intersect at right angles at the point $(2, - 1)$ .