Question

determine the point on the curve x^2+y^2=13 where the tangent are perpendicular to the line 3x-2y=0
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Answer 1

Given:

A curve x^2+y^2=13 and a line 3x-2y=0.

To Find:

A point from which tangent to curve will be perpendicular to curve.

Solution:

Firstly, we'll find slope of curve using derivatives

x^2+y^2=13

Taking derivative

2x + 2y.dy/dx = 0

dy/dx = -2x/2y = -x/y

Also, dy/dx is slope of curve

=> Slope of curve = -x/y

Now, Slope of line = 3/2

Since, tangent and line are perpendicular

=> slope of curve ×  Slope of line = -1

=> -x/y × 3/2 = -1

=> x/y = 2/3

=> x = 2y/3

Putting x = 2y/3 in curve

4y^2/9 + y^2 = 13

13y^2/9 = 13

y^2 = 9

y = ±3

Putting y = ±3 in curve, we get

x = ±2

Hence, Required points are ( ±2 , ±3 ).

Answer 2

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

TO DETERMINE

The point on the curve

$\sf{ {x}^{2} + {y}^{2} = 13 \: }$

where the tangent are perpendicular to the line 3x-2y=0

CALCULATION

The given equation of the curve is

$\sf{ {x}^{2} + {y}^{2} = 13 \: } \: \: ........(1)$

The given equation of the line is

$\sf{3x - 2y = 0} \: \: \: \: \: .....(2)$

Let ( h, k) be the required point

Differentiating both sides of Equation (1) with respect to x we get

$\displaystyle \sf{2x + 2y \frac{dy}{dx} = 0\: }$

$\implies \: \displaystyle \sf{\frac{dy}{dx} = - \frac{x}{y} \: } \: \:$

$\implies \: \displaystyle \sf{ \bigg[ \: \frac{dy}{dx} \: \bigg] _ {(h, k)}\: = - \frac{h}{k}} \: \:$

$\displaystyle \sf{ So \: the \: slope \: of \: the \: curve \: (1) \: at \: (h,k) = - \frac{h}{k}} \: \:$

Differentiating both sides of Equation (2) with respect to x we get

$\displaystyle \sf{3 - 2 \frac{dy}{dx} = 0\: }$

$\implies \: \displaystyle \sf{\frac{dy}{dx} = \frac{3}{2} \: } \: \:$

$\implies \: \displaystyle \sf{ \bigg[ \: \frac{dy}{dx} \: \bigg] _ {(h, k)}\: = \frac{3}{2}} \: \:$

$\displaystyle \sf{ So \: the \: slope \: of \: the \: line \: (1) \: at \: (h,k) = \frac{3}{2}} \: \:$

By the given condition

$\implies \: \displaystyle \sf{ \frac{3}{2} \times - \frac{h}{k}} \: = - 1 \:$

$\implies \: \displaystyle \sf{k = \frac{3h}{2} } \: \: \: ....(3)$

Since (h, k) is a point on the curve (1)

$\implies \: \displaystyle \sf{ \: {h}^{2} + { \bigg(\frac{3h}{2} \bigg)}^{2} = 13\: }$

$\implies \: \displaystyle \sf{ \: \frac{13 {h}^{2} }{4} = 13\: }$

$\implies \: \displaystyle \sf{ {h}^{2} = 4\: }$

$\implies \: \displaystyle \sf{ {h} = \pm \: 2\: }$

From Equation (3)

when h = 2 we have k = 3

Again From Equation (3)

when h = - 2 we have k = - 3

RESULT

The required points are ( 2 , 3 ) , ( -2 , - 3 )

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