Question

find the equation of tangent & normal to the curve 4x^2+9y^2=40 at point(1,2)​

Answer 1

your answer is in attachment please have a look

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Answer 2

Equation of tangent is 2x+9y=20 and

equation of normal is 9x-2y-5=0.

Given:

• A curve $4 {x}^{2} + 9 {y}^{2} = 40$
• A point (1,2)

To find:

• Find the equation of tangent & normal to the curve at given point.

Solution:

Formula\Concept to be used:

• Slope of tangent to the curve is m= dy/dx.
• Equation of line passing through point(x1,y1) having slope m:$\bf (y - y_1) = m(x - x_1)$
• Slope of two lines; perpendicular with each other must satisfy the equation $\bf m_1m_2 = - 1$

Step 1:

Find dy/dx and slope at point(1,2).

Differentiate the curve with respect to x.

$4 \times 2x + 9 \times 2y \frac{dy}{dx} = 0$

or

$4x + 9y \frac{dy}{dx} = 0$

or

$\bf \frac{dy}{dx} = \frac{ - 4x}{9y}$

Put the point in dy/dx.

$m = \frac{ - 4(1)}{9 \times 2}$

or

$\bf \red{m = \frac{ - 2}{9}}$

Step 2:

Find the equation of tangent at (1,2)

According to the formula.

$y - 2 = \frac{ - 2}{9} (x - 1)$

or

$9(y - 2) = - 2(x - 1)$

or

$\bf 2x + 9y = 20$

Thus,

Equation of tangent is 2x+9y=20.

Step 3:

Find equation of normal.

As normal is perpendicular to tangent, so slope of normal is 9/2.

Both normal and tangent passes through same point.

Equation of normal:$y - 2 = \frac{9}{2} (x - 1)$

or

$2y - 4 = 9x - 9$

or

$- 9x + 2y = - 9 + 4$

or

$\bf 9x - 2y - 5 = 0$

Thus,

Equation of normal is 9x-2y-5=0

Learn more:

1) Find the slope of tangent to the curve y=5x^2 at (-1;5)

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2) Consider the circle x^2+y^2-6x+4y-12=0.The equation of a tangent to this circle that is parallel to the line 4x+3y+5=0.

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