Find the equations of the tangent and normal to the curve 16x² + 9y² = 144 at (2, 1). x,y>0

Question

Also find the point of intersection where both t? 

 

 

Answer 1

The Given point is wrong.

Consider the point as (2, 3).

The curve is16x² + 9y² = 144 and the point is (2, 3), x, y >0.

Differentiate the curve with respect to x.

32x + 18yy ' = 0

18yy ' = - 32x

⇒ y ' = - 16x/9y.

When, x = 2 and y = 3 , y ' = - (16*2)/(9*3) = - 32/27.

y ' = - 32/27.

This is the slope (m ) of the tangent line to the implicit curve at (2, 3).

 

Slope - intercept form line equation is y = mx + b, where m is slope and b is y - intercept.

Now the tangent line equation is y = (- 32/27)x  + b.

Find the y - intercept by substituting the the point in the tangent line equation say (x, y) = (2, 3).

3 = (- 32/27)2  + b

b = 3 + 64/27

b = (81 + 64)/27

⇒ b = 145/27.

The tangent line equation  is y = (- 32/27)x + (145/27).

 

The normal line and tangent are perpendecular to each other.

Since the slopes of perpendecular lines are negative reciprocals the slope of nolmal line through the point  (2, 3) is 27/32.

Slope (m) = 27/32.

Now, the normal line equation is y = (27/32)x + b.

Find the y - intercept by substituting the the point in the normal line equation say (x, y) = (2, 3).

3 = (27/32)(2) + b

b = 3 - 27/16

b = (48 - 27)/16

⇒ b = 21/16.

The normal line equation  is y = (27/32)x + (21/16).

 

The tangent line is y = (- 32/27)x + (145/27) → (1)

The normal line is y = (27/32)x + (21/16)       → (2)

To find the intersection of the tangent and normal line, solve the above two equations.

From eq (1) & (2),

(- 32/27)x + (145/27) = (27/32)x + (21/16)

(27/32)x + (32/27)x = 145/27 - 21/16

x(27*27 + 32*32)/864 = (145*16 - 21*27)/432

x = [(2320 - 567)2]/[729 + 1024]

x = (1753 * 2)/1753

⇒ x = 2.

Substitute the value x = 2 in eq (2).

y = (27/32)2 + (21/16)

= 27/16 + 21/16

= (27 + 21)/16

= 48/16

⇒ y = 3.

Therefore, intersection point of the tangent and nomal lines is (2, 3).