Answer to Question #271762 in Analytic Geometry for Bebe

Question

Find the equations of the tangents and normal to the following curves:

1. y2 + 8x = 0, parallel to x + y + 4 = 0.

2. x2 = 3y, perpendicular to x – 2y + 7 = 0.

3. x2 + 9y2 = 25, parallel to 4x + 9y + 30 = 0.

4. 25x2 + 4y2 = 100, perpendicular to 8x – 15y + 4 = 0.

5. x2 – y2 = 15, parallel to 4x – y + 20 = 0.

Answer 1

equation of the tangent:

$y-y_0=f'(x_0)(x-x_0)$

equation of the normal:

$y-y_0=-(x-x_0)/f'(x_0)$

1.

$$y^2+8x=0$$ $$2yy'=-8$$ $$y'=\dfrac{-4}{y}$$ $$slope=m_1=y'=\dfrac{-4}{y}$$

Tangent is parallel to $x+y+4=0.$

$$y=-x-4, slope=m_2=-1$$

Then $m_1=m_2=>-\dfrac{4}{y}=-1=>y=4$

Find $x$

$$(4)^2+8x=0=>x=-2$$

Point $(-2, 4)$

Tangent line is

$$y-4=-(x-(-2))$$

The equation of the tangent line in slope-intercept form

$$y=-x+2$$

Find the equation of the normal at the point $(-2, 4)$

$$slope=m_3=-\dfrac{1}{m_1}=1$$ $$y-4=(x-(-2))$$

The equation of the normal line in slope-intercept form

$$y=x+6$$

2.

$$x^2=3y$$ $$2x=3y'$$ $$y'=\dfrac{2x}{3}$$ $$slope=m_1=y'=\dfrac{2x}{3}$$

Tangent is perpendicular to $x-2y+7=0.$

$$y=\dfrac{1}{2}x+\dfrac{7}{2}, slope=m_2=\dfrac{1}{2}$$

Then $m_1=-\dfrac{1}{m_2}=>\dfrac{2x}{3}=-2=>x=-3$

Find $y$

$$(-3)^2=3y=>y=3$$

Point $(-3, 3)$

Tangent line is

$$y-3=-2(x-(-3))$$

The equation of the tangent line in slope-intercept form

$$y=-2x-3$$

Find the equation of the normal at the point $(-3, 3)$

$$slope=m_3=-\dfrac{1}{m_1}=\dfrac{1}{2}$$ $$y-3=\dfrac{1}{2}(x-(-3))$$

The equation of the normal line in slope-intercept form

$$y=\dfrac{1}{2}x+\dfrac{9}{2}$$

3.

$$x^2+9y^2=25$$ $$2x+18yy'=0$$ $$y'=\dfrac{-x}{9y}$$ $$slope=m_1=y'=-\dfrac{x}{9y}$$

Tangent is parallel to $4x + 9y + 30 = 0.$

$$y=-\dfrac{4}{9}x-\dfrac{10}{3}, slope=m_2=-\dfrac{4}{9}$$

Then $m_1=m_2=>-\dfrac{x}{9y}=-\dfrac{4}{9}=>x=4y$

Find $y$

$$(4y)^2+9y^2=25=>y=\pm1$$

Point $(-4, -1),$ Point $(4, 1).$

Point $(-4, -1)$

Tangent line is

$$y-(-1)=-\dfrac{4}{9}(x-(-4))$$

The equation of the tangent line in slope-intercept form

$$y=-\dfrac{4}{9}x-\dfrac{25}{9}$$

Find the equation of the normal at the point $(-4, -1)$

$$slope=m_3=-\dfrac{1}{m_1}=\dfrac{9}{4}$$ $$y-(-1)=\dfrac{9}{4}(x-(-4))$$

The equation of the normal line in slope-intercept form

$$y=\dfrac{9}{4}x+8$$

Point $(4, 1)$

Tangent line is

$$y-1=-\dfrac{4}{9}(x-4)$$

The equation of the tangent line in slope-intercept form

$$y=-\dfrac{4}{9}x+\dfrac{25}{9}$$

Find the equation of the normal at the point $(4, 1)$

$$slope=m_3=-\dfrac{1}{m_1}=\dfrac{9}{4}$$ $$y-1=\dfrac{9}{4}(x-4)$$

The equation of the normal line in slope-intercept form

$$y=\dfrac{9}{4}x-8$$

4.

$$25x^2+4y^2=100$$ $$50x+8yy'=0$$ $$y'=\dfrac{-25x}{4y}$$ $$slope=m_1=y'=-\dfrac{25x}{4y}$$

Tangent is perpendicular to $8x – 15y + 4 = 0.$

$$y=\dfrac{8}{15}x+\dfrac{4}{15}, slope=m_2=\dfrac{8}{15}$$

Then $m_1=-\dfrac{1}{m_2}=>-\dfrac{25x}{4y}=-\dfrac{15}{8}=>x=\dfrac{3}{10}y$

Find $y$

$$25(\dfrac{3}{10}y)^2+4y^2=100=>y=\pm4$$

Point $(-\dfrac{6}{5}, -4),$ Point $(\dfrac{6}{5}, 4).$

Point $(-\dfrac{6}{5}, -4)$

Tangent line is

$$y-(-4)=-\dfrac{15}{8}(x-(-\dfrac{6}{5}))$$

The equation of the tangent line in slope-intercept form

$$y=-\dfrac{15}{8}x-\dfrac{25}{4}$$

Find the equation of the normal at the point $(-\dfrac{6}{5}, -4)$

$$slope=m_3=-\dfrac{1}{m_1}=\dfrac{8}{15}$$ $$y-(-4)=\dfrac{8}{15}(x-(-\dfrac{6}{5}))$$

The equation of the normal line in slope-intercept form

$$y=\dfrac{8}{15}x-\dfrac{84}{25}$$

Point $(\dfrac{6}{5}, 4)$

Tangent line is

$$y-4=-\dfrac{15}{8}(x-\dfrac{6}{5})$$

The equation of the tangent line in slope-intercept form

$$y=-\dfrac{15}{8}x+\dfrac{25}{4}$$

Find the equation of the normal at the point $(\dfrac{6}{5}, 4)$

$$slope=m_3=-\dfrac{1}{m_1}=\dfrac{8}{15}$$ $$y-4=\dfrac{8}{15}(x-\dfrac{6}{5})$$

The equation of the normal line in slope-intercept form

$$y=\dfrac{8}{15}x+\dfrac{84}{25}$$

5.

$$x^2-y^2=15$$ $$2x-2yy'=0$$ $$y'=\dfrac{x}{y}$$ $$slope=m_1=y'=\dfrac{x}{y}$$

Tangent is parallel to $4x – y + 20 = 0.$

$$y=4x+20, slope=m_2=4$$

Then $m_1=m_2=>\dfrac{x}{y}=4=>x=4y$

Find $y$

$$(4y)^2-y^2=15=>y=\pm1$$

Point $(-4, -1),$ Point $(4, 1).$

Point $(-4, -1)$

Tangent line is

$$y-(-1)=4(x-(-4))$$

The equation of the tangent line in slope-intercept form

$$y=4x+15$$

Find the equation of the normal at the point $(-4, -1)$

$$slope=m_3=-\dfrac{1}{m_1}=-\dfrac{1}{4}$$ $$y-(-1)=-\dfrac{1}{4}(x-(-4))$$

The equation of the normal line in slope-intercept form

$$y=-\dfrac{1}{4}x-2$$

Point $(4, 1)$

Tangent line is

$$y-1=4(x-4)$$

The equation of the tangent line in slope-intercept form

$$y=4x-15$$

Find the equation of the normal at the point $(4, 1)$

$$slope=m_3=-\dfrac{1}{m_1}=-\dfrac{1}{4}$$ $$y-1=-\dfrac{1}{4}(x-4)$$

The equation of the normal line in slope-intercept form

$$y=-\dfrac{1}{4}x+2$$