Answer to Question #270953 in Analytic Geometry for Akiyah

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

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

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