Question

find the tangents of the following curves x^2 + 16y^2 = 32 from (-6, 7/2). the answers are x+4y -8 and 41x + 4y + 232

Answer 1

if you\[\frac{ dy }{ dx }\]what does that get?

Answer 2

-x/6y

Answer 3

\[\frac{ d x^2 }{ dx }=2x...\frac{d9}{dx}=0\]
\[\frac{ dy }{ dx } = ??\]

Answer 4

2x + 32y dy/dx = 0

Answer 5

so you get
\[ y'= \frac{-x}{16y} \]

Answer 6

yes im sorry. typed it wrong.

Answer 7

y' will be the slope of the tangent line at point (x,y) on the ellipse

you want that slope to be the same as the slope between point (x,y) and point(-6,7/2)

in other words:
\[ \frac{y- \frac{7}{2}}{x+6}= \frac{-x}{16y} \]

Answer 8

yes thats what i did and i got stuck up

Answer 9

after cross multiplying you get
\[ 16y^2 -56y = -x^2 -6x \\
x^2 + 16y^2 -56y +6x= 0
\]
we can simplify that a bit... notice that the equation of the ellipse says \( x^2+16y^2=32\)
so you have
\[ 32 -56 y + 6x= 0 \\
3x= 28y-16
\]

Answer 10

remember that last equation, we will need it later.

meanwhile, to continue, we must make that an equation of 1 variable.
One way to proceed is square both sides (because we can know
\( x^2 = 32-16y^2 \) from the equation of the ellipse

squaring both sides:
\[ 9x^2 = (28y-16)^2 \]
replace x^2 to get
\[ 9(32-16y^2) = (28y-16)^2 \]

Answer 11

now solve for y....
use \( 3x= 28y-16\) to find x

use x and y to find the slope using \( m= \frac{-x}{16y} \)
now use the point - slope formula
\[ y - y_0 = m(x - x_0) \]
where \( (x_0,y_0) = ( -6,7/2) \)

change to standard form to match your answers.

Answer 12

thank you so much @phi !!