Circle (x-3)^2+(y-2)^2=4. At(4,2-3^½) , tangent line equation is? Let y’=y-2. Dilating y’ by 2 have ellipse. Equation of tangent line at(4,2-2(3^½))is?Original tangent line dilated by 2 from (3,2) is?

Question

Have circle (x-3)²+(y-2)²=4.

At point (4,2-√3) tangent line

is y=(√3)x/3+2-7(√3)/3.

Let y’=y-2. x’=x-3. Dilating y’ by 2 and x’ by 1

have ellipse.

What is equation of tangent line dilated by 2 from (3,2)

center such that it is tangent to ellipse

at x=4, y=2-2(√3)? Does it have slope of (√3)/3?

 

Answer 1

is the equation for the ellipse.

We get the tangent line for the ellipse through the point (4, 2 –2√3)

by vertically dilating the tangent line for circle through the point (4, 2 –√3)

(for the circle), vertically dilated by 2 becomes (for the ellipse)

which, after a bit of algebraic manipulation, becomes

The tangent line of the ellipse has a slope (vertically dilated by a factor of 2) of (2√3)/3.

Desmos graph: desmos.com/calculator/trnccq7yov

Answer 2

y=(√3)x/3+2-7(√3)/3 is tangent line to circle x=4.

call y'=y-2, recall center of circle and ellipse are (3,2)

dilate y' tangent line to circle equation

have y=2(√3)x/2)+2-14(√3)/3

If tangent line equation is determined

for (x-3)²/4+(y-2)²/16=1 ellipse

find y=2(√3)x/2)+2-14(√3)/3

No ,dilating line does not keep same slope

when working with functions