Question

Find the equation of the tangent through the external point (8,-1) to the ellipse 2x^2+5y^2=70

Answer 1

help..

Answer 2

@ganeshie8

Answer 3

Take derivative of the curve, then plug it into the tangent line equation.

Answer 4

4x + 10yy'=0
hence \(y'= \dfrac{-4x}{10y}\), at (8,-1), \(y'= 16/5\)

Answer 5

4x+10y(dy/dx)=0

4(8)+10(-1)dy/dx=0?

Answer 6

ok then wats next?

Answer 7

that is m, right? that is the slope of the tangent line at (8,-1),
y - y_0 = m (x -x_0) , that is it.

Answer 8

where \((x_0, y_0) =(8,-1)\)

Answer 9

we need to be careful, (8, -1) is not on the ellipse

Answer 10

Wow!! ok, let me check

Answer 11

oh, yea, so, the tangent line to the curve which passes through (8,-1) right? @ganeshie8

Answer 12

ok so y+1=16/5x-128/5
y=(16/5)x+128/5

Answer 13

Sorry AliLnn. I misread the question. hehehe... I guided you wrong until the derivative. Let @ganeshie8 take over. @ganeshie8 Please.

Answer 14

ok