Question

find the equation of the tangent to the ellipse 4x^2+7y^2=28 from the point (3,-2)​

Answer 1

The equation of tangent at point (3,-2) is given by

$6x-7y=32$

Step-by-step explanation:

Equation of ellipse

$4x^2+7y^2=28$

Point :(3,-2)

Differentiate w.r. t x

$8x+14y\frac{dy}{dx}=0$

Using formula :$\frac{dx^n}{dx}=nx^{n-1}$

$14y\frac{dy}{dx}=-8x$

$\frac{dy}{dx}=-\frac{8x}{14y}$

Substitute x=3 and y=-2

$\frac{dy}{dx}=-\frac{8(3)}{14(-2)}=\frac{24}{28}=\frac{6}{7}$

Slope of tangent=m=$\frac{6}{7}$

Point-slope of form:$y-y_1=m(x-x_1)$

Using the formula

The equation of tangent at point (3,-2) is given by

$y-(-2)=\frac{6}{7}(x-3)$

$y+2=\frac{6}{7}(x-3)$

$7y+14=6x-18$

$6x-7y=14+18=32$

Hence, the equation of tangent at point (3,-2) is given by

$6x-7y=32$

#Learns more:

https://brainly.in/question/13970752:Answered  by Lublana

Answer 2

Therefore the equation of tangent at (3,-2) is

6x-7y=32

Step-by-step explanation:

Given equation,

4x²+7y²=28

Differences with respect to x

$8x+14 y\frac{dy}{dx} =0$                                       $\frac{d}{dx} (x^n)=nx^{n-1}$          

$\Rightarrow\frac{dy}{dx} =-\frac{8x}{14y}$

$\Rightarrow[\frac{dy}{dx} ]_{(3,-2)}=-\frac{8.3}{14(-2)}=\frac{6}{7}$

Therefore the slope of the tangent is $=\frac{6}{7}$

The equation of line passes through the point $(x_1,y_2)$ is

$y-y_1=m(x-x_1)$                   Where m is the slope of the line.

Therefore the equation of tangent at (3,-2) is

y-(-2)=$\frac{6}{7}$ (x-3)

⇔7y+14=6x-18

⇔6x-7y=32