Question

Find the equation of tangents to the ellipse x^2/50+y^2/32=1 which passes through a point (15,-4)

Answer 1

Answer:

Equation of tangent line is, $12x-5y=200$  

Explanation:

Given: An ellipse equation $\frac{x^2}{50}+\frac{y^2}{32} =1$ or we can write it as: $32x^2+50y^2=1600$ or $16x^2+25y^2=800$

⇒$16x^2+25y^2-800=0$

The derivative of above equation is:

$32x+50y \cdot y'=0$

On simplify we get;

$y'=-\frac{32x}{50y}$ or

$y'=-\frac{16x}{25y}$ or $y'=\frac{dy}{dx} = -\frac{16x}{25y}$

Evaluate $\frac{dy}{dx}$ at (15, -4)

we get, $\frac{dy}{dx}=-\frac{16\cdot 15}{25\cdot( -4)}$

on simplify we get, $\frac{dy}{dx} =\frac{12}{5}$

The point $(x_{0}, y_{0})$  to the tangent line to the ellipse is (15, -4) and the tangent to the ellipse is given by:

$\frac{dy}{dx}= \frac{y-y_{0}}{x-x_{0}}$

then, we have

$\frac{12}{5}=\frac{y-(-4)}{x-15}$

⇒ $(y+4)= (\frac{12}{5})(x-15)$ or$5(y+4)=(12)(x-15)$

On simplifying we get,

$5y+20=12x-180$

on further simplifying we get;

$12x-5y=200$

therefore, the equation of tangent to the ellipse  $\frac{x^2}{50}+\frac{y^2}{32} =1$ which passes through a point (15, -4) is,  $12x-5y=200$

Answer 2

this is the write answer