Question

If the line $y=3x+c$ is a tangent to the ellipse $\frac{x^2}{16}+\frac{y^2}{25}=1$, then $c=?$

• A. $11$
• B. $13$
• C. $-11$
• D. $-13$

Answer 1

Answer: (B), (D)

$13$
$-13$

Equation of ellipse, $\frac{x^2}{16}+\frac{y^2}{25}=1⟶\left(1\right)$

finding slope of this ellipse,

$\frac{2x}{16}+\frac{2y}{25}\frac{dx}{dy}=0$

$\frac{dx}{dy}=-\frac{x\times 25}{16\times y}⟶\left(2\right)$

and $y=3x+c$ is a slope to ellipse

Therefore,

$\frac{-25x}{16y}=3$

$x=\frac{-48y}{25}⟶\left(3\right)$

putting value of x in $y=3x+c$

$y=3\times \frac{-48y}{25}+c$

$y=\frac{-144y}{25}+c⟶\left(4\right)$

putting x and y in equation (1)

$\frac{{\left(\frac{-4y}{2}\right)}^2}{16}+\frac{{(\frac{-144y}{25}+c)}^2}{25}=1$

$c=\pm 13$