Question

If the line $y=x+4$ is a tangent to the ellipse $\frac{x^2}{p^2}+\frac{x^2}{3p^2}=1$, then $p=?$

• A. $\pm 4$
• B. $\pm 2$
• C. $\pm 8$
• D. $\pm 16$

Answer 1

Answer: (B)

$\pm 2$

Equation of ellipse, $\frac{x^2}{p^2}+\frac{y^2}{{3p}^2}=1⟶\left(1\right)$

Slope of ellipse,

$\frac{2x}{P^2}+\frac{2y}{{3P}^2}\frac{dy}{dx}=0$

$\frac{dx}{dy}=\frac{-2x\times {3P}^2}{P^2\times 2y}\Rightarrow \frac{-3x}{y}$

$y=x+4$ is a tangent to this ellipse, $⟶\left(2\right)$

So,

$\frac{-3x}{y}=1$

$-3x=y⟶\left(3\right)$

putting this in equation (2)

$-3x=x+4$

$-4x=4$

$x=-1$

putting this in (3)

$-3\times -1=y$

$y=3$

putting value of x and y in eqn (1)

$\frac{{(-1)}^2}{P^2}+\frac{{\left(3\right)}^2}{{3P}^2}=1$

$\frac{1}{P^2}(1+\frac{9}{3})=1$

$P^2=\frac{12}{3}$

$P=\sqrt{4}$

$P=\pm 2$