If the line x+4y=√7, where R, touches the ellipse 3x^2+4y^2=1 at the point P in the first …

Question

If the line $\alpha x+4y=\sqrt7$, where $\alpha\in\mathbb R$, touches the ellipse $3x^2+4y^2=1$ at the point $P$ in the first quadrant, then one of the focal distances of $P$ is:

Accepted Answer

Write the ellipse as $\dfrac{x^2}{1/3}+\dfrac{y^2}{1/4}=1$, so $a^2=\tfrac13,\ b^2=\tfrac14$ (major axis along $x$), $e^2=1-\dfrac{3}{4}=\dfrac14\Rightarrow e=\dfrac12$.

The line $\alpha x+4y=\sqrt7$ touches the ellipse when $c^2=a^2m^2+b^2$ with $m=-\alpha/4,\ c=\sqrt7/4$:

$$\frac{7}{16}=\frac13\cdot\frac{\alpha^2}{16}+\frac14 \;\Rightarrow\; \alpha^2=9 \;\Rightarrow\; \alpha=\pm3.$$

For $P$ in the first quadrant take $\alpha=3$: line $3x+4y=\sqrt7$. Using the tangent-point formula $x_1=\dfrac{a^2 l}{n},\,y_1=\dfrac{b^2 m}{n}$ for $lx+my=n$:

$$x_1=\frac{(1/3)(3)}{\sqrt7}=\frac{1}{\sqrt7},\qquad y_1=\frac{(1/4)(4)}{\sqrt7}=\frac{1}{\sqrt7}.$$

So $P=\left(\dfrac{1}{\sqrt7},\dfrac{1}{\sqrt7}\right)$. Focal distances are $a\mp ex_1$:

$$r_1=\frac{1}{\sqrt3}-\frac12\cdot\frac{1}{\sqrt7},\qquad r_2=\frac{1}{\sqrt3}+\frac12\cdot\frac{1}{\sqrt7}.$$

$r_2=\dfrac{1}{\sqrt3}+\dfrac{1}{2\sqrt7}$ matches option D.

Solutions · 1