Find the equation of the conic with focus $(3,0)$, directrix $x=-2$, and eccentricity $\frac{3}{2}$. Determine the type of the conic.

Step-by-step solution

1. Apply the definition: distance to focus equals $e$ times distance to directrix: $\sqrt{(x-3{)}^2+y^2}=\frac{3}{2}|x+2|$.
2. Square both sides: $(x-3{)}^2+y^2=\frac{9}{4}(x+2{)}^2$.
3. Multiply by 4 to clear fractions: $4(x-3{)}^2+4y^2=9(x+2{)}^2$.
4. Expand and simplify: $5x^2+60x-4y^2=0$.
5. Rearrange and complete the square for $x$: $5[(x+6{)}^2-36]-4y^2=0$.
6. Obtain $5(x+6{)}^2-4y^2=180$ and divide by 180 to get the standard form: $\frac{(x+6{)}^2}{36}-\frac{y^2}{45}=1$.
7. Since the equation has the form $\frac{X^2}{a^2}-\frac{Y^2}{b^2}=1$, it is a hyperbola.

Concept tested: Focus-directrix – conic equation

Published 7 October 2026 · How this solution was prepared and checked