Conic Sections
Conic Section
star_batch_jee_advanced_2025
Grade 11
Question:
From any point $P$ lying in first quadrant on the ellipse $\frac{x^2}{25} + \frac{y^2}{16} = 1$, $PN$ is drawn perpendicular to the major axis and produced to $Q$ so that $NQ$ equals $PS$, where $S$ is the focus $(-3, 0)$. Then the locus of $Q$ is:
$5y - 3x - 25 = 0$
$3x + 5y + 25 = 0$
$3x - 5y - 25 = 0$
None of these
Step-by-Step Solution
Key Concept: The focal chord property relates the distance from a focus to the directrix through the eccentricity parameter.
Given ellipse with $a^2 = 25$ and $b^2 = 16$, the eccentricity is $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{16}{25}} = \frac{3}{5}$. For point $Q(h,k)$ where $k < 0$, using the focal chord property $|k| = SP = a + ex_1$ where $P(x_1, y_1)$ is on the ellipse. Substituting the ellipse equation and solving yields $3x + 5y + 25 = 0$ as the locus of point $Q$.
Correct Answer: 2