Ellipse
Normals and Tangents
JEE Main 2019
Grade 11
Question:
If the normal to the ellipse $3x^2 + 4y^2 = 12$ at point $P$ is parallel to the line $2x + y = 4$ and the tangent to the ellipse at $P$ passes through $Q(4, 4)$, then $|PQ|$ is:
(1) $\frac{5\sqrt{5}}{2}$
(2) $\frac{\sqrt{61}}{2}$
(3) $\frac{\sqrt{221}}{2}$
(4) $\frac{\sqrt{157}}{2}$
Step-by-Step Solution
Key Concept: Ellipse $x^2/4 + y^2/3 = 1$. Normal slope at $(x_1, y_1)$: $4y_1/3x_1 = -2 \Rightarrow y_1 = -3x_1/2$. Substituting into ellipse: $x_1 = \pm 1$. Tangent at $(-1, 3/2)$ passes through $(4, 4)$. $P = (-1, 3/2)$, $|PQ| = \sqrt{25 + 25/4} = 5\sqrt{5}/2$.
The detailed step-by-step mathematical proof is available inside the Mathbee app workspace.
Correct Answer: (1)