First slide

Ellipse

Question

$The normal at a point P on the ellipse x^{2} + 4 y^{2} = 16 meets the x -axis at Q . If M is the midpoint of the line segment$
$P Q, then the locus of M intersects the latus rectums of the given ellipse at points$

Moderate

A

$(\pm (3 \sqrt{5}) / 2, \pm 2 / 7)$
B

$(\pm (3 \sqrt{5}) / 2, \pm \sqrt{19} / 7)$
C

$(\pm 2 \sqrt{3}, \pm 1 / 7)$
D

$(\pm 2 \sqrt{3}, \pm 4 \sqrt{3} / 7)$

Solution

$Normal at P is given by 4 x \sec ϕ - 2 y cosec ϕ = 12$
$\begin{array}{l} ∴ Q \equiv (3 \cos ϕ, 0) \\ Let mid-point of P Q be M (α, β) . \\ ∴ α = \frac{3 \cos ϕ + 4 \cos ϕ}{2} = \frac{7}{2} \cos ϕ \\ or \cos ϕ = \frac{2}{7} α \\ and β = \sin ϕ \end{array}$
$Using \cos^{2} ϕ + \sin^{2} ϕ = 1, we have$
$\begin{array}{l} \frac{4}{49} α^{2} + β^{2} = 1 \\ or \frac{4}{49} x^{2} + y^{2} = 1 - - - - (1) \end{array}$
$\begin{array}{l} Now, the latus rectum to above ellipse is \\ x = \pm 2 \sqrt{3} - - - - - (2) \end{array}$
$Solving (1) and (2), we have$
$\begin{aligned} \frac{48}{49} + y^{2} = 1 \\ or y = \pm \frac{1}{7} \end{aligned}$
$The points of intersection are (\pm 2 \sqrt{3}, \pm 1 / 7) .$

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