Let $\mathrm{A}(\alpha, 0)$ and $\mathrm{B}(0, \beta)$ be the points on the line $5 \mathrm{x}+7 \mathrm{y}=50$. Let the point P divide the line segment AB internally in the ratio $7: 3$. Let $3 \mathrm{x}- 25=0$ be a directrix of the ellipse $E: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ and the corresponding focus be S . If from S , the perpendicular on the x -axis passes through P , then the length of the latus rectum of E is equal to

JEE Main · 2024 · 30 Jan 2024 · Shift 2 · Ellipse

Let

A (α, 0)

and

B (0, β)

be the points on the line

5 x + 7 y = 50

. Let the point P divide the line segment AB internally in the ratio

7 : 3

. Let

3 x - 25 = 0

be a directrix of the ellipse

E : \frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1

and the corresponding focus be S . If from S , the perpendicular on the x -axis passes through P , then the length of the latus rectum of E is equal to

(A) $\frac{25}{3}$
(B) $\frac{32}{9}$
(C) $\frac{25}{9}$
(D) $\frac{32}{5}$ ✓ Correct

Answer: (D)

Solution

A (α, 0)

lies on

5 x + 7 y = 50

, so

5 α = 50 \Rightarrow A = (10, 0)

B (0, β)

gives

7 β = 50 \Rightarrow B = (0, \frac{50}{7})

P

divides

A B

internally in the ratio

7 : 3

P = (\frac{7 \cdot 0 + 3 \cdot 10}{10}, \frac{7 \cdot \frac{50}{7} + 3 \cdot 0}{10}) = (3, 5)

.

The directrix

3 x - 25 = 0

, i.e.

x = \frac{25}{3}

, gives

\frac{a}{e} = \frac{25}{3}

. The focus

S

corresponding to this directrix is

(a e, 0)

; the perpendicular from

S

to the x-axis is the vertical line through

S

, and this passes through

P (3, 5)

only if

a e = 3

.

Then

a^{2} = (a e) (\frac{a}{e}) = 3 \cdot \frac{25}{3} = 25 \Rightarrow a = 5

, and

e = \frac{a e}{a} = \frac{3}{5}

b^{2} = a^{2} (1 - e^{2}) = 25 (1 - \frac{9}{25}) = 16

.

Length of latus rectum

= \frac{2 b ^{2}}{a} = \frac{32}{5}

JEE Main · 2024 · 30 Jan 2024 · Shift 2 · Single correct option

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JEE Main · 2024 · 30 Jan 2024 · Shift 2 · Ellipse

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