Let x-y=0 be the equation of tangent at the point (5, 5) lying on a parabola whose focus is at (2,1). The length of latus rectum of the parabola is

Question

Let x-y=0 be the equation of tangent at the point (5, 5) lying on a parabola whose focus is at (2,1). The length of latus rectum of the parabola is (A) (3/5) (B) (4/5) (C) (2/5) (D) none

Tardigrade Admin · Accepted Answer

Image of (2,1) in x = y is (1,2). Slope of directrix =-((5-1/5-2))=(-4/3) Equation of directrix =y-2=(-4/3)(x-1) ⇒ 4 x +3 y -10=0 Latus rectum =2 ×|(8+3-10/5)|=(2/5).

Q. Let $x - y = 0$ be the equation of tangent at the point $(5, 5)$ lying on a parabola whose focus is at $(2, 1)$ . The length of latus rectum of the parabola is

$\frac{3}{5}$

$\frac{4}{5}$

$\frac{2}{5}$

none

Image of $(2, 1)$ in $x = y$ is $(1, 2)$ .
Slope of directrix $= - (\frac{5 - 1}{5 - 2}) = \frac{- 4}{3}$
Equation of directrix $= y - 2 = \frac{- 4}{3} (x - 1)$
$\Rightarrow 4 x + 3 y - 10 = 0$
Latus rectum $= 2 \times ∣ ∣ \frac{8 + 3 - 10}{5} ∣ ∣ = \frac{2}{5}$ .

Q. Let x−y=0 be the equation of tangent at the point (5,5) lying on a parabola whose focus is at (2,1). The length of latus rectum of the parabola is

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54​

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