If the normal at one end of the latus rectum | vedclass

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(B) Given the parabola $y^2=16x$, we have $4a=16$, so $a=4$. The coordinates of the ends of the latus rectum are $(a, 2a)$ and $(a, -2a)$, which are $

Vedclass · Accepted Answer

$32$

Vedclass · Answer

(B) Given the parabola $y^2=16x$, we have $4a=16$, so $a=4$. The coordinates of the ends of the latus rectum are $(a, 2a)$ and $(a, -2a)$, which are $(4, 8)$ and $(4, -8)$. Let us consider the point $(4, 8)$. The slope of the tangent at $(4, 8)$ is given by $2y \frac{dy}{dx} = 16 \implies \frac{dy}{dx} = \frac{8}{y}$. At $(4, 8)$, the slope of the tangent is $m_t = \frac{8}{8} = 1$. The slope of the normal is $m_n = -\frac{1}{m_t} = -1$. The equation of the normal at $(4, 8)$ is $y - 8 = -1(x - 4) \implies y = -x + 12$. This normal meets the $X$-axis $(y=0)$ at $x=12$, so point $P$ is $(12, 0)$. The chord passes through $P(12, 0)$ and is perpendicular to the normal. Since the normal has slope $-1$, the chord has slope $m_c = 1$. The equation of the chord is $y - 0 = 1(x - 12) \implies y = x - 12$. Substituting $y = x - 12$ into $y^2 = 16x$: $(x - 12)^2 = 16x \implies x^2 - 24x + 144 = 16x \implies x^2 - 40x + 144 = 0$. $(x - 36)(x - 4) = 0$, so $x = 4$ or $x = 36$. For $x = 4$, $y = 4 - 12 = -8$. For $x = 36$, $y = 36 - 12 = 24$. The endpoints of the chord are $(4, -8)$ and $(36, 24)$. The length of the chord is $\sqrt{(36 - 4)^2 + (24 - (-8))^2} = \sqrt{32^2 + 32^2} = \sqrt{2 	imes 32^2} = 32 \sqrt{2}$.

If the normal at one end of the latus rectum of the parabola $y^2=16x$ meets the $X$-axis at the point $P$, then the length of the chord passing through $P$ and perpendicular to the normal is (in $\sqrt{2}$)

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