The latus rectum of the conic passing through | vedclass

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Question

(B) The slope of the tangent at any point $(x, y)$ is $\frac{dy}{dx}$.The slope of the normal at $(x, y)$ is $-\frac{dx}{dy}$.The equation of the norm

Vedclass · Accepted Answer

$2$

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(B) The slope of the tangent at any point $(x, y)$ is $\frac{dy}{dx}$.The slope of the normal at $(x, y)$ is $-\frac{dx}{dy}$.The equation of the normal at $(x, y)$ is given by:$Y - y = -\frac{dx}{dy}(X - x)$Since the normal intersects the $x$-axis at $(x + 1, 0)$,we substitute $X = x + 1$ and $Y = 0$:$0 - y = -\frac{dx}{dy}(x + 1 - x)$$-y = -\frac{dx}{dy}(1)$$y = \frac{dx}{dy}$$y \, dy = dx$Integrating both sides,we get:$\int y \, dy = \int dx$$\frac{y^2}{2} = x + C$Since the conic passes through the origin $(0, 0)$,we have $0 = 0 + C$,so $C = 0$.Thus,the equation of the conic is $y^2 = 2x$.Comparing this with the standard form $y^2 = 4ax$,we get $4a = 2$,which is the length of the latus rectum.Therefore,the length of the latus rectum is $2$.

The latus rectum of the conic passing through the origin and having the property that the normal at each point $(x, y)$ intersects the $x$-axis at $(x + 1, 0)$ is:

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