The line x cos alpha + y sin alpha = p will t | vedclass

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Question

(A) The equation of the line is $x \cos \alpha + y \sin \alpha = p$,which can be written as $x \cos \alpha + y \sin \alpha - p = 0$. The equation of t

Vedclass · Accepted Answer

$p \cos \alpha + a = 0$

Vedclass · Answer

(A) The equation of the line is $x \cos \alpha + y \sin \alpha = p$,which can be written as $x \cos \alpha + y \sin \alpha - p = 0$. The equation of the parabola is $y^2 = 4a(x + a)$. The condition for the line $y = mx + c$ to touch the parabola $y^2 = 4ax'$ (where $x' = x+a$) is $c = \frac{a'}{m}$. Rewriting the line equation: $y \sin \alpha = -x \cos \alpha + p \implies y = -x \cot \alpha + p \csc \alpha$. Here,$m = -\cot \alpha$ and $c = p \csc \alpha$. The parabola is $y^2 = 4a(x+a)$. Comparing with $Y^2 = 4aX$,we have $Y = y$,$X = x+a$,and the parameter $a$ remains $a$. The condition for tangency is $c = \frac{a}{m} + am$ (since the vertex is at $(-a, 0)$). Substituting the values: $p \csc \alpha = \frac{a}{-\cot \alpha} + a(-\cot \alpha)$. $p \csc \alpha = -a 	an \alpha - a \cot \alpha = -a (\frac{\sin \alpha}{\cos \alpha} + \frac{\cos \alpha}{\sin \alpha}) = -a (\frac{\sin^2 \alpha + \cos^2 \alpha}{\sin \alpha \cos \alpha}) = -\frac{a}{\sin \alpha \cos \alpha}$. $p \frac{1}{\sin \alpha} = -\frac{a}{\sin \alpha \cos \alpha} \implies p = -\frac{a}{\cos \alpha}$. Thus,$p \cos \alpha = -a$,or $p \cos \alpha + a = 0$.

The line $x \cos \alpha + y \sin \alpha = p$ will touch the parabola $y^2 = 4a(x + a)$ if:

Similar Questions

The locus of a point such that two tangents drawn from it to the parabola $y^2 = 4ax$ are such that the slope of one is double the other is:

Let $P(\alpha, \beta)$ be a point on the parabola $y^2 = 4x$. If $P$ also lies on the chord of the parabola $x^2 = 8y$ whose midpoint is $(1, 5/4)$,then $(\alpha - 28)(\beta - 8)$ is equal to:

Find the equation of the parabola whose axis is parallel to the $y$-axis and which passes through the points $(0,4), (1,9)$ and $(4,5)$.

If the equation of the directrix of a parabola is $3x + 4y + 15 = 0$ and the equation of the tangent at the vertex is $3x + 4y - 5 = 0$,then what is the length of the latus rectum?

Statement $(A)$: If the normal at the ends of the latus rectum of the parabola $y^2 = 4x$ meet the curve again at $P$ and $P'$,then $PP' = 12$ units.
Reason $(R)$: If the normal at $T_1$ to the parabola $y^2 = 4ax$ meets the parabola again at $T_2$,then $T_2 = -T_1 - \frac{2}{T_1}$.

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