If S(a, b) is a fixed point and P(alpha, beta | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The general equation of a parabola is defined as $PF^2 = e^2 PM^2$,where $e=1$ for a parabola.Given the equation $4[(x-a)^2+(y-b)^2]=(\alpha x+\be

Vedclass · Accepted Answer

$\alpha^2+\beta^2=4$

Vedclass · Answer

(B) The general equation of a parabola is defined as $PF^2 = e^2 PM^2$,where $e=1$ for a parabola.Given the equation $4[(x-a)^2+(y-b)^2]=(\alpha x+\beta y+7)^2$,we can rewrite it as:$(x-a)^2+(y-b)^2 = \left(\frac{\alpha x+\beta y+7}{2}\right)^2$To match the form $PF^2 = PM^2$,we normalize the line equation:$(x-a)^2+(y-b)^2 = \left(\frac{\alpha x+\beta y+7}{\sqrt{\alpha^2+\beta^2}}\right)^2 \cdot \frac{\alpha^2+\beta^2}{4}$For this to represent a parabola,the eccentricity $e$ must be $1$,which implies the coefficient $\frac{\alpha^2+\beta^2}{4} = 1^2$.Therefore,$\alpha^2+\beta^2=4$.

If $S(a, b)$ is a fixed point and $P(\alpha, \beta)$ is a variable point such that $4[(x-a)^2+(y-b)^2]=(\alpha x+\beta y+7)^2$ represents a parabola,then the locus of $P(\alpha, \beta)$ is

Similar Questions

If the normal to the parabola $y^2 = 4ax$ at the point $(a, 2a)$ meets the parabola again at the point $(at^2, 2at)$,then what is the value of $t$?

Let $P(x, y)$ be a variable point on the parabola $y = 4x^2 + 1$. Let $Q(c, c)$ be the foot of the perpendicular drawn from $P$ to the line $y = x$. If $R(h, k)$ is the mid-point of $PQ$,then the locus of $R$ is:

The straight line $y = 2x + \lambda$ does not meet the parabola $y^2 = 2x$,if

The equation of the normal to the parabola $x^2 = 8y$ at $x = 4$ is

For the parabola $y^2 = 8x$,what is the position of the point $(2, 5)$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module