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

Find the equation of the parabola with focus $(2, 0)$ and directrix $x = -2$.

If $(-1,-1)$ is the focus and $x+y+4=0$ is the directrix of a parabola,then its vertex is

The equation of a tangent line to the parabola $y^2 = 8x$,which passes through the point $(1, 3)$,is:

The value of $\lambda$ for which the curve $(7x+5)^{2}+(7y+3)^{2}=\lambda^{2}(4x+3y-24)^{2}$ represents a parabola is

The ordinates of the points $P$ and $Q$ on the parabola with focus $(3,0)$ and directrix $x = -3$ are in the ratio $3:1$. If $R(\alpha, \beta)$ is the point of intersection of the tangents to the parabola at $P$ and $Q$,then $\frac{\beta^2}{\alpha}$ is equal to $.............$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module